Mathematics

Limits and Continuity

49 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

What is the limit of a sequence?

  1. The value that the sequence approaches as the index approaches infinity

  2. The value that the sequence takes at infinity

  3. The value that the sequence takes at the origin

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The limit of a sequence is the value that the sequence approaches as the index approaches infinity.

Multiple choice

What is a limit?

  1. The value that a function approaches as the input approaches a certain value.

  2. The value that a function approaches as the output approaches a certain value.

  3. The value that a function approaches as the domain approaches a certain value.

  4. The value that a function approaches as the range approaches a certain value.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A limit is the value that a function approaches as the input approaches a certain value.

Multiple choice

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

  1. 0

  2. 2

  3. 4

  4. 6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To find the limit, we can factor the numerator and cancel the common factor of $(x - 2)$ in the numerator and denominator. This gives $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x + 2)(x - 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 2 + 2 = 4$.

Multiple choice

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

  1. 0

  2. 2

  3. 4

  4. 6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To find the limit, we can factor the numerator and cancel out the common factor (x - 2): (\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x + 2)(x - 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 2 + 2 = 4).

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).