📝 0 times infinity indeterminate form (15 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 15 questions available

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What is 0 times infinity indeterminate form?

Definition:
The product $0 \cdot \infty$ is an indeterminate form because the outcome depends on how fast each factor approaches its limit. To apply L'Hôpital's Rule, one must rewrite the product as a quotient, either $\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$ , transforming it into $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . This algebraic manipulation enables the use of derivative-based limit evaluation techniques.

Example:
Evaluate $\lim_{x\to 0^+} x \ln(x)$ . This is $0 \cdot (-\infty)$ . Rewrite as $\lim_{x\to 0^+} \frac{\ln(x)}{1/x}$ , which is $\frac{-\infty}{\infty}$ . Apply L'Hôpital's: $\lim_{x\to 0^+} \frac{1/x}{-1/x^2} = \lim_{x\to 0^+} (-x) = 0$ .

Reason:
Many physical and economic models involve products of vanishing and exploding quantities. Converting these to quotients allows for precise calculation of limits, revealing whether the zero or the infinity 'wins' in the interaction, which is critical for accurate modeling of boundary behaviors.

Easy

Medium

Hard

📝 All 0 times infinity indeterminate form MCQs

Q1. Which of the following best defines an indeterminate form of type $0/0$ ?

A.A limit expression where both numerator and denominator approach zero. ✅

B.A limit where numerator approaches a finite non‑zero number and denominator approaches zero.

C.A limit where both numerator and denominator approach infinity.

D.A limit where numerator approaches zero and denominator approaches a non‑zero constant.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: An indeterminate form of type $0/0$ occurs when, as the variable approaches a particular point, both the numerator and denominator tend to zero. In this situation the limit cannot be directly evaluated because the ratio could approach any finite number, infinity, or fail to exist, requiring further analysis such as algebraic simplification or L'Hospital's Rule.

Q2. L'Hospital's Rule can be applied to a limit of the form $0/0$ provided which of the following conditions holds?

A.Both functions are continuous at the point.

B.Both functions are differentiable near the point and the limit of their derivative ratio exists. ✅

C.Only the denominator needs to be differentiable.

D.The original limit must be equal to zero.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: L'Hospital's Rule requires that both the numerator and denominator be differentiable on an open interval containing the point of interest (except possibly at the point itself) and that the limit of the quotient of their derivatives exist (finite or infinite). Continuity alone is insufficient, and the original limit does not need to be zero; the rule resolves the indeterminate $0/0$ situation.

Q3. If $\lim_{x\to a} f(x)=0$ and $\lim_{x\to a} g(x)=0$ , and you know that $\lim_{x\to a} \frac{f'(x)}{g'(x)}=3$ , what is $\lim_{x\to a} \frac{f(x)}{g(x)}$ ?

A.0

B.1

C.3 ✅

D.Does not exist

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When both original functions tend to zero and the limit of the ratio of their derivatives exists, L'Hospital's Rule tells us that the original limit equals the limit of the derivative ratio. Since $\lim_{x\to a} \frac{f'(x)}{g'(x)}=3$ , the original limit $\lim_{x\to a} \frac{f(x)}{g(x)}$ is also 3. This direct application avoids further algebraic manipulation.

Q4. Suppose $\lim_{x\to 0} \frac{\sin x}{x}=1$ . Using this fact, evaluate $\lim_{x\to 0} \frac{1-\cos x}{x^2}$ .

A.0

\\frac{1}{2}

✅

C.1

D.2

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The identity $1-\cos x = 2\sin^2\!\left(\frac{x}{2}\right)$ allows us to rewrite the limit as $\frac{2\sin^2(x/2)}{x^2}=2\left(\frac{\sin(x/2)}{x/2}\right)^2\frac{1}{4}$ . As $x\to0$ , $\frac{\sin(x/2)}{x/2}\to1$ , so the expression approaches $\frac{2}{4}= \frac{1}{2}$ . This uses the known sine limit to resolve the $0/0$ form.

Q5. Given $\lim_{x\to 0} \frac{e^{x}-1}{\sin x}=L$ . Which of the following statements correctly describes the value of $L$ ?

A.L = 0 because numerator grows slower than denominator.

B.L = 1 because both numerator and denominator are approximately

x

for small

x

. ✅

C.L = \infty because exponential dominates.

D.L does not exist due to oscillation.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For small $x$ , the exponential can be expanded as $e^{x}=1+x+\frac{x^{2}}{2}+...$ and $\sin x = x-\frac{x^{3}}{6}+...$ . Keeping the leading terms, both numerator and denominator behave like $x$ , so their ratio approaches 1. Hence $L=1$ . This reasoning uses first‑order approximations to resolve the indeterminate form.

Q6. Consider the limit $\displaystyle \lim_{x\to 0} \frac{\ln(1+x^2)-x^2}{x^4}$ . Which of the following procedures correctly evaluates the limit?

A.Apply L'Hospital twice, obtaining

\frac{2/(1+x^2)-2}{4x^3}

and then evaluate at 0.

B.Rewrite numerator as a series,

\ln(1+x^2)=x^2-\frac{x^4}{2}+O(x^6)

, then simplify. ✅

C.Direct substitution yields 0/0, so the limit is undefined.

D.Factor

x^2

from numerator and cancel with denominator.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Expanding $\ln(1+x^2)$ as a Taylor series gives $\ln(1+x^2)=x^2-\frac{x^4}{2}+O(x^6)$ . Substituting into the numerator yields $(x^2-\frac{x^4}{2}+...)-x^2 = -\frac{x^4}{2}+O(x^6)$ . Dividing by $x^4$ leaves $-\frac12+O(x^2)$ , so the limit equals $-\frac12$ . This series method sidesteps repeated differentiation.

Q7. Which method is generally more efficient for evaluating $\displaystyle \lim_{x\to 0} \frac{1-\cos x}{x^2}$ : series expansion or L'Hospital's Rule?

A.Series expansion, because it avoids repeated differentiation. ✅

B.L'Hospital's Rule, because it always gives the answer in one step.

C.Both methods are equally efficient.

D.Neither method works for this limit.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Using the series $\cos x = 1-\frac{x^{2}}{2}+O(x^{4})$ directly gives $1-\cos x = \frac{x^{2}}{2}+O(x^{4})$ . Dividing by $x^{2}$ yields $\frac12+O(x^{2})$ , so the limit is $\frac12$ . While L'Hospital's Rule works, it requires differentiating twice, making the series approach quicker and more transparent for this particular indeterminate form.

Q8. Compare the outcomes of applying L'Hospital's Rule once versus twice to the limit $\displaystyle \lim_{x\to 0} \frac{e^{x}-1-x}{x^2}$ . Which statement is true?

A.Applying once yields the correct limit; applying twice leads to an indeterminate form again.

B.Applying once still gives

0/0

; applying twice yields the correct limit of

\tfrac12

. ✅

C.Both applications give the same limit value.

D.L'Hospital's Rule cannot be applied because the functions are not differentiable.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The first differentiation gives $\frac{e^{x}-1}{2x}$ , which is still $0/0$ as $x\to0$ . Applying L'Hospital a second time yields $\frac{e^{x}}{2}$ , and evaluating at $x=0$ gives $\frac12$ . Thus the limit is $\frac12$ , and two applications are necessary to resolve the indeterminate form.

Q9. Evaluate the limit $\displaystyle \lim_{x\to 0} \frac{\tan x - x}{x^3}$ using either series or L'Hospital. Which result is obtained?

A.0

\\frac{1}{3}

✅

\\frac{2}{3}

D.1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The series expansion of $\tan x$ is $x+\frac{x^{3}}{3}+O(x^{5})$ . Subtracting $x$ leaves $\frac{x^{3}}{3}+O(x^{5})$ . Dividing by $x^{3}$ gives $\frac13+O(x^{2})$ , so the limit approaches $\frac13$ . This calculation resolves the $0/0$ form without needing differentiation.

Q10. For the limit $\displaystyle L=\lim_{x\to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$ , determine $L$ without using L'Hospital's Rule. Which reasoning leads to the correct value?

A.Rationalize numerator to obtain

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

and simplify.

B.Apply L'Hospital twice to get

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

evaluated at 0.

C.Use the series

\sqrt{1\pm x}=1\pm \tfrac{x}{2}-\tfrac{x^{2}}{8}+...

and subtract. ✅

D.Observe the function is odd, so limit equals 0.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Expanding each square root gives $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+O(x^{3})$ and $\sqrt{1-x}=1-\frac{x}{2}-\frac{x^{2}}{8}+O(x^{3})$ . Their difference is $x+O(x^{3})$ . Dividing by $x$ yields $1+O(x^{2})$ , so the limit is $1$ . This series approach correctly handles the $0/0$ situation.

Q11. Consider $\displaystyle \lim_{x\to 0} \frac{\ln(1+x)-x}{x^2}$ . Which of the following steps correctly leads to the limit value?

A.Expand both numerator functions as series up to

x^2

terms and subtract. ✅

B.Apply L'Hospital three times directly.

C.Factor

x

from numerator and cancel.

D.Use the identity

\ln(1+x)=\sin x

for small

x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using the series $\ln(1+x)=x-\frac{x^{2}}{2}+O(x^{3})$ , the numerator becomes $(x-\frac{x^{2}}{2}+...)-x = -\frac{x^{2}}{2}+O(x^{3})$ . Dividing by $x^{2}$ gives $-\frac12+O(x)$ , so the limit equals $-\frac12$ . This method resolves the indeterminate form efficiently.

Q12. Explain why the limit $\displaystyle \lim_{x\to 0} \frac{\sin x}{x}$ is not an indeterminate form, even though direct substitution gives $0/0$ .

A.Because both functions are continuous at the point.

B.Because

\sin x

and

x

have the same first‑order behavior near 0, yielding a finite ratio. ✅

C.Because L'Hospital's Rule cannot be applied.

D.Because the numerator approaches 1.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Although substituting $x=0$ yields $0/0$ , the ratio $\frac{\sin x}{x}$ approaches a well‑defined finite value because both numerator and denominator vanish at the same rate. The leading term of the sine series is $x$ , so the quotient tends to 1. This known limit removes the indeterminate nature without further manipulation.

Q13. Synthesize the relationship between the definition of derivative as a limit and the appearance of the $0/0$ indeterminate form.

A.The derivative limit always yields a

0/0

form, which must be resolved. ✅

B.When the function is linear, the limit avoids

0/0

C.The derivative

\displaystyle f&#039;(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

often produces

0/0

, requiring techniques like L'Hospital or algebraic manipulation.

D.Derivatives never involve indeterminate forms.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The definition of the derivative involves the quotient $\frac{f(a+h)-f(a)}{h}$ . As $h\to0$ , both the numerator and denominator typically approach zero, creating a $0/0$ form. Resolving this indeterminate expression—through algebraic simplification, series expansion, or L'Hospital's Rule—yields the derivative value, linking limits and rates of change.

Q14. Which of the following statements correctly captures the principle behind transforming a $0/0$ limit into a determinate one by factoring common terms?

A.Factoring removes the zero in the numerator only.

B.Factoring can cancel a common factor that causes both numerator and denominator to vanish, leaving a finite limit. ✅

C.Factoring always changes the limit value.

D.Factoring is only useful for polynomial functions.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When both numerator and denominator share a factor that becomes zero at the limit point, factoring allows that common factor to be canceled, eliminating the $0/0$ situation. The remaining expression typically yields a finite value, providing a straightforward way to evaluate limits that initially appear indeterminate.

Q15. If $\displaystyle \lim_{x\to 0} \frac{f(x)}{x}=2$ and $\displaystyle \lim_{x\to 0} \frac{g(x)}{x}=3$ , what is $\displaystyle \lim_{x\to 0} \frac{f(x)}{g(x)}$ ?

A.0

B.2

C.3

\\frac{2}{3}

✅

💡 Difficulty: medium | ✅ Correct: D

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