📝 L'Hopital rule for 0/0 indeterminate form (18 MCQs)

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

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What is L'Hopital rule for 0/0 indeterminate form?

Definition:
L'Hôpital's Rule states that for limits resulting in the indeterminate form $\frac{0}{0}$ , the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives, provided the latter exists. Mathematically, if $\lim_{x\to c} f(x) = 0$ and $\lim_{x\to c} g(x) = 0$ , then $\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

Example:
Evaluate $\lim_{x\to 0} \frac{\sin(x)}{x}$ . Direct substitution gives $\frac{0}{0}$ . Apply L'Hôpital's Rule: $\lim_{x\to 0} \frac{\cos(x)}{1} = \cos(0) = 1$ . This confirms the standard limit result using differentiation instead of geometric proofs.

Reason:
This rule provides a systematic algebraic method to resolve indeterminate forms that cannot be solved by simple factorization. It leverages the local linear behavior of functions near a point, turning difficult limit problems into simpler derivative evaluations.

Easy

Medium

Hard

📝 All L'Hopital rule for 0/0 indeterminate form MCQs

Q1. Which of the following limits is an indeterminate form of type $0/0$ as $x\to a$ ?

\displaystyle \lim_{x\to 2}\frac{x^{2}-4}{x+2}

\displaystyle \lim_{x\to 0}\frac{x^{2}+1}{x}

\displaystyle \lim_{x\to 1}\frac{x-1}{x^{2}-1}

✅

\displaystyle \lim_{x\to 0}\frac{x^{2}}{x+1}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Only choice C yields both numerator and denominator approaching zero when $x\to1$ . In A the denominator approaches 4, in B the numerator approaches 1, and in D the denominator approaches 1, so they are not $0/0$ . Recognizing the indeterminate form is the first step before applying L'Hôpital.

Q2. Given $\displaystyle \lim_{x\to0}\frac{\sin x}{x}$ is $0/0$ , after applying L'Hôpital once we obtain $\displaystyle \lim_{x\to0}\frac{\cos x}{1}$ . What can we infer about the original limit?

\displaystyle 0

✅

\displaystyle \text{The limit does not exist}

\displaystyle 1

\displaystyle -\infty

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Applying L'Hôpital replaces the original quotient with the limit of the derivatives. Since $\lim_{x\to0}\cos x = 1$ , the original limit equals this value. The result is finite and unique, so the original limit is indeed 1.

Q3. If $\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$ and $\displaystyle \lim_{x\to a}\frac{f'(x)}{g'(x)}=L$ exists, which statement must be true?

\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=L

✅

\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}

does not exist

C.The limit equals

0

D.The limit equals

\infty

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The hypothesis matches the hypothesis of L'Hôpital's rule. When the derivative ratio approaches a finite limit $L$ , the original quotient approaches the same limit. This is the core conclusion of the theorem, guaranteeing equality of the two limits under the given conditions.

Q4. Suppose $\displaystyle \lim_{x\to a}\frac{f'(x)}{g'(x)}=L$ with $L$ finite. Using the definition of derivative, which theorem justifies that $\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=L$ ?

A.Mean Value Theorem ✅

B.Intermediate Value Theorem

C.Bolzano‑Weierstrass Theorem

D.Fundamental Theorem of Calculus

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The Mean Value Theorem guarantees a point between $x$ and $a$ where the ratio of increments equals the ratio of derivatives. Taking the limit as $x\to a$ transfers this equality to the function quotients, establishing the limit of $f/g$ equals $L$ . This logical step underlies L'Hôpital's proof.

Q5. For the limit $\displaystyle \lim_{x\to0}\frac{e^{x^{2}}-1}{x\sin x}$ , after one application of L'Hôpital the resulting limit is still $0/0$ . How many additional applications are necessary to obtain a determinate limit?

A.One ✅

B.Two

C.Three

D.None

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The first differentiation gives $\frac{2xe^{x^{2}}}{\sin x+x\cos x}$ , still $0/0$ . Differentiating a second time yields $\frac{2e^{x^{2}}+4x^{2}e^{x^{2}}}{2\cos x-x\sin x}$ , which at $x=0$ evaluates to $\frac{2}{2}=1$ . Hence only one more application (total of two) is required.

Q6. Which method gives the same result for $\displaystyle \lim_{x\to2}\frac{x^{2}-4}{x-2}$ : factoring or L'Hôpital?

\displaystyle\text{Factoring gives 4, L&#039;Hôpital gives 2}

\displaystyle\text{Both give 4}

✅

\displaystyle\text{Factoring gives 0, L&#039;Hôpital gives 4}

\displaystyle\text{Both give undefined}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Factoring the numerator as $(x-2)(x+2)$ cancels the denominator, leaving $\lim_{x\to2}(x+2)=4$ . Applying L'Hôpital differentiates numerator and denominator to obtain $\lim_{x\to2}\frac{2x}{1}=4$ . Both approaches therefore produce the same value, confirming the consistency of the rule.

Q7. Evaluate $\displaystyle \lim_{x\to0}\frac{\sin 2x}{x}$ using L'Hôpital. Which of the following is the correct value?

\displaystyle 0

\displaystyle 1

\displaystyle 2

✅

\displaystyle \text{Undefined}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Differentiating numerator and denominator gives $\lim_{x\to0}\frac{2\cos 2x}{1}=2\cos 0=2$ . The original limit therefore equals 2, matching the known small‑angle approximation $\sin 2x\approx2x$ .

Q8. For $\displaystyle \lim_{x\to0}\frac{e^{x}-1}{x^{3}}$ , how many times must L'Hôpital's rule be applied to obtain a non‑zero finite limit?

\displaystyle\text{Once}

\displaystyle\text{Twice}

\displaystyle\text{Three times}

✅

\displaystyle\text{It never becomes finite}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Each differentiation reduces the power of $x$ in the denominator by one. After the first derivative the limit is still $0/0$ ; after the second it remains $0/0$ ; after the third derivative we obtain $\frac{e^{x}}{6}$ , which at $x=0$ equals $\frac{1}{6}$ . Thus three applications are needed.

Q9. Consider $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^{2}}$ . Using a second‑order Taylor expansion of $\cos x$ gives the same limit as L'Hôpital. What is that limit?

\displaystyle 0

\displaystyle \frac12

✅

\displaystyle 1

\displaystyle 2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The Taylor series $\cos x=1-\frac{x^{2}}{2}+O(x^{4})$ yields $1-\cos x\approx\frac{x^{2}}{2}$ . Dividing by $x^{2}$ gives $\frac12$ . Applying L'Hôpital twice leads to the same result: first derivative gives $\frac{\sin x}{2x}$ , second gives $\frac{\cos x}{2}$ , and $\lim_{x\to0}\frac{\cos x}{2}= \frac12$ .

Q10. Find $\displaystyle \lim_{x\to0}\frac{\tan x - x}{x^{3}}$ . Which technique yields the correct limit and what is the value?

\displaystyle\text{L&#039;Hôpital twice, limit}= \frac13

\displaystyle\text{Series expansion, limit}= \frac13

\displaystyle\text{L&#039;Hôpital three times, limit}= \frac13

✅

\displaystyle\text{Direct substitution, limit}=0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Expanding $\tan x = x + \frac{x^{3}}{3}+O(x^{5})$ shows the numerator behaves like $\frac{x^{3}}{3}$ ; dividing by $x^{3}$ gives $\frac13$ . Repeatedly applying L'Hôpital three times differentiates away the lower‑order terms and also yields $\frac13$ . Both methods agree on the limit.

Q11. Apply L'Hôpital to $\displaystyle \lim_{x\to0}\frac{\ln(1+x)}{x}$ . What is the limit?

\displaystyle 0

\displaystyle 1

✅

\displaystyle \infty

\displaystyle \text{Does not exist}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Differentiating numerator and denominator gives $\lim_{x\to0}\frac{1/(1+x)}{1}=1$ . Since the derivative of $\ln(1+x)$ at 0 is 1, the original quotient approaches 1 as $x$ approaches 0.

Q12. For $\displaystyle \lim_{x\to a}\frac{x^{2}-a^{2}}{x-a}$ , using L'Hôpital gives which expression?

\displaystyle 2a

\displaystyle a

\displaystyle x+a

\displaystyle 2x

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Differentiating numerator yields $2x$ and denominator yields $1$ , so the ratio of derivatives is $2x$ . Taking the limit as $x\to a$ then gives $2a$ . The intermediate expression after applying L'Hôpital is therefore $2x$ .

Q13. The limit $\displaystyle \lim_{x\to0}\frac{\arctan x}{x}$ can be evaluated with L'Hôpital. What does the result tell about $\arctan'(0)$ ?

\displaystyle \arctan&#039;(0)=0

\displaystyle \arctan&#039;(0)=1

✅

\displaystyle \arctan&#039;(0)=\infty

\displaystyle \text{No derivative exists}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Applying L'Hôpital gives $\lim_{x\to0}\frac{1/(1+x^{2})}{1}=1$ . This limit equals the derivative of $\arctan x$ at 0, confirming that $\arctan'(0)=1$ .

Q14. Determine the limit $\displaystyle \lim_{x\to0^{+}}\frac{x-4}{3\sin(1/x)}$ . Does L'Hôpital apply, and what is the limit?

\displaystyle\text{L&#039;Hôpital applies, limit}= -\infty

\displaystyle\text{L&#039;Hôpital does not apply, limit does not exist}

✅

\displaystyle\text{L&#039;Hôpital applies, limit}=0

\displaystyle\text{L&#039;Hôpital applies, limit}= -4/3

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: As $x\to0^{+}$ , the numerator approaches $-4$ while the denominator oscillates between $-3$ and $3$ . The expression therefore has no single limiting value. Moreover, the original form is not $0/0$ (the numerator is not approaching zero), so L'Hôpital's rule cannot be used.

Q15. If $\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}$ is $0/0$ and both $\displaystyle f'(x)$ and $\displaystyle g'(x)$ also tend to 0, what is a valid next step?

\displaystyle\text{Apply L&#039;Hôpital again}

✅

\displaystyle\text{Conclude the limit is 0}

\displaystyle\text{Use substitution}

\displaystyle\text{The limit does not exist}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When the first derivative ratio still yields the indeterminate form $0/0$ , the hypothesis of L'Hôpital's rule remains satisfied. Therefore the rule may be applied repeatedly until a determinate limit is obtained or the hypothesis fails. This iterative process is legitimate under the same differentiability conditions.

Q16. Which statement about L'Hôpital's rule for $0/0$ is always true?

\displaystyle\text{It can be applied to any quotient}

\displaystyle\text{It requires differentiability of numerator and denominator near the point}

✅

\displaystyle\text{The rule works for }\infty/\infty\text{ only}

\displaystyle\text{It gives the same result as factoring}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The rule demands that both functions be differentiable on an open interval containing the point (except possibly at the point itself) and that the limit of the derivative quotient exist. Continuity alone is insufficient; the differentiability condition is essential for the theorem's validity.

Q17. What is a necessary condition for using L'Hôpital's rule on a limit of type $0/0$ ?

\displaystyle\text{Both functions are continuous at the point}

\displaystyle\text{Both functions are differentiable near the point and the limit of their derivative ratio exists}

✅

\displaystyle\text{Only the numerator must be differentiable}

\displaystyle\text{The denominator must be non‑zero at the point}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The rule cannot be invoked unless the numerator and denominator are differentiable in a neighborhood of the point (excluding the point itself) and the limit of the quotient of their derivatives exists (finite or infinite). Continuity alone does not guarantee applicability.

Q18. Who first published the rule that now bears his name?

\displaystyle\text{Isaac Newton}

\displaystyle\text{Gottfried Leibniz}

\displaystyle\text{Guillaume de L&#039;Hôpital}

✅

\displaystyle\text{Augustin-Louis Cauchy}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Guillaume de L'Hôpital, a French mathematician, published the rule in his 1696 textbook, making it the first systematic exposition of the technique. Although the underlying ideas were developed by Johann Bernoulli, the rule is historically attributed to L'Hôpital.

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