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Limits and Derivatives – CBSE NCERT Study Resources

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11th

11th - Mathematics

Limits and Derivatives

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Overview

This chapter introduces the fundamental concepts of limits and derivatives, which form the basis of calculus. Students will learn about the intuitive idea of limits, algebra of limits, and the definition of derivatives. The chapter also covers standard derivative formulas and their applications.

Limits

A limit describes the value that a function approaches as the input approaches some value.

Key concepts include:

  • Left-hand limit and right-hand limit
  • Algebra of limits (sum, difference, product, quotient)
  • Limits of polynomial and rational functions
  • Evaluation of simple limits

Derivatives

The derivative of a function represents the rate of change of the function with respect to its variable.

Topics covered:

  • Definition of derivative as a limit
  • Derivative of simple functions from first principles
  • Derivative of sum, difference, product and quotient of functions
  • Derivative of polynomial and trigonometric functions

Standard Derivatives

Important derivative formulas include:

  • d/dx(xn) = nxn-1
  • d/dx(sin x) = cos x
  • d/dx(cos x) = -sin x
  • d/dx(tan x) = sec2 x

Applications

Basic applications of derivatives include:

  • Finding rate of change of quantities
  • Determining increasing/decreasing nature of functions
  • Finding approximate values

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define the limit of a function at a point.
Answer:

The value a function approaches as the input approaches a point.

Question 2:
Find limx→2 (3x + 1).
Answer:
Numeric answer:
7
Question 3:
What is the derivative of f(x) = x2?
Answer:

f'(x) = 2x

Question 4:
State the power rule for derivatives.
Answer:

If f(x) = xn, then f'(x) = nxn-1.

Question 5:
Compute limx→0 (sin x)/x.
Answer:
Numeric answer:
1
Question 6:
Differentiate f(x) = 5 with respect to x.
Answer:

f'(x) = 0

Question 7:
What is the derivative of f(x) = ex?
Answer:

f'(x) = ex

Question 8:
Find limx→∞ (1 + 1/x)x.
Answer:
Numeric answer:
e
Question 9:
Differentiate f(x) = √x using first principles.
Answer:

f'(x) = 1/(2√x)

Question 10:
What is the derivative of f(x) = ln x?
Answer:

f'(x) = 1/x

Question 11:
Find limx→3 (x2 - 9)/(x - 3).
Answer:
Numeric answer:
6
Question 12:
State the sum rule for derivatives.
Answer:

(f + g)' = f' + g'

Question 13:
Define the term limit of a function at a point.
Answer:

The limit of a function f(x) at a point x = a is the value that f(x) approaches as x gets arbitrarily close to a, but not necessarily equal to a.

Question 14:
What is the derivative of the function f(x) = 5 with respect to x?
Answer:

The derivative of a constant function like f(x) = 5 is always 0 because the slope of a horizontal line is zero.

Question 15:
State the power rule for differentiation.
Answer:

The power rule states that if f(x) = xn, then its derivative is f'(x) = n·xn-1, where n is any real number.

Question 16:
Find the limit: limx→2 (3x + 1).
Answer:

Substitute x = 2 directly into the function:
3(2) + 1 = 6 + 1 = 7.
Thus, the limit is 7.

Question 17:
Differentiate f(x) = sin(x) with respect to x.
Answer:

The derivative of sin(x) with respect to x is cos(x).

Question 18:
What is the derivative of f(x) = ex?
Answer:

The derivative of ex is ex itself, as the exponential function is unique in being its own derivative.

Question 19:
Evaluate: limx→0 (sin(x)/x.
Answer:

This is a standard limit:
limx→0 (sin(x)/x) = 1.

Question 20:
Find the derivative of f(x) = √x using the first principles.
Answer:

Using first principles:
f'(x) = limh→0 [√(x + h) - √x]/h
Rationalize the numerator:
= limh→0 [(√(x + h) - √x)(√(x + h) + √x)] / [h(√(x + h) + √x)]
Simplify:
= limh→0 h / [h(√(x + h) + √x)]
= 1 / (2√x).

Question 21:
What does the derivative of a function represent geometrically?
Answer:

Geometrically, the derivative represents the slope of the tangent line to the curve of the function at any given point. It measures the rate of change of the function at that point.

Question 22:
Find the derivative of f(x) = x3 + 2x2 - 5x + 7.
Answer:

Apply the power rule to each term:
f'(x) = 3x2 + 4x - 5.

Question 23:
What is the derivative of sin(x) with respect to x?
Answer:

The derivative of sin(x) with respect to x is cos(x).

Question 24:
Differentiate ex with respect to x.
Answer:

The derivative of ex with respect to x is ex itself.

Question 25:
What is the derivative of a constant function f(x) = c?
Answer:

The derivative of a constant function f(x) = c is 0, as the slope of a horizontal line is zero.

Question 26:
Evaluate: limx→0 (sin x)/x.
Answer:

The limit limx→0 (sin x)/x is a standard result and equals 1.

Question 27:
What is the derivative of tan(x) with respect to x?
Answer:

The derivative of tan(x) with respect to x is sec2(x).

Question 28:
State whether the following limit exists: limx→0 (1/x).
Answer:

The limit limx→0 (1/x) does not exist because the function approaches +∞ from the right and -∞ from the left, leading to a discontinuity.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define limit of a function at a point.
Answer:

The limit of a function f(x) at a point x = a is the value that f(x) approaches as x approaches a from both sides (left and right).
Mathematically, if limx→a f(x) = L, then f(x) gets arbitrarily close to L as x approaches a.

Question 2:
Find the derivative of f(x) = 3x2 + 5x - 2 using first principles.
Answer:

Using first principles (definition of derivative):
f'(x) = limh→0 [f(x + h) - f(x)] / h
Given f(x) = 3x2 + 5x - 2,
f(x + h) = 3(x + h)2 + 5(x + h) - 2
Expanding and simplifying:
f'(x) = limh→0 [6xh + 3h2 + 5h] / h
= limh→0 (6x + 3h + 5)
= 6x + 5

Question 3:
State the algebra of limits theorem for sum of two functions.
Answer:

If limx→a f(x) = L and limx→a g(x) = M, then:
limx→a [f(x) + g(x)] = L + M
Algebra of limits allows us to break down complex limits into simpler parts for evaluation.

Question 4:
Differentiate y = sin(x) + cos(x) with respect to x.
Answer:

Using standard derivatives:
dy/dx = d/dx [sin(x)] + d/dx [cos(x)]
= cos(x) - sin(x)

Remember: The derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x).

Question 5:
Evaluate limx→2 (x2 - 4)/(x - 2).
Answer:

Factorize numerator:
(x2 - 4) = (x - 2)(x + 2)
Cancel common term:
limx→2 (x + 2) = 4

This is an example of resolving an indeterminate form (0/0) by factorization.

Question 6:
What is the derivative of a constant function? Give an example.
Answer:

The derivative of any constant function is 0.
Example: For f(x) = 5, f'(x) = 0.

This is because a constant function has no rate of change (its graph is a horizontal line).

Question 7:
Find the derivative of f(x) = √x using first principles.
Answer:

Using first principles:
f'(x) = limh→0 [√(x + h) - √x] / h
Multiply numerator and denominator by conjugate:
= limh→0 h / [h(√(x + h) + √x)]
Simplify:
= 1 / (2√x)

This shows how roots can be differentiated using limits.

Question 8:
State the Sandwich Theorem for limits.
Answer:

If f(x) ≤ g(x) ≤ h(x) for all x near a (except possibly at a), and
limx→a f(x) = limx→a h(x) = L, then
limx→a g(x) = L.

This theorem is useful for evaluating limits of functions that are 'squeezed' between two others.

Question 9:
Differentiate y = ex + logex with respect to x.
Answer:

Using standard derivatives:
dy/dx = d/dx [ex] + d/dx [logex]
= ex + 1/x

Note: ex is unique as it is its own derivative, while the derivative of natural log is 1/x.

Question 10:
Evaluate limx→0 sin(x)/x.
Answer:

This is a standard limit:
limx→0 sin(x)/x = 1

This result is fundamental in calculus and is used to derive many trigonometric derivatives. The limit exists even though both numerator and denominator approach 0.

Question 11:
Find the derivative of f(x) = 3x2 + 2x + 1 using first principles.
Answer:

Using first principles:
f'(x) = limh→0 [f(x + h) - f(x)] / h
= limh→0 [3(x + h)2 + 2(x + h) + 1 - (3x2 + 2x + 1)] / h
= limh→0 [3x2 + 6xh + 3h2 + 2x + 2h + 1 - 3x2 - 2x - 1] / h
= limh→0 (6xh + 3h2 + 2h) / h
= limh→0 (6x + 3h + 2)
= 6x + 2.

Question 12:
Evaluate limx→2 (x2 - 4) / (x - 2).
Answer:

Simplify the expression:
limx→2 (x2 - 4) / (x - 2) = limx→2 (x - 2)(x + 2) / (x - 2)
= limx→2 (x + 2)
= 4.

Question 13:
Differentiate f(x) = sin x + cos x with respect to x.
Answer:

Using standard derivatives:
f'(x) = d/dx (sin x) + d/dx (cos x)
= cos x - sin x.

Question 14:
What is the derivative of a constant function? Explain with an example.
Answer:

The derivative of a constant function is always zero because the rate of change is constant.
Example: If f(x) = 5, then f'(x) = 0.

Question 15:
Find limx→0 (sin 3x) / x.
Answer:

Using the standard limit limx→0 (sin ax) / x = a:
limx→0 (sin 3x) / x = 3.

Question 16:
Differentiate f(x) = ex log x with respect to x.
Answer:

Using the product rule:
f'(x) = ex · d/dx (log x) + log x · d/dx (ex)
= ex · (1/x) + log x · ex
= ex (1/x + log x).

Question 17:
Evaluate limx→1 (x3 - 1) / (x - 1).
Answer:

Using factorization:
limx→1 (x3 - 1) / (x - 1) = limx→1 (x - 1)(x2 + x + 1) / (x - 1)
= limx→1 (x2 + x + 1)
= 3.

Question 18:
State the power rule of differentiation with an example.
Answer:

The power rule states that if f(x) = xn, then f'(x) = n xn-1.
Example: For f(x) = x4, f'(x) = 4x3.

Question 19:
Find the derivative of f(x) = √x using first principles.
Answer:

Using first principles:
f'(x) = limh→0 [√(x + h) - √x] / h
= limh→0 [(√(x + h) - √x)(√(x + h) + √x)] / [h(√(x + h) + √x)]
= limh→0 (x + h - x) / [h(√(x + h) + √x)]
= limh→0 h / [h(√(x + h) + √x)]
= limh→0 1 / (√(x + h) + √x)
= 1 / (2√x).

Question 20:
Find the derivative of f(x) = 3x2 + 2x + 1 using first principles.
Answer:

Using the first principle of derivatives:
f'(x) = limh→0 [f(x + h) - f(x)] / h
Substitute f(x) = 3x2 + 2x + 1:
= limh→0 [3(x + h)2 + 2(x + h) + 1 - (3x2 + 2x + 1)] / h
= limh→0 [3x2 + 6xh + 3h2 + 2x + 2h + 1 - 3x2 - 2x - 1] / h
= limh→0 (6xh + 3h2 + 2h) / h
= limh→0 (6x + 3h + 2)
= 6x + 2

Question 21:
Differentiate sin(x) + cos(x) with respect to x.
Answer:

Using standard derivatives:
d/dx [sin(x)] = cos(x)
d/dx [cos(x)] = -sin(x)
Thus,
d/dx [sin(x) + cos(x)] = cos(x) - sin(x)

Question 22:
What is the derivative of a constant function? Justify your answer.
Answer:

The derivative of a constant function f(x) = c is 0.
Reason: The slope of a horizontal line (constant function) is zero, and the derivative represents the slope of the tangent at any point.

Question 23:
Find limx→0 sin(x)/x.
Answer:

This is a standard limit:
limx→0 sin(x)/x = 1
Note: This result is derived using the squeeze theorem or geometric interpretation of the sine function.

Question 24:
Differentiate ex + logex with respect to x.
Answer:

Using standard derivatives:
d/dx [ex] = ex
d/dx [logex] = 1/x
Thus,
d/dx [ex + logex] = ex + 1/x

Question 25:
Evaluate limx→3 (2x + 1).
Answer:

Since the function f(x) = 2x + 1 is continuous at x = 3, we can substitute directly:
limx→3 (2x + 1) = 2(3) + 1 = 7

Question 26:
State the product rule for differentiation.
Answer:

The product rule states that if u(x) and v(x) are differentiable functions, then:
d/dx [u(x) · v(x)] = u'(x) · v(x) + u(x) · v'(x)
In words: Derivative of the product is the derivative of the first times the second, plus the first times the derivative of the second.

Question 27:
Find the derivative of tan(x) using the quotient rule.
Answer:

Express tan(x) as sin(x)/cos(x) and apply the quotient rule:
d/dx [tan(x)] = d/dx [sin(x)/cos(x)]
= [cos(x) · cos(x) - sin(x) · (-sin(x))] / cos2(x)
= [cos2(x) + sin2(x)] / cos2(x)
= 1 / cos2(x)
= sec2(x)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define the limit of a function at a point and explain its significance in calculus.
Answer:

The limit of a function f(x) as x approaches a point a is the value that f(x) gets arbitrarily close to as x approaches a. Mathematically, it is written as limx→a f(x) = L.


Significance in calculus:

  • Limits form the foundation for defining derivatives and integrals.
  • They help analyze the behavior of functions near specific points, even where the function might not be defined.
  • Essential for understanding continuity, rates of change, and areas under curves.
Question 2:
Differentiate between the left-hand limit and right-hand limit of a function with an example.
Answer:

The left-hand limit (LHL) of f(x) as x approaches a is the value approached when x comes from values less than a (denoted as limx→a⁻ f(x)).
The right-hand limit (RHL) is the value approached when x comes from values greater than a (denoted as limx→a⁺ f(x)).


Example: For f(x) = |x|/x at x = 0:
LHL: limx→0⁻ f(x) = -1 (since x is negative).
RHL: limx→0⁺ f(x) = 1 (since x is positive).
The limit does not exist at x = 0 because LHL ≠ RHL.

Question 3:
State the algebra of limits and apply it to find limx→2 (3x² - 4x + 1).
Answer:

The algebra of limits states that if limx→a f(x) and limx→a g(x) exist, then:

  • limx→a [f(x) ± g(x)] = lim f(x) ± lim g(x)
  • limx→a [f(x) · g(x)] = lim f(x) · lim g(x)
  • limx→a [k · f(x)] = k · lim f(x) (for constant k)


Solution for limx→2 (3x² - 4x + 1):
= 3 · limx→2 x² - 4 · limx→2 x + limx→2 1
= 3 · (2)² - 4 · (2) + 1
= 12 - 8 + 1
= 5.

Question 4:
Explain the concept of derivative as a rate of change and provide its geometrical interpretation.
Answer:

The derivative of a function f(x) at a point x = a measures the instantaneous rate of change of f(x) with respect to x. It is defined as:
f'(a) = limh→0 [f(a + h) - f(a)] / h.


Geometrical interpretation:

  • The derivative represents the slope of the tangent to the curve y = f(x) at the point (a, f(a)).
  • It gives the steepness or direction of the curve at that point.
  • If the derivative is positive/negative, the function is increasing/decreasing at that point.

Question 5:
Find the derivative of f(x) = 5x³ + 2x - 7 using the first principles method.
Answer:

Using first principles (definition of derivative):
f'(x) = limh→0 [f(x + h) - f(x)] / h.


Steps:
1. Compute f(x + h):
5(x + h)³ + 2(x + h) - 7 = 5(x³ + 3x²h + 3xh² + h³) + 2x + 2h - 7.
2. Subtract f(x):
[5x³ + 15x²h + 15xh² + 5h³ + 2x + 2h - 7] - [5x³ + 2x - 7] = 15x²h + 15xh² + 5h³ + 2h.
3. Divide by h:
(15x²h + 15xh² + 5h³ + 2h) / h = 15x² + 15xh + 5h² + 2.
4. Take limit as h→0:
f'(x) = 15x² + 2.

Question 6:
Define limit of a function at a point and explain its significance in calculus.
Answer:

The limit of a function f(x) at a point x = a is the value that f(x) approaches as x gets arbitrarily close to a, without necessarily reaching it. Mathematically, it is written as limx→a f(x) = L.

Significance in calculus:

  • Limits form the foundation for defining derivatives and integrals.
  • They help analyze the behavior of functions near points where they might not be explicitly defined.
  • Used to study continuity, differentiability, and convergence in higher mathematics.
Question 7:
Differentiate between left-hand limit and right-hand limit with an example.
Answer:

The left-hand limit (LHL) of f(x) at x = a is the value approached by f(x) when x approaches a from values less than a (denoted as limx→a⁻ f(x)).

The right-hand limit (RHL) is the value approached when x approaches a from values greater than a (denoted as limx→a⁺ f(x)).

Example: For f(x) = |x|/x at x = 0:
LHL: limx→0⁻ f(x) = -1 (since x < 0).
RHL: limx→0⁺ f(x) = 1 (since x > 0).

Question 8:
State the algebra of limits and apply it to evaluate limx→2 (x² + 3x - 5).
Answer:

The algebra of limits states that if limx→a f(x) and limx→a g(x) exist, then:
1. Sum rule: limx→a [f(x) + g(x)] = limx→a f(x) + limx→a g(x).
2. Product rule: limx→a [f(x) × g(x)] = limx→a f(x) × limx→a g(x).
3. Constant multiple rule: limx→a [k·f(x)] = k·limx→a f(x).

Evaluating limx→2 (x² + 3x - 5):
Step 1: Apply sum rule: limx→2 x² + limx→2 3x - limx→2 5.
Step 2: Direct substitution (since polynomial is continuous):
(2)² + 3(2) - 5 = 4 + 6 - 5 = 5.

Question 9:
Explain the first principle of derivatives and use it to find the derivative of f(x) = x².
Answer:

The first principle of derivatives (or definition of derivative) states that the derivative of f(x) at x is:
f'(x) = limh→0 [f(x + h) - f(x)] / h.

For f(x) = x²:
Step 1: Compute f(x + h) = (x + h)² = x² + 2xh + h².
Step 2: Substitute into definition:
f'(x) = limh→0 [(x² + 2xh + h²) - x²] / h
= limh→0 (2xh + h²) / h
= limh→0 (2x + h).
Step 3: Evaluate limit as h→0: f'(x) = 2x.

Question 10:
What is the physical interpretation of a derivative? Illustrate with an example.
Answer:

The derivative represents the instantaneous rate of change of a function with respect to its variable.

Physical interpretation examples:
1. In kinematics, the derivative of position s(t) with respect to time t gives velocity v(t).
2. In economics, the derivative of cost function C(x) gives marginal cost.

Illustration for s(t) = t²:
Velocity v(t) = ds/dt = 2t.
At t = 3 sec, velocity is 6 m/s, showing how fast position changes at that instant.

Question 11:
Differentiate between the left-hand limit and right-hand limit with an example.
Answer:

The left-hand limit (LHL) of f(x) at x = a is the value approached by f(x) as x approaches a from the left (i.e., x → a-).
The right-hand limit (RHL) is the value approached as x approaches a from the right (i.e., x → a+).

Example: For f(x) = |x|/x at x = 0:
LHL: limx→0- f(x) = -1 (since |x| = -x for x < 0).
RHL: limx→0+ f(x) = 1 (since |x| = x for x > 0).

Question 12:
State the algebra of limits and apply it to evaluate limx→2 (x2 - 3x + 2)/(x - 2).
Answer:

The algebra of limits states that if limx→a f(x) and limx→a g(x) exist, then:
1. Sum rule: lim(f + g) = lim f + lim g
2. Product rule: lim(f · g) = lim f · lim g
3. Quotient rule: lim(f/g) = lim f / lim g (if lim g ≠ 0).

Solution:
limx→2 (x2 - 3x + 2)/(x - 2)
= limx→2 (x - 1)(x - 2)/(x - 2) (factorizing numerator)
= limx→2 (x - 1) (canceling common factor)
= 2 - 1 = 1.

Question 13:
Explain the concept of derivative as a limit and find the derivative of f(x) = x2 + 1 using first principles.
Answer:

The derivative of a function f(x) at a point x = a is defined as the limit:
f'(a) = limh→0 [f(a + h) - f(a)]/h.
It represents the instantaneous rate of change of f(x) with respect to x.

For f(x) = x2 + 1:
f'(x) = limh→0 [(x + h)2 + 1 - (x2 + 1)]/h
= limh→0 [x2 + 2xh + h2 + 1 - x2 - 1]/h
= limh→0 (2xh + h2)/h
= limh→0 (2x + h)
= 2x.

Question 14:
What is the chain rule in differentiation? Demonstrate its application to find the derivative of sin(2x + 3).
Answer:

The chain rule is used to differentiate composite functions. If y = f(g(x)), then:
dy/dx = f'(g(x)) · g'(x).

For y = sin(2x + 3):
Let u = 2x + 3, so y = sin(u).
dy/du = cos(u) and du/dx = 2.
By chain rule:
dy/dx = dy/du · du/dx
= cos(u) · 2
= 2cos(2x + 3).

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Using the first principles of derivatives, find the derivative of the function f(x) = x2 + 3x. Explain the significance of this method in calculus.
Answer:
Theoretical Framework

We studied that the derivative of a function using first principles is defined as the limit of the difference quotient. For f(x) = x2 + 3x, we apply the formula: f'(x) = limh→0 [f(x+h) - f(x)]/h.


Evidence Analysis
  • Step 1: Compute f(x+h) = (x+h)2 + 3(x+h).
  • Step 2: Simplify the difference quotient to [2xh + h2 + 3h]/h.
  • Step 3: Take the limit as h→0 to get f'(x) = 2x + 3.

Critical Evaluation

This method is foundational as it derives the derivative from basic limits, reinforcing the connection between limits and derivatives.

Question 2:
Evaluate the limit: limx→2 (x3 - 8)/(x - 2). Discuss why this limit represents the derivative of a function at a point.
Answer:
Theoretical Framework

Our textbook shows that limits of the form limx→a [f(x) - f(a)]/(x - a) define the derivative f'(a). Here, the limit resembles the derivative of f(x) = x3 at x = 2.


Evidence Analysis
  • Step 1: Factorize the numerator as (x - 2)(x2 + 2x + 4).
  • Step 2: Cancel (x - 2) and substitute x = 2 to get 12.

Critical Evaluation

This confirms that the limit equals f'(2) where f(x) = x3, illustrating how limits underpin derivative definitions.

Question 3:
A car's position is modeled by s(t) = t2 + 5t (meters). Find its velocity at t = 3s using derivatives. Explain the real-world significance of this calculation.
Answer:
Theoretical Framework

We know velocity is the derivative of position. For s(t) = t2 + 5t, the derivative v(t) = ds/dt = 2t + 5.


Evidence Analysis
  • Step 1: Compute v(t) = 2t + 5.
  • Step 2: Substitute t = 3 to get v(3) = 11 m/s.

Critical Evaluation

This shows how derivatives model real-world motion, linking calculus to physics. The result helps predict the car's speed at any time.

Question 4:
Using the first principles of derivatives, find the derivative of the function f(x) = x3 + 2x. Explain the significance of the limit approach in calculus.
Answer:
Theoretical Framework

We studied that the derivative of a function f(x) at a point x=a is defined as the limit of the difference quotient as h approaches 0. For f(x) = x3 + 2x, we apply the first principles formula: f'(x) = limh→0 [(f(x+h) - f(x))/h].


Evidence Analysis
  • Step 1: Expand f(x+h) = (x+h)3 + 2(x+h) = x3 + 3x2h + 3xh2 + h3 + 2x + 2h
  • Step 2: Compute f(x+h) - f(x) = 3x2h + 3xh2 + h3 + 2h
  • Step 3: Divide by h → 3x2 + 3xh + h2 + 2
  • Step 4: Take limit h→0 → f'(x) = 3x2 + 2

Critical Evaluation

The limit approach ensures precise calculation of instantaneous rates of change. Our textbook shows this method works for all polynomial functions, confirming its universality.

Question 5:
Evaluate limx→2 (x2 - 4)/(x - 2) using algebraic manipulation. Discuss why direct substitution fails and how limits resolve indeterminate forms.
Answer:
Theoretical Framework

We encounter the 0/0 indeterminate form when directly substituting x=2 in limx→2 (x2-4)/(x-2). Our textbook shows factorization resolves this by canceling common terms.


Evidence Analysis
  • Step 1: Factor numerator → (x+2)(x-2)
  • Step 2: Cancel (x-2) → limx→2 (x+2)
  • Step 3: Now substitute x=2 → 2+2 = 4

Critical Evaluation

Limits help analyze functions at points where they're undefined. This technique is vital for calculus as seen in derivative definitions. [Diagram: Graph showing hole at x=2 with y approaching 4]

Question 6:
A particle's position is given by s(t) = t2 + 5t (meters). Find its instantaneous velocity at t=3s using derivatives. Relate this to real-world motion analysis.
Answer:
Theoretical Framework

We know velocity is the derivative of position. For s(t) = t2 + 5t, we first find v(t) = ds/dt.


Evidence Analysis
  • Step 1: Compute derivative → v(t) = 2t + 5
  • Step 2: Evaluate at t=3 → v(3) = 2(3) + 5 = 11 m/s

Critical Evaluation

This matches our textbook's kinematic equations. Derivatives transform position-time data into velocity information, crucial for physics applications like motion sensors. [Diagram: Position-time parabola with tangent at t=3]

Question 7:
Using the first principles of derivatives, find the derivative of the function \( f(x) = x^2 + 3x \). Verify your result using standard differentiation rules. Discuss the significance of limits in this derivation.
Answer:
Theoretical Framework

We studied that the derivative of a function at a point is defined as the limit of the difference quotient. For \( f(x) = x^2 + 3x \), we apply the first principle: \( f'(x) = \\lim_{h \ o 0} \ rac{f(x+h) - f(x)}{h} \).


Evidence Analysis
  • Step 1: Expand \( f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h \).
  • Step 2: Substitute into the limit: \( f'(x) = \\lim_{h \ o 0} \ rac{(2xh + h^2 + 3h)}{h} = 2x + 3 \).
  • Verification: Using standard rules, \( \ rac{d}{dx}(x^2) = 2x \) and \( \ rac{d}{dx}(3x) = 3 \), confirming the result.

Critical Evaluation

Limits ensure precision in defining instantaneous rates of change. Without them, derivatives would lack rigor.

Question 8:
A car's position function is \( s(t) = t^3 - 6t^2 + 9t \) (meters). Find its velocity and acceleration at \( t = 2 \) seconds. Interpret the results in a real-world context.
Answer:
Theoretical Framework

We learned that velocity \( v(t) \) is the first derivative of \( s(t) \), and acceleration \( a(t) \) is the second derivative. For \( s(t) = t^3 - 6t^2 + 9t \), we differentiate twice.


Evidence Analysis
  • Velocity: \( v(t) = 3t^2 - 12t + 9 \). At \( t = 2 \), \( v(2) = 3(4) - 24 + 9 = -3 \) m/s.
  • Acceleration: \( a(t) = 6t - 12 \). At \( t = 2 \), \( a(2) = 0 \) m/s².

Critical Evaluation

The negative velocity indicates reverse motion. Zero acceleration means constant velocity at \( t = 2 \), useful for predicting motion in physics.

Question 9:
Using the first principles of derivatives, find the derivative of the function f(x) = √x. Verify your result using standard differentiation rules.
Answer:
Theoretical Framework

We studied that the derivative of a function using first principles is given by: f'(x) = limh→0 [f(x+h) - f(x)]/h. For f(x) = √x, we substitute and simplify.


Evidence Analysis
  • Substituting: f'(x) = limh→0 [√(x+h) - √x]/h
  • Rationalizing numerator: Multiply by (√(x+h) + √x)/(√(x+h) + √x)
  • Simplifies to: limh→0 h/[h(√(x+h) + √x)] = 1/(2√x)

Critical Evaluation

Our textbook shows the standard derivative of √x as 1/(2√x), confirming our result. This validates the first principles approach.

Question 10:
A car's position is modeled by s(t) = t³ - 6t² + 9t (meters). Analyze its velocity and acceleration at t = 2s. Interpret the results physically.
Answer:
Theoretical Framework

We know velocity v(t) is the derivative of s(t), and acceleration a(t) is the derivative of v(t). We differentiate the given function.


Evidence Analysis
  • v(t) = ds/dt = 3t² - 12t + 9. At t=2, v(2) = -3 m/s
  • a(t) = dv/dt = 6t - 12. At t=2, a(2) = 0 m/s²

Critical Evaluation

The negative velocity indicates motion opposite to the positive direction. Zero acceleration means constant velocity at that instant, suggesting a possible turning point.

Question 11:
A particle moves along a straight line with position function s(t) = t³ - 6t² + 9t. Analyze its velocity and acceleration at t = 2 seconds. Interpret the physical significance of these values.
Answer:
Theoretical Framework

We know that velocity v(t) is the first derivative of s(t), and acceleration a(t) is the second derivative. For s(t) = t³ - 6t² + 9t, we differentiate twice.


Evidence Analysis
  • First derivative: v(t) = 3t² - 12t + 9. At t = 2, v(2) = -3 m/s (negative indicates reverse direction).
  • Second derivative: a(t) = 6t - 12. At t = 2, a(2) = 0 m/s² (constant velocity).

Critical Evaluation

The negative velocity shows the particle is moving backward, while zero acceleration means no change in speed at t = 2. This aligns with real-world motion where acceleration affects velocity.

Question 12:
A car's position is modeled by s(t) = t³ - 6t² + 9t (meters). Analyze its velocity and acceleration at t=2s using derivatives. Discuss the physical interpretation.
Answer:
Theoretical Framework

We know velocity v(t) is the first derivative of s(t), and acceleration a(t) is the second derivative. We differentiate s(t) = t³ - 6t² + 9t.


Evidence Analysis
  • v(t) = ds/dt = 3t² - 12t + 9 → v(2) = 3(4) - 24 + 9 = -3 m/s
  • a(t) = dv/dt = 6t - 12 → a(2) = 0 m/s²

Critical Evaluation

The negative velocity at t=2s indicates reverse motion. Zero acceleration means constant velocity at that instant, which matches real-world scenarios where objects momentarily maintain speed.

Question 13:
A particle moves along a straight line with position function s(t) = t³ - 6t² + 9t. Analyze its velocity and acceleration at t = 2 seconds. Interpret the physical significance.
Answer:
Theoretical Framework

We know that velocity v(t) is the first derivative of s(t), and acceleration a(t) is the second derivative. For s(t) = t³ - 6t² + 9t, we differentiate twice.


Evidence Analysis
  • Velocity: v(t) = 3t² - 12t + 9. At t = 2, v(2) = -3 m/s
  • Acceleration: a(t) = 6t - 12. At t = 2, a(2) = 0 m/s²

Critical Evaluation

The negative velocity indicates motion opposite to the positive direction. Zero acceleration implies constant velocity at t = 2, a critical point in the particle's motion.

Question 14:
Using the first principles of derivatives, find the derivative of the function f(x) = x2 + 3x. Verify your result using standard differentiation rules.
Answer:
Theoretical Framework

We studied that the derivative of a function using first principles is defined as: f'(x) = limh→0 [f(x+h) - f(x)]/h. For f(x) = x2 + 3x, we apply this definition.


Evidence Analysis
  • Step 1: f(x+h) = (x+h)2 + 3(x+h) = x2 + 2xh + h2 + 3x + 3h
  • Step 2: f(x+h) - f(x) = 2xh + h2 + 3h
  • Step 3: Divide by h and take limit: f'(x) = limh→0 (2x + h + 3) = 2x + 3

Critical Evaluation

Using standard rules, the derivative of x2 is 2x and of 3x is 3, confirming our result. This consistency validates the first principles approach.

Question 15:
A particle moves along a straight line with position function s(t) = t3 - 6t2 + 9t. Analyze its velocity and acceleration at t = 2 seconds. Interpret the physical meaning.
Answer:
Theoretical Framework

We know velocity v(t) is the first derivative of s(t), and acceleration a(t) is the second derivative. For s(t) = t3 - 6t2 + 9t, we differentiate twice.


Evidence Analysis
  • Velocity: v(t) = 3t2 - 12t + 9. At t=2, v(2) = 3(4) - 24 + 9 = -3 m/s
  • Acceleration: a(t) = 6t - 12. At t=2, a(2) = 0 m/s2

Critical Evaluation

The negative velocity indicates backward motion, while zero acceleration means constant velocity at that instant. This aligns with real-world kinematics where derivatives model motion dynamics.

Question 16:
Using the first principles of derivatives, find the derivative of the function f(x) = √x. Verify your result using standard differentiation rules and discuss its significance in calculus.
Answer:
Theoretical Framework

We studied that the derivative of a function using first principles is defined as: f'(x) = limh→0 [f(x+h) - f(x)]/h. For f(x) = √x, substituting into the formula gives f'(x) = limh→0 [√(x+h) - √x]/h.


Evidence Analysis
  • Rationalizing the numerator: Multiply by [√(x+h) + √x]/[√(x+h) + √x].
  • Simplifies to limh→0 h/[h(√(x+h) + √x)] = 1/(2√x).
  • Standard rule confirms d/dx(√x) = 1/(2√x).

Critical Evaluation

This derivation reinforces the power of first principles in validating standard results. It also highlights the continuity and differentiability of root functions.

Question 17:
A car's position is modeled by s(t) = t³ - 6t² + 9t (meters). Analyze its velocity and acceleration at t=2s using derivatives. Interpret the physical meaning of your results.
Answer:
Theoretical Framework

We know velocity v(t) is the first derivative of s(t), and acceleration a(t) is the second derivative. For s(t) = t³ - 6t² + 9t, we compute these derivatives algebraically.


Evidence Analysis
  • v(t) = ds/dt = 3t² - 12t + 9 → v(2) = -3 m/s (moving backward).
  • a(t) = d²s/dt² = 6t - 12 → a(2) = 0 m/s² (constant velocity).

Critical Evaluation

The negative velocity indicates reversed direction, while zero acceleration shows no change in speed at t=2s. This models real-world motion where derivatives describe instantaneous rates of change.

Question 18:
Using the first principles of derivatives, find the derivative of the function f(x) = √x. Explain each step with proper reasoning.
Answer:
Theoretical Framework

We studied that the derivative of a function f(x) using first principles is given by the limit: f'(x) = limh→0 [f(x+h) - f(x)] / h. For f(x) = √x, we substitute the function into the formula.


Evidence Analysis
  • Substitute f(x+h) = √(x+h) and f(x) = √x into the limit formula.
  • Rationalize the numerator by multiplying by (√(x+h) + √x).
  • Simplify to get f'(x) = limh→0 1 / (√(x+h) + √x).
  • Evaluate the limit as h→0 to obtain f'(x) = 1 / (2√x).

Critical Evaluation

This derivation confirms the standard result from our textbook. The rationalization step is crucial to eliminate the indeterminate form.

Question 19:
A particle moves along a straight line with position given by s(t) = t³ - 6t² + 9t + 4. Analyze its velocity and acceleration at t = 2 seconds. Interpret the physical meaning.
Answer:
Theoretical Framework

We know that velocity v(t) is the first derivative of s(t), and acceleration a(t) is the second derivative. We differentiate s(t) twice to find these.


Evidence Analysis
  • First derivative: v(t) = 3t² - 12t + 9. At t=2, v(2) = -3 m/s (negative indicates reverse direction).
  • Second derivative: a(t) = 6t - 12. At t=2, a(2) = 0 m/s² (constant velocity).

Critical Evaluation

The results show the particle is moving backward at t=2 with zero acceleration, meaning no change in velocity at that instant. This aligns with our textbook's examples of motion analysis.

Question 20:
Define the concept of a limit in calculus and explain its significance in understanding the behavior of functions. Illustrate with an example of a function approaching a specific limit.
Answer:

The limit of a function is a fundamental concept in calculus that describes the value a function approaches as the input (or variable) approaches a certain point. It helps us understand the behavior of functions, especially near points where they might not be explicitly defined or exhibit abrupt changes.


Significance of limits:

  • Limits allow us to analyze functions at points where they are not directly evaluable.
  • They form the foundation for defining continuity, derivatives, and integrals.
  • Limits help in understanding asymptotic behavior and trends in real-world applications like physics and engineering.

Example: Consider the function f(x) = (x² - 1)/(x - 1). At x = 1, the function is undefined (0/0 form). However, by simplifying, we get f(x) = x + 1 for x ≠ 1. Thus, the limit as x approaches 1 is 2. This shows how limits help fill gaps in functions.

Question 21:
Explain the first principles of derivatives (using limits) and derive the derivative of the function f(x) = x² from first principles.
Answer:

The first principles of derivatives involve computing the derivative of a function using the limit definition. It measures the instantaneous rate of change of the function at any point.


Steps to derive the derivative of f(x) = x²:


1. Start with the definition of the derivative using limits:
f'(x) = limh→0 [f(x + h) - f(x)] / h


2. Substitute f(x) = x² into the formula:
f'(x) = limh→0 [(x + h)² - x²] / h


3. Expand (x + h)²:
f'(x) = limh→0 [x² + 2xh + h² - x²] / h


4. Simplify the numerator:
f'(x) = limh→0 [2xh + h²] / h


5. Factor out h:
f'(x) = limh→0 h(2x + h) / h


6. Cancel h and take the limit as h→0:
f'(x) = 2x


Thus, the derivative of f(x) = x² is 2x, derived purely from first principles.

Question 22:
Define the concept of a limit in calculus and explain its significance in determining the behavior of functions. Illustrate with an example of a function approaching a specific limit.
Answer:

The limit of a function at a point is a fundamental concept in calculus that describes the value a function approaches as the input approaches a certain point. It helps analyze the behavior of functions, especially near points where they might not be explicitly defined.


Significance: Limits allow us to study:

  • Continuity of functions
  • Derivatives (rates of change)
  • Behavior near undefined points (e.g., holes or asymptotes)

Example: Consider the function f(x) = (x² - 1)/(x - 1). At x = 1, the function is undefined, but we can evaluate its limit as x approaches 1:


Step 1: Simplify the function:
f(x) = (x - 1)(x + 1)/(x - 1)
Step 2: Cancel the common term (x - 1):
f(x) = x + 1 (for x ≠ 1)
Step 3: Now, take the limit as x approaches 1:
lim (x→1) f(x) = 1 + 1 = 2


Thus, the function approaches 2 as x approaches 1, even though it is undefined at x = 1.

Question 23:
Differentiate between the first principle of derivatives and standard differentiation rules. Derive the derivative of f(x) = x² using the first principle and verify it using standard rules.
Answer:

First Principle of Derivatives (Definition): It is the fundamental method to find derivatives using limits. It defines the derivative as:
f'(x) = lim (h→0) [f(x + h) - f(x)] / h


Standard Differentiation Rules: These are shortcut formulas derived from the first principle, such as the power rule, product rule, etc., for faster calculations.


Derivation using First Principle (for f(x) = x²):
Step 1: Write the difference quotient:
[f(x + h) - f(x)] / h = [(x + h)² - x²] / h
Step 2: Expand (x + h)²:
= [x² + 2xh + h² - x²] / h
Step 3: Simplify:
= (2xh + h²) / h = h(2x + h) / h = 2x + h
Step 4: Take the limit as h→0:
f'(x) = lim (h→0) (2x + h) = 2x


Verification using Standard Rules:
Using the power rule (d/dx [x^n] = nx^(n-1)):
f'(x) = d/dx (x²) = 2x^(2-1) = 2x


Both methods yield the same result, confirming the correctness of the derivative.

Question 24:
Define the concept of a limit in calculus and explain its significance with an example. Also, discuss the difference between left-hand limit and right-hand limit with a graphical representation.
Answer:

The limit of a function at a point is a fundamental concept in calculus that describes the behavior of the function as the input approaches that point. It is denoted as limx→a f(x) = L, meaning that as x gets arbitrarily close to a, the function f(x) approaches the value L.


Significance: Limits help in understanding continuity, derivatives, and integrals. For example, the derivative of a function is defined using limits.


Example: Consider f(x) = (x2 - 1)/(x - 1). As x approaches 1, f(x) approaches 2, even though the function is undefined at x = 1.


Left-hand limit (LHL) is the value the function approaches as x approaches a from the left (x → a-). Right-hand limit (RHL) is the value as x approaches a from the right (x → a+). For a limit to exist, LHL and RHL must be equal.


Graphical Representation: Imagine a piecewise function with a jump discontinuity at x = a. The LHL and RHL would be different, showing the function approaches different values from either side.

Question 25:
Differentiate between average rate of change and instantaneous rate of change using the concept of derivatives. Provide a real-life application where this distinction is crucial.
Answer:

The average rate of change of a function f(x) over an interval [a, b] is the slope of the secant line joining the points (a, f(a)) and (b, f(b)). It is calculated as (f(b) - f(a))/(b - a).


The instantaneous rate of change at a point x = a is the derivative f'(a), representing the slope of the tangent line at that point. It is the limit of the average rate of change as the interval approaches zero.


Key Difference: The average rate gives a broad overview over an interval, while the instantaneous rate provides precise information at a specific point.


Real-life Application: In speed monitoring, the average speed over a trip is the average rate of change of distance with time. However, the instantaneous speed (derivative) is crucial for traffic police to determine if a vehicle is speeding at a particular moment.

Question 26:
Evaluate the limit:
limx→0 (sin(3x) / tan(5x))
Justify each step using appropriate limit theorems and trigonometric identities.
Answer:

To evaluate the given limit, we will use standard trigonometric identities and the fundamental limit limθ→0 (sinθ/θ) = 1.


Step 1: Express the limit in terms of sine and cosine functions:
limx→0 (sin(3x) / tan(5x)) = limx→0 (sin(3x) / (sin(5x)/cos(5x)))
= limx→0 (sin(3x) * cos(5x) / sin(5x))


Step 2: Split the limit into simpler parts using the product rule of limits:
= limx→0 (sin(3x)/sin(5x)) * limx→0 cos(5x)


Step 3: Evaluate limx→0 cos(5x) directly:
Since cosine is continuous, limx→0 cos(5x) = cos(0) = 1


Step 4: Rewrite the remaining limit using the identity limθ→0 (sinθ/θ) = 1:
limx→0 (sin(3x)/sin(5x)) = limx→0 [(sin(3x)/3x * 3x] / [sin(5x)/5x * 5x]
= (3/5) * limx→0 [(sin(3x)/3x) / (sin(5x)/5x)]
= (3/5) * [1 / 1] = 3/5


Final Answer: Combining all steps, the limit evaluates to (3/5) * 1 = 3/5.


Key Concepts Used:

  • Trigonometric identity for tan(x)
  • Product rule of limits
  • Continuity of cosine function
  • Standard limit for sin(x)/x

Question 27:
Using the first principles of derivatives, find the derivative of the function f(x) = x2 + 3x + 2. Show all steps clearly and explain the significance of the derivative in this context.
Answer:

To find the derivative of f(x) = x2 + 3x + 2 using the first principles, we apply the definition of the derivative:

f'(x) = limh→0 [f(x + h) - f(x)] / h

Step 1: Compute f(x + h):
f(x + h) = (x + h)2 + 3(x + h) + 2
= x2 + 2xh + h2 + 3x + 3h + 2

Step 2: Subtract f(x) from f(x + h):
f(x + h) - f(x) = (x2 + 2xh + h2 + 3x + 3h + 2) - (x2 + 3x + 2)
= 2xh + h2 + 3h

Step 3: Divide by h:
[f(x + h) - f(x)] / h = (2xh + h2 + 3h) / h
= 2x + h + 3

Step 4: Take the limit as h→0:
f'(x) = limh→0 (2x + h + 3)
= 2x + 3

Significance of the derivative: The derivative f'(x) = 2x + 3 represents the instantaneous rate of change of the function f(x) at any point x. For example, if x represents time, the derivative gives the velocity of an object whose position is described by f(x).

Question 28:
Using the first principles of derivatives, find the derivative of the function f(x) = x2 + 3x + 2. Show all steps clearly and explain the significance of the limit concept in this process.
Answer:

To find the derivative of f(x) = x2 + 3x + 2 using first principles, we apply the definition of the derivative:


f'(x) = limh→0 [f(x + h) - f(x)] / h


Step 1: Compute f(x + h):


f(x + h) = (x + h)2 + 3(x + h) + 2
= x2 + 2xh + h2 + 3x + 3h + 2


Step 2: Subtract f(x) from f(x + h):


f(x + h) - f(x) = [x2 + 2xh + h2 + 3x + 3h + 2] - [x2 + 3x + 2]
= 2xh + h2 + 3h


Step 3: Divide by h:


[f(x + h) - f(x)] / h = (2xh + h2 + 3h) / h
= 2x + h + 3


Step 4: Take the limit as h→0:


f'(x) = limh→0 (2x + h + 3)
= 2x + 0 + 3
= 2x + 3


Significance of the Limit Concept: The limit ensures that we find the instantaneous rate of change of the function by making h infinitesimally small. Without the limit, we would only get an average rate of change over an interval, not the exact slope at a point.

Question 29:
Explain the concept of limits in calculus with a suitable example. Discuss its significance in finding the derivative of a function.
Answer:

The concept of limits is fundamental in calculus and helps us understand the behavior of a function as the input approaches a certain value. A limit answers the question: 'What value does the function approach as the variable gets arbitrarily close to a specific point?'


Example: Consider the function f(x) = (x² - 1)/(x - 1). To find the limit as x approaches 1, we observe that direct substitution gives an indeterminate form (0/0). However, simplifying the function:


f(x) = (x - 1)(x + 1)/(x - 1) = x + 1 (for x ≠ 1)

Thus, the limit as x → 1 is 2, even though f(1) is undefined.


Significance in Derivatives: The derivative of a function at a point is defined using limits:


f'(x) = limh→0 [f(x + h) - f(x)]/h

This formula calculates the instantaneous rate of change (slope of the tangent) by examining the behavior of the difference quotient as h approaches 0. Without limits, we could only approximate derivatives using average rates of change over intervals.


Key Insight: Limits allow us to analyze functions at points where they might not be explicitly defined, bridging gaps in continuity and enabling precise calculations in calculus.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A car's speed is modeled by the function s(t) = 3t² + 2t (in km/h), where t is time in hours. Using limits, find the instantaneous rate of change of speed at t = 2. Explain the physical meaning of this result.
Answer:
Problem Interpretation

We need to find the derivative of s(t) at t=2, representing the car's acceleration at that moment.

Mathematical Modeling

The derivative s'(t) is the limit of the difference quotient as h→0:

  • s'(t) = limh→0 [s(t+h) - s(t)]/h
Solution

Substituting s(t):

  • s'(t) = 6t + 2
  • At t=2, s'(2) = 14 km/h²

This means the car's speed increases by 14 km/h every hour at t=2.

Question 2:
Prove using first principles that the derivative of f(x) = √x is 1/(2√x). Include all limit steps and rationalization.
Answer:
Problem Interpretation

We must derive f'(x) for f(x)=√x using the limit definition of derivatives.

Mathematical Modeling

The difference quotient is:

  • f'(x) = limh→0 [√(x+h) - √x]/h
Solution

Rationalizing the numerator:

  • Multiply numerator and denominator by (√(x+h) + √x)
  • Simplifies to: limh→0 h/[h(√(x+h) + √x)]
  • Cancel h: 1/(√(x+h) + √x)

Taking limit as h→0 gives 1/(2√x), completing the proof.

Question 3:
A car's position function is given by s(t) = 2t² + 3t + 5 (in meters). Using limits, find its instantaneous velocity at t = 2s. Also, derive the velocity function using derivatives.
Answer:
Problem Interpretation

We need to find the car's velocity at t=2s using limits and confirm it via differentiation.


Mathematical Modeling
  • Instantaneous velocity = limh→0 [s(2+h)-s(2)]/h
  • Derivative approach: v(t) = ds/dt

Solution

Limit method: Substituting s(t), we get limh→0 [2(2+h)²+3(2+h)+5-19]/h = 11 m/s. Derivative method: v(t)=4t+3 → v(2)=11 m/s. Both methods agree.

Question 4:
Prove that the derivative of f(x) = sin(x) is cos(x) using the first principles method. Include all key steps.
Answer:
Problem Interpretation

Our textbook shows derivatives can be computed using limh→0 [f(x+h)-f(x)]/h.


Mathematical Modeling
  • Start with derivative definition
  • Apply trigonometric identity for sin(x+h)

Solution

f'(x)=limh→0 [sin(x+h)-sinx]/h = limh→0 [2cos(x+h/2)sin(h/2)]/h. Using limθ→0 sinθ/θ=1, this simplifies to cosx.

Question 5:
A car's speed is modeled by the function v(t) = 3t² + 2t (in km/h) where t is time in hours. Using limits, find the instantaneous acceleration at t = 2 hours.
Answer:
Problem Interpretation

We need to find the car's acceleration at t = 2 using the derivative of v(t).

Mathematical Modeling

Acceleration is the derivative of velocity. We compute v'(t) using first principles:

Solution
  • Limit definition: v'(t) = limh→0 (v(t+h) - v(t))/h
  • Substitute v(t) = 3t² + 2t: v'(t) = 6t + 2
  • At t = 2, v'(2) = 6(2) + 2 = 14 km/h²
Question 6:
Prove using limits that the derivative of f(x) = sin x is cos x by first principles.
Answer:
Problem Interpretation

Our textbook shows derivatives can be computed using limits. We'll prove d/dx(sin x) = cos x.

Mathematical Modeling

Using the limit definition: f'(x) = limh→0 (sin(x+h) - sin x)/h.

Solution
  • Expand sin(x+h) using identity: sin x cos h + cos x sin h
  • Simplify limit: limh→0 [sin x(cos h - 1)/h + cos x(sin h/h)]
  • Apply standard limits: sin x(0) + cos x(1) = cos x
Question 7:
A car's position function is s(t) = 2t² + 3t + 5 (in meters). Find its instantaneous velocity at t = 2s using limits. Also, derive the velocity function using first principles.
Answer:
Problem Interpretation

We need to find the car's velocity at t=2s by calculating the derivative of s(t) using limits.

Mathematical Modeling
  • Velocity v(t) = limh→0 [s(t+h) - s(t)]/h
  • Substitute s(t) = 2t² + 3t + 5
Solution

Expanding s(t+h): 2(t+h)² + 3(t+h) + 5 = 2t² + 4th + 2h² + 3t + 3h + 5. The difference quotient simplifies to (4th + 2h² + 3h)/h = 4t + 2h + 3. Taking limit h→0 gives v(t) = 4t + 3. At t=2, v(2) = 11 m/s.

Question 8:
The population P(t) of bacteria grows as P(t) = 500e0.1t. Using limits, find its growth rate at t=5 hours. Verify by differentiation rules.
Answer:
Problem Interpretation

We must determine how fast the bacteria population is growing at t=5h by finding P'(5) using limits.

Mathematical Modeling
  • Growth rate = limh→0 [P(t+h) - P(t)]/h
  • P(t) = 500e0.1t
Solution

The difference quotient is 500(e0.1(t+h) - e0.1t)/h = 500e0.1t(e0.1h - 1)/h. Using limx→0(ex-1)/x = 1, we get P'(t) = 50e0.1t. At t=5, P'(5) ≈ 82.4 bacteria/hour. Differentiation confirms this result.

Question 9:
A car's speed is modeled by the function s(t) = 3t² + 2t (in km/h), where t is time in hours. Using limits, find the instantaneous rate of change of speed at t = 2. Interpret the result.
Answer:
Problem Interpretation

We need to find the derivative of s(t) at t = 2 to determine the car's acceleration at that moment.


Mathematical Modeling

The derivative s'(t) represents the instantaneous rate of change. We compute it using the limit definition:


Solution
  • s'(t) = limh→0 (s(t+h) - s(t))/h
  • Substitute s(t) = 3t² + 2t and simplify to get s'(t) = 6t + 2.
  • At t = 2, s'(2) = 6(2) + 2 = 14 km/h².

This means the car's speed increases by 14 km/h every hour at t = 2.

Question 10:
Prove using first principles that the derivative of f(x) = √x is f'(x) = 1/(2√x). Show all steps.
Answer:
Problem Interpretation

We must derive f'(x) for f(x) = √x using the limit definition of derivatives.


Mathematical Modeling

The derivative is f'(x) = limh→0 (f(x+h) - f(x))/h.


Solution
  • Substitute f(x) = √x: f'(x) = limh→0 (√(x+h) - √x)/h.
  • Multiply numerator and denominator by (√(x+h) + √x) to rationalize.
  • Simplify to f'(x) = limh→0 h/(h(√(x+h) + √x)).
  • Cancel h and take limit: f'(x) = 1/(2√x).

Our textbook shows this confirms the derivative formula.

Question 11:
A car's speed is given by the function s(t) = 3t2 + 2t + 5 (in km/h), where t is time in hours. Using the concept of limits, find the instantaneous speed of the car at t = 2 hours. Show all steps clearly.
Answer:

To find the instantaneous speed at t = 2, we compute the derivative of s(t) using the first principles of derivatives (limit definition).


The function is: s(t) = 3t2 + 2t + 5

Step 1: Apply the limit definition of the derivative:
s'(t) = limh→0 [s(t + h) - s(t)] / h

Step 2: Substitute t = 2 and expand s(2 + h):
s(2 + h) = 3(2 + h)2 + 2(2 + h) + 5 = 12 + 12h + 3h2 + 4 + 2h + 5 = 3h2 + 14h + 21
s(2) = 3(2)2 + 2(2) + 5 = 12 + 4 + 5 = 21

Step 3: Compute the difference quotient:
[s(2 + h) - s(2)] / h = [3h2 + 14h + 21 - 21] / h = (3h2 + 14h) / h = 3h + 14

Step 4: Take the limit as h→0:
s'(2) = limh→0 (3h + 14) = 14

Thus, the instantaneous speed at t = 2 hours is 14 km/h.
Question 12:
A function f(x) is defined as f(x) = (x2 - 4) / (x - 2) for x ≠ 2. Using the concept of limits, determine whether the function is continuous at x = 2. If not, identify the type of discontinuity.
Answer:

To check continuity at x = 2, we evaluate the limit of f(x) as x→2 and compare it with f(2) (if defined).


Step 1: Simplify the function for x ≠ 2:
f(x) = (x2 - 4) / (x - 2) = [(x - 2)(x + 2)] / (x - 2) = x + 2 (for x ≠ 2)

Step 2: Compute the limit as x→2:
limx→2 f(x) = limx→2 (x + 2) = 4

Step 3: Check f(2):
The original function is undefined at x = 2 (denominator becomes zero).

Conclusion:
Since the limit exists (4) but f(2) is undefined, the function has a removable discontinuity (hole) at x = 2.

Note: The discontinuity can be removed by defining f(2) = 4, making the function continuous.
Question 13:
A car's speed is given by the function s(t) = 3t² + 2t + 5, where t is time in seconds and s(t) is speed in m/s.

(a) Find the instantaneous rate of change of speed at t = 2 seconds.

(b) Interpret the physical meaning of this result.

Answer:

(a) To find the instantaneous rate of change of speed, we compute the derivative of s(t) with respect to t.


s(t) = 3t² + 2t + 5
s'(t) = d/dt (3t²) + d/dt (2t) + d/dt (5)
s'(t) = 6t + 2
At t = 2 seconds:
s'(2) = 6(2) + 2 = 14 m/s²

(b) The result 14 m/s² represents the acceleration of the car at t = 2 seconds. It indicates how quickly the speed is increasing at that exact moment.

Question 14:
A function f(x) is defined as f(x) = (x² - 4)/(x - 2) for x ≠ 2 and f(2) = k.

(a) Find the limit of f(x) as x approaches 2.

(b) Determine the value of k that makes f(x) continuous at x = 2.

Answer:

(a) To find the limit of f(x) as x approaches 2, simplify the function:


f(x) = (x² - 4)/(x - 2)
Factorize the numerator:
f(x) = (x - 2)(x + 2)/(x - 2)
Cancel (x - 2) (valid since x ≠ 2):
f(x) = x + 2
Now, take the limit:
limx→2 f(x) = 2 + 2 = 4

(b) For f(x) to be continuous at x = 2, f(2) must equal the limit as x approaches 2. Thus, k = 4.

Question 15:
A car is moving along a straight road. The distance s (in meters) covered by the car in time t (in seconds) is given by s(t) = 2t2 + 3t. Using the concept of limits, find the instantaneous velocity of the car at t = 2 seconds.
Answer:

To find the instantaneous velocity at t = 2 seconds, we need to compute the derivative of s(t) with respect to t, which represents the velocity function v(t).


Given: s(t) = 2t2 + 3t


Step 1: Find the derivative s'(t) using the power rule of differentiation.


s'(t) = d/dt (2t2 + 3t)
s'(t) = 4t + 3

Step 2: Substitute t = 2 into s'(t) to find the instantaneous velocity.


v(2) = 4(2) + 3 = 8 + 3 = 11 m/s

Thus, the instantaneous velocity of the car at t = 2 seconds is 11 m/s.

Question 16:
A function f(x) is defined as f(x) = (x2 - 4)/(x - 2) for x ≠ 2. Using the concept of limits, determine whether the function is continuous at x = 2. If not, redefine the function to make it continuous.
Answer:

To check the continuity of f(x) at x = 2, we need to evaluate the limit of f(x) as x approaches 2.


Given: f(x) = (x2 - 4)/(x - 2)


Step 1: Simplify the function for x ≠ 2.


f(x) = (x - 2)(x + 2)/(x - 2) = x + 2 (for x ≠ 2)

Step 2: Find the limit as x approaches 2.


limx→2 f(x) = limx→2 (x + 2) = 4

Step 3: Check if f(2) is defined. The original function is undefined at x = 2.


Since the limit exists but the function is undefined at x = 2, the function is not continuous at x = 2.


To make it continuous, redefine f(x) as:


f(x) = { (x2 - 4)/(x - 2), x ≠ 2
4, x = 2 }

Now, the function is continuous at x = 2 because limx→2 f(x) = f(2) = 4.

Question 17:
A function f(x) is defined as f(x) = (x2 - 4) / (x - 2) for x ≠ 2 and f(2) = k. Determine the value of k such that f(x) is continuous at x = 2 using the concept of limits.
Answer:

For f(x) to be continuous at x = 2, the limit of f(x) as x→2 must exist and be equal to f(2) = k.


First, simplify the function for x ≠ 2:


f(x) = (x2 - 4) / (x - 2)

Factorize the numerator:


f(x) = (x - 2)(x + 2) / (x - 2)

Cancel the common factor (x - 2):


f(x) = x + 2 (for x ≠ 2)

Now, find the limit as x→2:


limx→2 f(x) = limx→2 (x + 2) = 2 + 2 = 4

For continuity, f(2) = k must equal this limit:


k = 4

Thus, the value of k that makes f(x) continuous at x = 2 is 4.

Question 18:
A car's speed is given by the function s(t) = 3t² + 2t + 5 (in km/h), where t is time in hours. Using the concept of limits, find the instantaneous speed of the car at t = 2 hours. Show all steps clearly.
Answer:

To find the instantaneous speed at t = 2 hours, we use the limit definition of the derivative of s(t).


The derivative s'(t) represents the instantaneous speed:

s'(t) = limh→0 [s(t + h) - s(t)] / h

Substitute s(t) = 3t² + 2t + 5:
s(t + h) = 3(t + h)² + 2(t + h) + 5 = 3t² + 6th + 3h² + 2t + 2h + 5

Now, compute the difference quotient:
[s(t + h) - s(t)] / h = [6th + 3h² + 2h] / h = 6t + 3h + 2

Take the limit as h→0:
s'(t) = 6t + 2

At t = 2:
s'(2) = 6(2) + 2 = 14 km/h

Thus, the car's instantaneous speed at t = 2 hours is 14 km/h.

Question 19:
A function f(x) is defined as f(x) = (x² - 4)/(x - 2) for x ≠ 2. Using the concept of limits, determine whether f(x) has a limit as x approaches 2. If yes, find the limit and explain if the function is continuous at x = 2.
Answer:

To find the limit of f(x) = (x² - 4)/(x - 2) as x→2, simplify the expression:


f(x) = (x² - 4)/(x - 2) = [(x - 2)(x + 2)]/(x - 2)

For x ≠ 2, cancel (x - 2):
f(x) = x + 2

Now, take the limit as x→2:
limx→2 f(x) = limx→2 (x + 2) = 4

The limit exists and equals 4. However, the function is not continuous at x = 2 because f(2) is undefined (division by zero).


For continuity, three conditions must be met:

  • f(2) must exist (fails here).
  • The limit as x→2 must exist (passes).
  • The limit must equal f(2) (cannot be checked).
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