Mathematics

Limits and Continuity

45 Questions

Limits and continuity explore the foundational behavior of mathematical functions as they approach specific values or infinity. The syllabus covers evaluating sequence limits, trigonometric boundaries, and logarithmic expressions. These rigorous quantitative aptitude questions are standard in collegiate and engineering assessments.

Sequence limit evaluationInfinity limit functionsTrigonometric limit valuesLogarithmic limitsContinuity theorems

Limits and Continuity Questions

Multiple choice

Evaluate the limit of (\sqrt{x^2 + 1} - x) as (x) approaches (\infty).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As (x) becomes very large, the term (x^2) dominates the expression, and the other terms become insignificant. Therefore, the limit is (\lim_{x \to \infty} \sqrt{x^2 + 1} - x = \lim_{x \to \infty} x\left(\sqrt{1 + \frac{1}{x^2}} - 1\right) = \infty).

Multiple choice

Find the limit of (\frac{\sin(3x)}{x}) as (x) approaches (0).

  1. 0

  2. 1

  3. 3

  4. Does not exist
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{3\cos(3x)}{1} = 3).

Multiple choice

Determine the limit of (\frac{e^{2x} - 1}{x}) as (x) approaches (0).

  1. 0

  2. 1

  3. 2

  4. \infty
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can rewrite the expression as (\frac{e^{2x} - 1}{x} = \frac{e^{2x} - e^0}{x - 0} = \frac{e^{2x} - e^0}{x} \cdot \frac{e^0 + e^0}{e^0 + e^0} = \frac{e^{2x} + 1}{x(e^0 + e^0)}). As (x) approaches (0), the numerator approaches (e^0 + 1 = 2), and the denominator approaches (2e^0 = 2). Therefore, the limit is (2).

Multiple choice

Evaluate the limit of (\frac{\ln(x + 1)}{x}) as (x) approaches (\infty).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using L'Hopital's rule, we find (\lim_{x \to \infty} \frac{\ln(x + 1)}{x} = \lim_{x \to \infty} \frac{\frac{1}{x + 1}}{1} = \lim_{x \to \infty} \frac{1}{x + 1} = 0).

Multiple choice

Find the limit of (\frac{x^3 - 8}{x - 2}) as (x) approaches (2).

  1. 6

  2. 8

  3. 10

  4. 12
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can factor the numerator as (x^3 - 8 = (x - 2)(x^2 + 2x + 4)). Substituting (x = 2) into the expression, we get (\frac{(2)^3 - 8}{2 - 2} = \frac{8 - 8}{0} = \frac{0}{0}). Since this is an indeterminate form, we can use L'Hopital's rule to find the limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 2} \frac{x^3 - 8}{x - 2} = \lim_{x \to 2} \frac{3x^2}{1} = 12).

Multiple choice

Determine the limit of (\frac{\tan(2x)}{\sin(3x)}) as (x) approaches (0).

  1. 0

  2. 1

  3. 2

  4. Does not exist
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\tan(2x)}{\sin(3x)} = \lim_{x \to 0} \frac{2\sec^2(2x)}{3\cos(3x)} = \frac{2}{3}).

Multiple choice

Find the limit of (\frac{\sqrt{x^2 + 9} - 3}{x}) as (x) approaches (\infty).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can rewrite the expression as (\frac{\sqrt{x^2 + 9} - 3}{x} = \frac{\sqrt{x^2 + 9} - 3}{x} \cdot \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} = \frac{(x^2 + 9) - 9}{x(\sqrt{x^2 + 9} + 3)} = \frac{x^2}{x(\sqrt{x^2 + 9} + 3)}). As (x) approaches (\infty), the term (x^2) dominates the expression, and the other terms become insignificant. Therefore, the limit is (\lim_{x \to \infty} \frac{\sqrt{x^2 + 9} - 3}{x} = \lim_{x \to \infty} \frac{x^2}{x(\sqrt{x^2 + 9} + 3)} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 9} + 3} = 0).

Multiple choice

Evaluate the limit of (\frac{x^2 - 4x + 3}{x - 3}) as (x) approaches (3).

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can factor the numerator as (x^2 - 4x + 3 = (x - 3)(x - 1)). Substituting (x = 3) into the expression, we get (\frac{(3)^2 - 4(3) + 3}{3 - 3} = \frac{9 - 12 + 3}{0} = \frac{0}{0}). Since this is an indeterminate form, we can use L'Hopital's rule to find the limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 3} \frac{x^2 - 4x + 3}{x - 3} = \lim_{x \to 3} \frac{2x - 4}{1} = 2).

Multiple choice

Find the limit of (\frac{\sin^2(x)}{x}) as (x) approaches (0).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\sin^2(x)}{x} = \lim_{x \to 0} \frac{2\sin(x)\cos(x)}{1} = \lim_{x \to 0} 2\sin(x)\cos(x) = 0).

Multiple choice

Determine the limit of (\frac{\log(x + 1)}{x}) as (x) approaches (\infty).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using L'Hopital's rule, we find (\lim_{x \to \infty} \frac{\log(x + 1)}{x} = \lim_{x \to \infty} \frac{\frac{1}{x + 1}}{1} = \lim_{x \to \infty} \frac{1}{x + 1} = 0).

Multiple choice

Find the limit of (\frac{x^3 + 2x^2 - 3x + 4}{x^2 - 1}) as (x) approaches (2).

  1. 5

  2. 7

  3. 9

  4. 11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can factor the numerator as (x^3 + 2x^2 - 3x + 4 = (x - 1)(x^2 + 3x - 4)). Substituting (x = 2) into the expression, we get (\frac{(2)^3 + 2(2)^2 - 3(2) + 4}{(2)^2 - 1} = \frac{8 + 8 - 6 + 4}{4 - 1} = \frac{14}{3} = 11).

Multiple choice

Evaluate the limit of (\frac{e^{2x} - e^{-2x}}{e^x}) as (x) approaches (\infty).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can rewrite the expression as (\frac{e^{2x} - e^{-2x}}{e^x} = \frac{e^{2x}}{e^x} - \frac{e^{-2x}}{e^x} = e^x - e^{-x}). As (x) approaches (\infty), the term (e^x) dominates the expression, and the other term becomes insignificant. Therefore, the limit is (\lim_{x \to \infty} \frac{e^{2x} - e^{-2x}}{e^x} = \lim_{x \to \infty} (e^x - e^{-x}) = \infty).

Multiple choice

Find the limit of (\frac{\sqrt{x^2 + 4x + 4} - x}{x + 2}) as (x) approaches (-2).

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can simplify the expression by rationalizing the numerator: (\frac{\sqrt{x^2 + 4x + 4} - x}{x + 2} = \frac{\sqrt{(x + 2)^2} - x}{x + 2} = \frac{x + 2 - x}{x + 2} = \frac{2}{x + 2}). Substituting (x = -2) into the expression, we get (\frac{2}{-2 + 2} = \frac{2}{0}), which is an indeterminate form. Therefore, we can use L'Hopital's rule to find the limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to -2} \frac{\sqrt{x^2 + 4x + 4} - x}{x + 2} = \lim_{x \to -2} \frac{\frac{1}{2\sqrt{x^2 + 4x + 4}}(2x + 4)}{1} = \lim_{x \to -2} \frac{2x + 4}{2\sqrt{x^2 + 4x + 4}} = 1).

Multiple choice

Determine the limit of (\frac{\sin(x) - x}{x^3}) as (x) approaches (0).

  1. 0

  2. 1

  3. \infty

  4. Does not exist
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\sin(x) - x}{x^3} = \lim_{x \to 0} \frac{\cos(x) - 1}{3x^2} = \lim_{x \to 0} \frac{-\sin(x)}{6x} = 0).

Multiple choice

What is the limit of the function (f(x) = \frac{x^2 - 4}{x - 2}) as (x) approaches (2)?

  1. 0

  2. 1

  3. 2

  4. 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can use L'Hopital's rule to evaluate the limit. Taking the derivative of the numerator and denominator, we get (f'(x) = \frac{2x}{1} = 2x) and (g'(x) = \frac{1}{1} = 1). Substituting (x = 2), we get (f'(2) = 4) and (g'(2) = 1). Therefore, the limit of (f(x)) as (x) approaches (2) is (\frac{4}{1} = 4).