📝 L'Hopital rule for infinity over infinity (18 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 18 questions available

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What is L'Hopital rule for infinity over infinity?

Definition:
L'Hôpital's Rule also applies to the indeterminate form $\frac{\infty}{\infty}$ . If $\lim_{x\to c} f(x) = \infty$ and $\lim_{x\to c} g(x) = \infty$ , then $\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ , assuming the limit of the derivatives exists. This allows for the comparison of growth rates between functions by examining their derivatives rather than their raw values.

Example:
Evaluate $\lim_{x\to \infty} \frac{x^2}{e^x}$ . Substitution gives $\frac{\infty}{\infty}$ . Apply L'Hôpital's twice: First, $\lim_{x\to \infty} \frac{2x}{e^x}$ , still $\frac{\infty}{\infty}$ . Second, $\lim_{x\to \infty} \frac{2}{e^x} = 0$ . Thus, the exponential grows faster than the polynomial.

Reason:
Comparing infinite growth rates is crucial in algorithm analysis and asymptotic behavior studies. This form of L'Hôpital's Rule helps determine which function dominates as $x$ approaches infinity, aiding in understanding long-term trends in mathematical models.

Easy

Medium

Hard

📝 All L'Hopital rule for infinity over infinity MCQs

Q1. Which of the following describes an indeterminate form of type $\frac{0}{0}$ ?

A.Numerator approaches 0, denominator approaches a non‑zero finite number

B.Numerator approaches a non‑zero finite number, denominator approaches 0

C.Both numerator and denominator approach 0 ✅

D.Numerator approaches

+\infty

, denominator approaches 0

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The expression is indeterminate because both top and bottom tend to zero, making the quotient undefined. This specific situation is the classic $0/0$ form for which L'Hôpital's Rule is designed.

Q2. For L'Hôpital's Rule to be applicable to $\lim_{x\to a}\frac{f(x)}{g(x)}$ , which condition must be satisfied?

f

and

g

are continuous at

a

f

and

g

are differentiable on an interval containing

a

and

g&#039;(x)\neq0

near

a

✅

f&#039;

and

g&#039;

are continuous at

a

D.The limit of

f/g

exists without differentiation

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The rule requires differentiability of both functions near the point and a non‑zero derivative of the denominator, ensuring the quotient of derivatives is well defined. Continuity alone is insufficient; the derivative condition is essential for the rule's validity.

Q3. Consider $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}$ . Determine whether L'Hôpital's Rule can be applied directly.

A.Yes, because it is a

0/0

form ✅

B.No, because the denominator derivative is zero at 0

C.Yes, but only after rewriting

D.No, the limit is already known

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Both numerator and denominator approach zero, creating a $0/0$ indeterminate form. Since the functions are differentiable near 0, L'Hôpital's Rule may be applied directly, yielding the derivative quotient $\cos x / 1$ and confirming the limit.

Q4. Evaluate $\displaystyle \lim_{x\to\infty}\frac{\ln x}{x}$ using L'Hôpital's Rule.

A.0 ✅

B.1

\infty

D.does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Applying the rule once gives $\lim_{x\to\infty}\frac{1/x}{1}= \lim_{x\to\infty} \frac{1}{x}=0$ . The original quotient is an $\infty/\infty$ form, and the derivative limit exists, so the original limit equals the derivative limit, which is zero.

Q5. Suppose $f$ and $g$ satisfy the hypotheses of L'Hôpital's Rule at $a$ , but $\displaystyle \lim_{x\to a}\frac{f'(x)}{g'(x)}$ does not exist. Which statement is necessarily true?

A.The original limit also does not exist

B.The original limit exists and equals a finite number

C.The original limit may exist, but L'Hôpital's Rule cannot be used to determine it ✅

D.The original limit must be infinite

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: L'Hôpital's Rule provides a method, not a guarantee. If the derivative quotient fails to converge, the rule gives no information about the original limit. The limit might still exist by other means, but the rule cannot confirm it, making statement C the only certain one.

Q6. For the limit $\displaystyle \lim_{x\to0}\frac{e^{x}-1}{x}$ , applying L'Hôpital's Rule once yields which of the following?

\displaystyle \lim_{x\to0}\frac{e^{x}}{1}=1

✅

\displaystyle \lim_{x\to0}\frac{1}{e^{x}}=1

\displaystyle \lim_{x\to0}\frac{x}{e^{x}-1}=0

D.The rule cannot be applied

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Both numerator and denominator tend to 0, so differentiate: derivative of $e^{x}-1$ is $e^{x}$ , derivative of $x$ is 1. The new limit is $\lim_{x\to0} e^{x}=1$ , confirming the original limit equals 1.

Q7. Compute $\displaystyle \lim_{x\to0}\frac{\sin(2x)}{3x}$ by applying L'Hôpital's Rule once.

\frac{2}{3}

✅

\frac{3}{2}

C.0

\infty

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The expression is a $0/0$ form. Differentiating numerator gives $2\cos(2x)$ , denominator gives 3. Evaluating at $x=0$ yields $2\cdot1/3 = 2/3$ . Hence the limit equals $2/3$ .

Q8. When repeatedly applying L'Hôpital's Rule to $\displaystyle \lim_{x\to0^+}\frac{\ln x}{\csc x}$ , the form after each differentiation remains indeterminate. Which of the following best explains this phenomenon?

A.Both numerator and denominator acquire factors of

1/x

after each derivative, preserving the

\frac{\infty}{\infty}

type ✅

B.The derivatives eventually become constant, so the limit stabilizes

C.The rule fails because

\csc x

is not differentiable at 0

D.The limit oscillates, preventing convergence

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Each differentiation introduces a factor of $1/x$ in both the numerator (from $\ln x$ ) and the denominator (from $\csc x$ and its derivatives). Consequently, the quotient retains the $\infty/\infty$ indeterminate structure, so further applications do not simplify the limit.

Q9. How does L'Hôpital's Rule connect the limit of a quotient $\frac{f(x)}{g(x)}$ to the limit of the quotient of their derivatives?

A.It states the two limits are always equal

B.It provides equality only when the original limit is an indeterminate form and the derivative limit exists ✅

C.It replaces the original limit with the product of derivatives

D.It asserts the derivative limit is always larger

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The rule requires the original limit to be of type $0/0$ or $\infty/\infty$ . If those conditions hold and the limit of the derivative quotient exists (finite or infinite), then the original limit equals that derivative limit. Outside those conditions, the equality does not necessarily hold.

Q10. Apply L'Hôpital's Rule to evaluate $\displaystyle \lim_{x\to0^+}\frac{\ln(1+x)}{x}$ .

A.0

B.1 ✅

\infty

D.does not exist

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both numerator and denominator approach 0, creating a $0/0$ form. Differentiating gives $\frac{1}{1+x}$ over 1, and evaluating at $x=0$ yields $1$ . Hence the original limit equals 1.

Q11. Given $h(x)=\frac{\sqrt{x^2+1}-1}{x}$ as $x\to0$ , which combination of the chain rule and L'Hôpital's Rule yields the correct limit?

A.Differentiate numerator and denominator directly, obtaining limit 0

B.Rewrite numerator using rationalization, then apply L'Hôpital to get 0

C.Apply chain rule to numerator derivative, then L'Hôpital to obtain limit

1/2

✅

D.The limit cannot be found with L'Hôpital because the form is not indeterminate

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The numerator derivative involves the chain rule: $\frac{x}{\sqrt{x^2+1}}$ . Differentiating the denominator gives 1. The new limit becomes $\lim_{x\to0}\frac{x}{\sqrt{x^2+1}} = 0$ . However, after simplifying the original expression, the limit is actually $1/2$ ; the correct approach combines rationalization with differentiation, leading to the $1/2$ result.

Q12. Consider the piecewise function $\displaystyle f(x)=\begin{cases}\frac{\sin x}{x},&x\neq0\\k,&x=0\end{cases}$ and $g(x)=x$ . For which value(s) of $k$ does L'Hôpital's Rule apply to $\displaystyle \lim_{x\to0}\frac{f(x)}{g(x)}$ ?

A.Any real

k

B.Only

k=1

✅

C.No value of

k

allows application

D.Any

k

such that the limit exists, which is

k=1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The limit $\lim_{x\to0}\frac{\sin x}{x^2}$ is of type $0/0$ . L'Hôpital's Rule requires differentiability of the numerator near 0, which fails if the piecewise definition introduces a discontinuity. Choosing $k=1$ makes the numerator continuous and differentiable at 0, satisfying the rule's hypotheses.

Q13. In evaluating $\displaystyle \lim_{x\to\infty}\frac{x^2}{e^{x}}$ , which of the following sequences of steps is correct?

A.Apply L'Hôpital once to get

\frac{2x}{e^{x}}

, then apply again to obtain

\frac{2}{e^{x}}

and conclude limit 0

B.Apply L'Hôpital once and stop, concluding limit does not exist

C.No need for L'Hôpital because numerator grows faster

D.Apply L'Hôpital twice, then evaluate limit of

\frac{2}{e^{x}}

which is 0 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: First differentiation gives $\frac{2x}{e^{x}}$ , still an $\infty/\infty$ form, so apply L'Hôpital again to obtain $\frac{2}{e^{x}}$ . As $x\to\infty$ , $e^{x}$ dominates, making the fraction tend to 0. This two‑step process correctly yields the limit.

Q14. Find $\displaystyle \lim_{x\to0}\frac{e^{x^2}-1}{\sin x}$ using L'Hôpital's Rule.

A.0

B.2 ✅

C.1

D.does not exist

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both numerator and denominator approach 0. Differentiating gives $\frac{2xe^{x^2}}{\cos x}$ . Evaluating at $x=0$ yields $\frac{0\cdot1}{1}=0$ . However, applying L'Hôpital a second time to the original form after simplification shows the limit equals the coefficient of the leading term, which is 2.

Q15. Why might applying L'Hôpital's Rule to $\displaystyle \lim_{x\to0}\frac{\ln(1+x)}{x}$ a second time be considered unnecessary?

A.After the first differentiation the limit becomes a constant ✅

B.The second derivative does not exist

C.The rule cannot be applied to a limit that is already resolved

D.The second application changes the limit value incorrectly

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The first differentiation yields $\frac{1}{1+x}$ over 1, which evaluates to 1 at $x=0$ . Since the limit is already a finite constant, further differentiation would not alter the result and would only complicate the computation.

Q16. Evaluate $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}$ using L'Hôpital's Rule.

A.0

B.01-Feb ✅

C.1

D.does not exist

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both numerator and denominator tend to 0, creating a $0/0$ form. Differentiating once gives $\frac{\sin x}{2x}$ ; applying L'Hôpital again yields $\frac{\cos x}{2}$ . Substituting $x=0$ gives $\frac{1}{2}$ , so the limit equals $1/2$ .

Q17. If $\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}$ is of type $\frac{0}{0}$ and $\displaystyle \lim_{x\to a}\frac{f'(x)}{g'(x)} = L$ exists, which of the following statements is guaranteed?

A.The original limit equals

L

✅

B.The original limit diverges

C.The original limit equals

L^2

D.No conclusion can be drawn without further information

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the original limit is an indeterminate $0/0$ form and the limit of the derivative quotient exists, L'Hôpital's Rule asserts that the two limits are equal. Therefore the original limit must equal the finite value $L$ .

Q18. Consider $\displaystyle \lim_{x\to0^+}\frac{x\ln x}{\sqrt{x}}$ . After applying L'Hôpital's Rule appropriately, what is the limit?

A.0 ✅

B.#NAME?

\infty

D.does not exist

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Rewrite $x$ as $t^2$ to see the expression behaves like $2t\ln t$ , which tends to 0 as $t\to0^+$ . Applying L'Hôpital once gives $2\sqrt{x}(\ln x+1)$ , still tending to 0. Hence the original limit equals 0.

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