Limits and Continuity of Functions Practice Questions — JAMB UTME

Evaluate the limit $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 4

Answer

The value of the limit is

4

.

Direct substitution of

x = 2

produces the indeterminate form

\frac{0}{0}

. Factoring the numerator gives

\frac{(x-2)(x+2)}{x-2}

. Canceling the non-zero factor

(x-2)

simplifies the expression to

x+2

. Evaluating the limit as

x

approaches

2

yields

2 + 2 = 4

.

Step-by-Step Solution

1

Check the form by direct substitution of

x = 2

Obtained the indeterminate form

\frac{0}{0}

Direct substitution results in division by zero, requiring algebraic simplification.

2

Factor the polynomial in the numerator

x^2 - 4 = (x - 2)(x + 2)

Difference of two squares factorization allows canceling common terms.

3

Cancel the common factor

(x - 2)

\frac{(x - 2)(x + 2)}{x - 2} = x + 2

For

x \neq 2

, division by

(x - 2)

is valid.

4

Evaluate the simplified limit as

x \to 2

2 + 2 = 4

Substitute

x = 2

directly into the continuous polynomial

x + 2

.

Key Concept

Evaluating indeterminate limits of the form

\frac{0}{0}

via factorization

Question 2Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + x - 6}{x - 2}, & x \neq 2 \\ 2k - 1, & x = 2 \end{cases}

If

f(x)

is continuous at

x = 2

, what is the value of the constant

k

?

3

2

5

-1

Show answer & explanation

Answer: 3

Answer

The value of the constant

k

is

3

.

By definition, a function

f(x)

is continuous at

x = c

if

\lim_{x \to c} f(x) = f(c)

. Factoring the numerator gives

x^2 + x - 6 = (x - 2)(x + 3)

. Canceling the common factor

(x - 2)

for

x \neq 2

, the limit as

x \to 2

is

2 + 3 = 5

. Equating

f(2) = 2k - 1

to 5 yields

2k - 1 = 5

, which solves to

k = 3

.

Step-by-Step Solution

1

Evaluate the limit of

f(x)

as

x

approaches

2

.

\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2} = \lim_{x \to 2} (x + 3) = 5

Direct substitution gives the indeterminate form

\frac{0}{0}

, so factor the numerator to simplify.

2

Apply the definition of continuity at a point.

f(2) = \lim_{x \to 2} f(x) \implies 2k - 1 = 5

For

f(x)

to be continuous at

x = 2

, the value of the function at

x = 2

must equal its limit as

x \to 2

.

3

Solve the linear equation for

k

.

2k = 6 \implies k = 3

Add 1 to both sides and divide by 2.

Key Concept

Continuity of a Piecewise Function at a Point

Alternative Method

Alternatively, use L'Hôpital's rule to evaluate the limit:

\lim_{x \to 2} \frac{\frac{d}{dx}(x^2+x-6)}{\frac{d}{dx}(x-2)} = \lim_{x \to 2} \frac{2x+1}{1} = 5

. Then set

2k - 1 = 5

to find

k = 3

.

Estimated Time:1m 30s

Question 3Question →

Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 0.25

Answer

The numerical value of the limit is 0.25 (or

\frac{1}{4}

).

When direct substitution into

\frac{\sqrt{x + 4} - 2}{x}

yields the indeterminate form

\frac{0}{0}

, rationalizing the numerator by multiplying by its conjugate

\sqrt{x + 4} + 2

simplifies the expression to

\frac{1}{\sqrt{x + 4} + 2}

. Taking the limit as

x \to 0

gives

\frac{1}{4} = 0.25

.

Step-by-Step Solution

1

Check direct substitution

Indeterminate form

\frac{0}{0}

Directly evaluating at

x = 0

yields zero in both numerator and denominator.

2

Multiply by the conjugate of the numerator

\frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)} = \frac{x}{x(\sqrt{x + 4} + 2)}

Rationalizing the radical in the numerator allows cancellation of the term causing the zero denominator.

3

Cancel common factors and evaluate limit

\frac{1}{\sqrt{0 + 4} + 2} = 0.25

Canceling

x

removes the zero factor, permitting direct evaluation.

Key Concept

Limits of indeterminate algebraic expressions using surd rationalization

Question 4Question →

What is the numerical value of the limit

\lim_{x \to 4} \frac{x - 4}{\sqrt{x + 5} - 3}

?

6

3

0

\frac{1}{6}

Show answer & explanation

Answer: $6$

Answer

The value of the limit is

6

.

Multiplying both the numerator and the denominator by the conjugate of the denominator

(\sqrt{x + 5} + 3)

allows the factor

(x - 4)

to cancel out, leaving

\sqrt{x + 5} + 3

. Evaluating this expression as

x \to 4

yields

3 + 3 = 6

.

Step-by-Step Solution

1

Check for direct substitution.

Substituting

x = 4

yields

\frac{4 - 4}{\sqrt{4 + 5} - 3} = \frac{0}{3 - 3} = \frac{0}{0}

, which is an indeterminate form.

Direct substitution gives

\frac{0}{0}

, requiring algebraic simplification such as rationalization.

2

Rationalize the denominator by multiplying the numerator and denominator by the conjugate

(\sqrt{x + 5} + 3)

.

\lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{(\sqrt{x + 5} - 3)(\sqrt{x + 5} + 3)} = \lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{(x + 5) - 9}

Using the difference of squares formula

(a - b)(a + b) = a^2 - b^2

eliminates the square root in the denominator.

3

Simplify the denominator and cancel out the common factor

(x - 4)

.

\lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{x - 4} = \lim_{x \to 4} (\sqrt{x + 5} + 3)

Since

x \neq 4

when evaluating the limit, the indeterminate factor

(x - 4)

cancels out.

4

Substitute

x = 4

into the simplified expression.

\sqrt{4 + 5} + 3 = \sqrt{9} + 3 = 3 + 3 = 6

Evaluates the limit after removing the zero-denominator condition.

Key Concept

Limits of indeterminate forms

\frac{0}{0}

involving radicals (Rationalization Technique)

Estimated Time:1m 30s

Question 5Question →

Evaluate the algebraic limit:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 16}

What is the exact value of this limit?

\frac{1}{32}

\frac{1}{16}

\frac{1}{8}

0

Show answer & explanation

Answer: $\frac{1}{32}$

Answer

The exact value of the limit is

\frac{1}{32}

.

The limit presents an indeterminate form

\frac{0}{0}

at

x = 4

. Factoring

x^2 - 16

into

(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)

allows the factor

(\sqrt{x} - 2)

to be cancelled from both the numerator and denominator. Substituting

x = 4

into the simplified expression

\frac{1}{(\sqrt{x} + 2)(x + 4)}

yields

\frac{1}{(2+2)(4+4)} = \frac{1}{32}

.

Step-by-Step Solution

1

Identify the form of the limit by direct substitution.

Substituting

x = 4

into

\frac{\sqrt{4} - 2}{4^2 - 16}

gives

\frac{0}{0}

, which is an indeterminate form.

Direct substitution yields

\frac{0}{0}

, requiring algebraic simplification.

2

Factor the denominator

x^2 - 16

.

x^2 - 16 = (x - 4)(x + 4)

Use the difference of two squares identity

a^2 - b^2 = (a - b)(a + b)

.

3

Factor

(x - 4)

as a difference of squares involving square roots.

x - 4 = (\sqrt{x})^2 - 2^2 = (\sqrt{x} - 2)(\sqrt{x} + 2)

This exposes the vanishing factor

(\sqrt{x} - 2)

in the denominator.

4

Cancel the common factor

(\sqrt{x} - 2)

and evaluate the limit.

\lim_{x \to 4} \frac{\sqrt{x} - 2}{(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)} = \lim_{x \to 4} \frac{1}{(\sqrt{x} + 2)(x + 4)} = \frac{1}{(\sqrt{4} + 2)(4 + 4)} = \frac{1}{4 \times 8} = \frac{1}{32}

Cancelling the factor removing the

\frac{0}{0}

condition allows direct evaluation.

Key Concept

Resolution of indeterminate limits of the form 0/0 using algebraic factorization and conjugate radical identities.

Estimated Time:2m 0s

Question 6Question →

Evaluate the trigonometric limit:

\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}

What is the numerical value of this limit?

Show answer & explanation

Answer: -4

Answer

The numerical value of the limit is -4.

Using either the sum-to-product identity

\cos(3x) - \cos(x) = -2 \sin(2x) \sin(x)

along with standard limits

\lim_{x \to 0} \frac{\sin(kx)}{x} = k

, or applying L'Hôpital's rule twice on the

\frac{0}{0}

form, yields the exact value

-4

.

Step-by-Step Solution

1

Check the form of the limit by direct substitution

Substituting

x = 0

gives

\frac{\cos(0) - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}

, an indeterminate form.

Determines whether algebraic transformation or L'Hôpital's rule is required.

2

Transform the numerator using the sum-to-product formula

\cos(3x) - \cos(x) = -2 \sin\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = -2 \sin(2x) \sin(x)

Converts difference of cosines into product of sines to utilize standard trigonometric limits.

3

Rewrite the fractional expression and apply limit laws

\lim_{x \to 0} \frac{-2 \sin(2x) \sin(x)}{x^2} = -2 \cdot \left(\lim_{x \to 0} \frac{\sin(2x)}{x}\right) \cdot \left(\lim_{x \to 0} \frac{\sin(x)}{x}\right)

Splits

x^2

into

x \cdot x

under each sine function.

4

Evaluate the individual standard limits

\lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin(x)}{x} = 1

Applies the known fundamental trigonometric limit rule

\lim_{u \to 0} \frac{\sin(au)}{u} = a

.

5

Calculate the final product

-2 \times 2 \times 1 = -4

Combines all factors to reach the evaluated value.

Key Concept

Trigonometric Limits and Indeterminate Forms

Alternative Method

Alternatively, apply L'Hôpital's rule twice. First derivative of numerator over denominator yields

\lim_{x \to 0} \frac{-3\sin(3x) + \sin(x)}{2x}

(still

\frac{0}{0}

). Differentiating a second time yields

\lim_{x \to 0} \frac{-9\cos(3x) + \cos(x)}{2} = \frac{-9(1) + 1}{2} = \frac{-8}{2} = -4

.

Estimated Time:1m 30s

Question 7Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + kx - 10}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases}

If

f(x)

is continuous at

x = 2

, what is the numerical value of the constant

k

?

Show answer & explanation

Answer: 3

Answer

The numerical value of the constant

k

is 3.

By definition of continuity,

f(x)

is continuous at

x = 2

if

\lim_{x \to 2} f(x) = f(2) = 7

. As

x \to 2

, the denominator

x - 2

approaches

0

. For the quotient to have a finite limit, the numerator

x^2 + kx - 10

must also evaluate to

0

at

x = 2

, yielding

2^2 + 2k - 10 = 0

. Solving this gives

2k = 6

, so

k = 3

. Substituting

k = 3

gives

\lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = 7

, confirming that

k = 3

is correct.

Step-by-Step Solution

1

Apply the definition of continuity at

x = 2

\lim_{x \to 2} f(x) = f(2) = 7

A function

f(x)

is continuous at

x = a

if and only if

\lim_{x \to a} f(x) = f(a)

.

2

Set the numerator to zero at the point of discontinuity

x = 2

2^2 + k(2) - 10 = 0

Because the denominator

(x - 2) \to 0

as

x \to 2

, the limit can only exist if the numerator also approaches

0

, forming an indeterminate form

\frac{0}{0}

that can be simplified.

3

Solve for the unknown parameter

k

4 + 2k - 10 = 0 \implies 2k - 6 = 0 \implies k = 3

Linear algebraic equation solving.

4

Verify that the simplified limit equals

f(2)

\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 5)}{x - 2} = \lim_{x \to 2} (x + 5) = 7

Canceling the common factor

(x - 2)

yields

7

, which matches

f(2) = 7

.

Key Concept

Continuity of a Piecewise Function and Limit Existence

Question 8Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3 \\ k + 2, & x = 3 \end{cases}

If

f(x)

is continuous at

x = 3

, what is the value of the constant

k

?

3

5

7

-2

Show answer & explanation

Answer: $5$

Answer

The constant value is

k = 5

.

For the function to be continuous at

x = 3

, the limit as

x \to 3

must equal the value of the function at

x = 3

, which is

f(3) = k + 2

. Factoring the numerator gives

2x^2 - 5x - 3 = (2x + 1)(x - 3)

. Canceling the common factor

(x - 3)

leaves

\lim_{x \to 3}(2x + 1) = 7

. Setting

k + 2 = 7

yields

k = 5

.

Step-by-Step Solution

1

Evaluate the limit of

f(x)

as

x

approaches

3

\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3} = \lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3} = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7

Direct substitution yields the indeterminate form

\frac{0}{0}

, so the numerator must be factored to cancel the common term

(x - 3)

.

2

Apply the definition of continuity at

x = 3

f(3) = \lim_{x \to 3} f(x) \implies k + 2 = 7

For a function to be continuous at a point

x = c

, the function value

f(c)

must equal the limit of

f(x)

as

x \to c

.

3

Solve for the constant

k

k = 7 - 2 = 5

Subtract

2

from both sides of the equation.

Key Concept

Continuity of a Piecewise Function at a Point

Question 9Question →

A piecewise function

f(x)

is defined by

f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & \text{for } x \neq 3 \ a^2 - 2, & \text{for } x = 3 \end{cases}

If

f(x)

is continuous at

x = 3

and

a > 0

, what is the numerical value of

a

?

Show answer & explanation

Answer: 3

Answer

The numerical value of a is 3.

For

f(x)

to be continuous at

x = 3

, the defined value

f(3) = a^2 - 2

must equal

\lim_{x \to 3} f(x)

. Factoring the numerator gives

\frac{(2x + 1)(x - 3)}{x - 3} = 2x + 1

for

x \neq 3

. Taking the limit as

x \to 3

yields

2(3) + 1 = 7

. Setting

a^2 - 2 = 7

leads to

a^2 = 9

, which gives

a = 3

under the constraint

a > 0

.

Step-by-Step Solution

1

Evaluate the limit of

f(x)

as

x \to 3

Factor the numerator

2x^2 - 5x - 3 = (2x + 1)(x - 3)

. For

x \neq 3

,

f(x) = 2x + 1

. Thus,

\lim_{x \to 3} f(x) = 2(3) + 1 = 7

.

Direct substitution gives an indeterminate form

\frac{0}{0}

, so canceling the common factor

(x - 3)

allows direct evaluation of the limit.

2

Apply the definition of continuity at

x = 3

f(3) = a^2 - 2 = 7

.

For a function to be continuous at a point

c

, the function value

f(c)

must equal the limit

\lim_{x \to c} f(x)

.

3

Solve for the parameter

a

a^2 = 9 \implies a = 3

(since

a > 0

).

Solving

a^2 = 9

gives solutions

3

and

-3

. The condition

a > 0

specifies the positive root.

Key Concept

Continuity of a Piecewise Function at a Point

Question 10Question →

Evaluate the limit $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x - 3}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 1

Answer

The numerical value of the limit is 1.

Substituting x = 3 into the original expression produces the indeterminate form 0/0. Factoring the numerator gives (x - 3)(x - 2). Canceling the common factor (x - 3) yields x - 2. Substituting x = 3 into x - 2 gives 3 - 2 = 1.

Step-by-Step Solution

1

Check the expression using direct substitution at x = 3.

The substitution yields \frac{3^2 - 5(3) + 6}{3 - 3} = \frac{0}{0}, an indeterminate form.

Direct substitution results in 0/0, indicating that algebraic factorization is needed.

2

Factorize the quadratic polynomial in the numerator.

x^2 - 5x + 6 = (x - 3)(x - 2)

Finding factors of +6 that sum to -5 allows cancellation of the denominator term.

3

Cancel the factor (x - 3) and evaluate the remaining linear expression at x = 3.

\lim_{x \to 3} (x - 2) = 3 - 2 = 1

The factor causing zero in the denominator is removed, leaving a continuous function.

Key Concept

Evaluating indeterminate limits (0/0) by algebraic factorization

Question 11Question →

A piecewise function

f(x)

is defined by

f(x) = \begin{cases} \frac{\sin(3x) + \tan(5x)}{2x}, & x \neq 0 \\ a^2 - 5, & x = 0 \end{cases}

If

f(x)

is continuous at

x = 0

, where

a > 0

, determine the numerical value of

a

.

Show answer & explanation

Answer: 3

Answer

The numerical value of

a

is 3.

For the function to be continuous at

x = 0

, the limit as

x \to 0

must equal the function value

f(0)

. Splitting the trigonometric limit gives

\frac{3}{2} + \frac{5}{2} = 4

. Setting

a^2 - 5 = 4

yields

a^2 = 9

. Because

a > 0

, taking the positive square root gives

a = 3

.

Step-by-Step Solution

1

Evaluate the limit of the trigonometric expression as

x

approaches 0.

\lim_{x \to 0} \frac{\sin(3x) + \tan(5x)}{2x} = 4

Using standard trigonometric limit principles: \lim_{x \to 0} \frac{\sin(kx)}{x} = k and \lim_{x \to 0} \frac{\tan(kx)}{x} = k.

2

Equate the limit value to

f(0)

to ensure continuity at

x = 0

.

a^2 - 5 = 4

By definition, a function is continuous at x = c if and only if \lim_{x \to c} f(x) = f(c).

3

Solve the resulting quadratic equation for the positive constant

a

.

a = 3

Adding 5 to both sides gives a^2 = 9; taking the principal square root gives a = 3 since a > 0.

Key Concept

Limits and Continuity of Functions

Question 12Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{1 - \cos(6x)}{x \tan(3x)}, & x \neq 0 \\ p + 4, & x = 0 \end{cases}

If

f(x)

is continuous at

x = 0

, what is the value of the constant

p

?

6

2

-1

-2

Show answer & explanation

Answer: 2

Answer

The value of the constant

p

is

2

.

For the piecewise function to be continuous at

x = 0

, the limit

\lim_{x \to 0} f(x)

must exist and equal

f(0) = p + 4

. By applying the double-angle identity

1 - \cos(6x) = 2\sin^2(3x)

and using the standard limit

\lim_{u \to 0} \frac{\sin u}{u} = 1

, the limit evaluates to

6

. Equating

p + 4 = 6

yields

p = 2

.

Step-by-Step Solution

1

State the continuity condition at

x = 0

\lim_{x \to 0} f(x) = f(0) = p + 4

For a function to be continuous at a point, its limit at that point must equal the defined function value.

2

Apply the trigonometric identity

1 - \cos(6x) = 2\sin^2(3x)

\frac{1 - \cos(6x)}{x \tan(3x)} = \frac{2\sin^2(3x)}{x \cdot \frac{\sin(3x)}{\cos(3x)}} = 2\cos(3x) \cdot \frac{\sin(3x)}{x}

Rewriting

\tan(3x)

as

\frac{\sin(3x)}{\cos(3x)}

allows cancellation of one

\sin(3x)

factor.

3

Evaluate the limit as

x \to 0

\lim_{x \to 0} \left[2\cos(3x) \cdot 3 \cdot \frac{\sin(3x)}{3x}\right] = 2(1)(3)(1) = 6

Using the standard trigonometric limit \lim_{u \to 0} \frac{\sin(u)}{u} = 1 and \cos(0) = 1.

4

Equate the limit to

f(0)

and solve for

p

p + 4 = 6 \implies p = 2

Subtracting 4 from both sides isolates the parameter

p

.

Key Concept

Continuity of a Piecewise Function using Trigonometric Limits

Estimated Time:2m 0s

Question 13Question →

Evaluate the limit:

\lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x + 2} - 2}

What is the numerical value of this limit?

Show answer & explanation

Answer: 48

Answer

The numerical value of the limit is 48.

Evaluating the limit of

\frac{x^3 - 8}{\sqrt{x + 2} - 2}

as

x \to 2

gives an indeterminate form

\frac{0}{0}

. Factorizing the numerator gives

(x - 2)(x^2 + 2x + 4)

, and rationalizing the denominator by multiplying numerator and denominator by

(\sqrt{x + 2} + 2)

converts the denominator to

x - 2

. Canceling

(x - 2)

leaves

(x^2 + 2x + 4)(\sqrt{x + 2} + 2)

. Evaluating at

x = 2

gives

(4 + 4 + 4)(\sqrt{4} + 2) = 12 \times 4 = 48

.

Step-by-Step Solution

1

Identify the limit form via direct substitution

Substituting

x = 2

yields

\frac{0}{0}

.

Direct evaluation results in an indeterminate form, requiring algebraic manipulation to eliminate the zero factor.

2

Factorize the numerator using the difference of cubes formula

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Exposing the factor

(x - 2)

is essential to resolving the zero denominator.

3

Rationalize the denominator using its algebraic conjugate

Multiply top and bottom by

(\sqrt{x + 2} + 2)

to get denominator

(x + 2) - 4 = x - 2

.

Applying

(a-b)(a+b) = a^2 - b^2

eliminates the square root from the denominator.

4

Cancel the common factor and compute the final value

\lim_{x \to 2} (x^2 + 2x + 4)(\sqrt{x + 2} + 2) = (12)(4) = 48.

With

(x - 2)

cancelled for

x \neq 2

, direct substitution now yields a defined real number.

Key Concept

Limits of Indeterminate Forms using Difference of Cubes and Surd Rationalization

Estimated Time:2m 30s

Question 14Question →

Evaluate the limit:

\lim_{x \to \infty} (\sqrt{x^2 + 6x} - x)

What is the numerical value of this limit?

3

6

0

Undefined

Show answer & explanation

Answer: 3

Answer

3

To evaluate the limit of

\sqrt{x^2 + 6x} - x

as

x \to \infty

, multiply and divide by its conjugate

\sqrt{x^2 + 6x} + x

. The numerator simplifies to

(x^2 + 6x) - x^2 = 6x

. Dividing both the numerator and denominator by

x

yields

\frac{6}{\sqrt{1 + 6/x} + 1}

. Taking the limit as

x \to \infty

reduces

\frac{6}{x}

to

0

, resulting in

\frac{6}{\sqrt{1} + 1} = 3

.

Step-by-Step Solution

1

Identify the indeterminate form

Direct evaluation gives

\infty - \infty

, which is an indeterminate form.

Substitution cannot be applied directly when subtracting infinite limits.

2

Multiply and divide by the algebraic conjugate

\lim_{x \to \infty} \frac{(\sqrt{x^2 + 6x} - x)(\sqrt{x^2 + 6x} + x)}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{6x}{\sqrt{x^2 + 6x} + x}

The identity

(a-b)(a+b) = a^2 - b^2

eliminates the square root in the numerator.

3

Factor

x

out of the denominator

\lim_{x \to \infty} \frac{6x}{x \left(\sqrt{1 + \frac{6}{x}} + 1\right)} = \lim_{x \to \infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1}

Dividing the numerator and denominator by

x

allows evaluation at infinity.

4

Compute the limit as

x \to \infty

Since

\lim_{x \to \infty} \frac{6}{x} = 0

, the expression becomes

\frac{6}{\sqrt{1 + 0} + 1} = \frac{6}{2} = 3

.

Terms with

x

in the denominator approach zero as

x

grows arbitrarily large.

Key Concept

Limits at infinity involving radical indeterminate forms of type

\infty - \infty

Estimated Time:2m 0s

Question 15Question →

What is the numerical value of the limit:

\lim_{x \to 0} \frac{1 - \cos(4x)}{x \sin(2x)}

?

4

2

8

0

Show answer & explanation

Answer: 4

Answer

The numerical value of the limit is 4.

Applying the double-angle trigonometric identity

1 - \cos(4x) = 2\sin^2(2x)

reduces the expression to

\frac{2\sin(2x)}{x}

. Rewriting this as

4 \cdot \frac{\sin(2x)}{2x}

and taking the limit as

x \to 0

using the standard limit

\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1

yields 4.

Step-by-Step Solution

1

Identify the form of the limit

Direct substitution of

x = 0

gives

\frac{1 - \cos(0)}{0 \cdot \sin(0)} = \frac{0}{0}

, which is an indeterminate form.

Indeterminate forms require algebraic simplification or trigonometric identities before evaluating the limit.

2

Apply trigonometric identity

Use

1 - \cos(4x) = 2\sin^2(2x)

to rewrite the numerator.

This transforms the numerator into a form containing sine terms matching the denominator.

3

Simplify the algebraic expression

\lim_{x \to 0} \frac{2\sin^2(2x)}{x\sin(2x)} = \lim_{x \to 0} \frac{2\sin(2x)}{x}

Cancel out the common

\sin(2x)

factor from numerator and denominator for

x \neq 0

.

4

Evaluate using the standard trigonometric limit

\lim_{x \to 0} 4 \cdot \frac{\sin(2x)}{2x} = 4 \cdot 1 = 4

Since

\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1

, setting

\theta = 2x

gives

\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1

.

Key Concept

Limits of Trigonometric Functions and Indeterminate Forms

Question 16Question →

What is the numerical value of the limit

\lim_{x \to 0} \frac{\sqrt{4 + x} - \sqrt{4 - x}}{x}

?

\frac{1}{2}

0

2

1

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The numerical value of the limit is $\frac{1}{2}$.

Rationalizing the numerator by multiplying with its conjugate gives $\frac{2x}{x(\sqrt{4+x} + \sqrt{4-x})}$. Canceling $x$ yields $\frac{2}{\sqrt{4+x} + \sqrt{4-x}}$, which evaluates to $\frac{2}{4} = \frac{1}{2}$ as $x \to 0$.

Step-by-Step Solution

1

Identify the indeterminate form

Substituting $x = 0$ directly yields $\frac{\sqrt{4} - \sqrt{4}}{0} = \frac{0}{0}$, which is an indeterminate form requiring rationalization.

Direct evaluation cannot give the true limit value when an indeterminate form is encountered.

2

Rationalize the numerator

Multiply the numerator and denominator by the conjugate $\sqrt{4 + x} + \sqrt{4 - x}$:

\frac{(\sqrt{4 + x} - \sqrt{4 - x})(\sqrt{4 + x} + \sqrt{4 - x})}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{(4 + x) - (4 - x)}{x(\sqrt{4 + x} + \sqrt{4 - x})}

The difference of squares identity $(a - b)(a + b) = a^2 - b^2$ eliminates the radical signs in the numerator.

3

Simplify the numerator and cancel common terms

$\frac{4 + x - 4 + x}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{2x}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{2}{\sqrt{4 + x} + \sqrt{4 - x}}$

Canceling the common factor $x$ eliminates the zero-causing term in the denominator.

4

Evaluate the simplified limit as $x \to 0$

$\frac{2}{\sqrt{4 + 0} + \sqrt{4 - 0}} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$

Direct substitution is now valid since the expression is continuous at $x = 0$.

Key Concept

Evaluation of algebraic limits of indeterminate form 0/0 using radical rationalization.

Question 17Question →

Evaluate the algebraic limit:

\lim_{x \to 2} \frac{x^3 - 8}{x^2 + x - 6}

What is the value of this limit?

\frac{4}{5}

\frac{8}{5}

\frac{12}{5}

0

Show answer & explanation

Answer: $\frac{12}{5}$

Answer

The correct value of the limit is

\frac{12}{5}

.

Evaluating the limit by direct substitution gives the indeterminate form

\frac{0}{0}

. Factoring the numerator

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

and denominator

x^2 + x - 6 = (x - 2)(x + 3)

allows cancellation of

(x - 2)

. Evaluating

\frac{x^2 + 2x + 4}{x + 3}

at

x = 2

yields

\frac{12}{5}

.

Step-by-Step Solution

1

Check for direct substitution

Substituting

x = 2

gives

\frac{2^3 - 8}{2^2 + 2 - 6} = \frac{0}{0}

, which is an indeterminate form.

Direct substitution results in

\frac{0}{0}

, requiring algebraic factorization.

2

Factor the numerator and the denominator

Numerator:

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Denominator:

x^2 + x - 6 = (x - 2)(x + 3)

Use the difference of cubes formula

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

and quadratic factorization.

3

Cancel the common factor and compute the limit

\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 3)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 3} = \frac{2^2 + 2(2) + 4}{2 + 3} = \frac{12}{5}

Eliminating the factor

(x - 2)

removes the removable discontinuity at

x = 2

.

Key Concept

Limits of Indeterminate Forms (0/0) using Factorization

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