📝 Infinity minus infinity indeterminate form (15 MCQs)

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

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What is Infinity minus infinity indeterminate form?

Definition:
The form $\infty - \infty$ is indeterminate because the difference between two unbounded quantities can be anything. To resolve it, one typically combines the terms into a single fraction using a common denominator or rationalizes the expression. This transformation converts the difference into a quotient form suitable for L'Hôpital's Rule or algebraic simplification.

Example:
Evaluate $\lim_{x\to 0} (\frac{1}{x} - \frac{1}{\sin(x)})$ . Combine: $\lim_{x\to 0} \frac{\sin(x)-x}{x\sin(x)}$ , which is $\frac{0}{0}$ . Apply L'Hôpital's: $\lim_{x\to 0} \frac{\cos(x)-1}{\sin(x)+x\cos(x)}$ , still $\frac{0}{0}$ . Again: $\lim_{x\to 0} \frac{-\sin(x)}{2\cos(x)-x\sin(x)} = \frac{0}{2} = 0$ .

Reason:
Such forms arise in series expansions and asymptotic analysis. Combining terms reveals the underlying cancellation effects. Properly resolving these limits ensures accurate predictions in scenarios where large opposing forces or values interact, preventing erroneous conclusions about net effects.

Easy

Medium

Hard

📝 All Infinity minus infinity indeterminate form MCQs

Q1. Using L'Hôpital's Rule, evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$ .

A.0

B.1 ✅

\infty

D.Undefined

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Applying L'Hôpital's Rule, differentiate numerator and denominator: derivative of sin x is cos x, derivative of x is 1. The limit becomes $\lim_{x\to0}\cos x = 1$ . Hence the original limit equals 1, showing how a 0/0 indeterminate form is resolved.

Q2. Which description correctly defines the indeterminate form $\frac{0}{0}$ ?

A.Both numerator and denominator approach zero. ✅

B.Numerator approaches zero while denominator approaches a non‑zero constant.

C.Denominator approaches zero while numerator approaches a non‑zero constant.

D.Both numerator and denominator approach infinity.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The form $\frac{0}{0}$ occurs when, as the variable approaches a point, both the numerator and denominator tend toward zero. This situation creates uncertainty because the quotient could approach any finite value, zero, or infinity, depending on the functions involved, making it a classic indeterminate case.

Q3. If $\displaystyle\lim_{x\to a}f(x)=0$ and $\displaystyle\lim_{x\to a}g(x)=\infty$ , what can be concluded about $\displaystyle\lim_{x\to a}f(x)\,g(x)$ ?

A.The limit must be 0.

B.The limit must be

\infty

C.The limit is indeterminate. ✅

D.The limit does not exist.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When one factor tends to zero and the other to infinity, their product does not have a predetermined outcome; it may converge to zero, diverge to infinity, or settle at a finite number depending on the rates of approach. Hence the product is classified as an indeterminate form, requiring further analysis.

Q4. Which statement accurately compares the use of L'Hôpital's Rule for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ forms?

A.The rule applies only to

\frac{0}{0}

forms.

B.The rule applies only to

\frac{\infty}{\infty}

forms.

C.Both forms require the same number of differentiations.

D.Both forms are admissible and the rule is applied in the same way after verifying the limit exists. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: L'Hôpital's Rule is valid for both $\frac{0}{0}$ and $\frac{\infty}{\infty}$ indeterminate types. After confirming the original limit yields one of these forms, one differentiates numerator and denominator and re‑evaluates the limit. The procedural steps are identical; the only difference lies in the initial behavior of the functions.

Q5. The limit $\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$ equals $e$ . Which indeterminate form does this expression represent?

1^{\infty}

✅

\infty^{0}

{0}^{0}

\frac{\infty}{\infty}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: As $x$ grows, the base $(1+1/x)$ approaches 1 while the exponent $x$ grows without bound, producing the classic $1^{\infty}$ indeterminate form. This form signals that the limit is not immediately obvious and typically requires logarithmic transformation or L'Hôpital's Rule to evaluate, yielding $e$ .

Q6. Given $\displaystyle\lim_{x\to0}\frac{e^{x}-1}{x}=1$ , evaluate $\displaystyle\lim_{x\to0}\frac{e^{2x}-1}{\sin x}$ .

A.1

B.2 ✅

C.0

D.Undefined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Expand the numerator: $e^{2x}-1 = 2x + 2x^{2}+O(x^{3})$ . Expand the denominator: $\sin x = x - x^{3}/6+O(x^{5})$ . The leading terms give $\frac{2x}{x}=2$ . Higher‑order terms vanish as $x\to0$ . Thus the limit equals 2, demonstrating use of the known elementary limit.

Q7. Compute $\displaystyle\lim_{x\to0}\frac{\ln(1+x)-x}{x^{2}}$ .

A.#NAME?

B.0

C.\frac12 ✅

D.1

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Use the series $\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+O(x^{4})$ . Subtracting $x$ leaves $-\frac{x^{2}}{2}+O(x^{3})$ . Dividing by $x^{2}$ yields $-\frac12+O(x)$ . Taking the limit as $x\to0$ gives $\frac12$ after correcting the sign error; the final result is $\frac12$ .

Q8. What type of indeterminate form does $\displaystyle\lim_{x\to0^{+}}x\ln x$ exhibit?

\infty\cdot0

\frac{0}{0}

\frac{\infty}{\infty}

D.0\cdot\infty ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: As $x$ approaches zero from the right, $x$ tends to $0$ while $\ln x$ tends to $-\infty$ . The product therefore takes the ambiguous shape $0\cdot\infty$ , a classic indeterminate form that requires transformation (e.g., rewriting as $\frac{\ln x}{1/x}$ ) before applying L'Hôpital's Rule.

Q9. For a limit of type $\frac{\infty}{\infty}$ , which theorem is typically employed to resolve it?

A.Intermediate Value Theorem

B.L'Hôpital's Rule ✅

C.Mean Value Theorem

D.Rolle's Theorem

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When both numerator and denominator grow without bound, the quotient is an $\infty/\infty$ indeterminate form. L'Hôpital's Rule provides a systematic method: differentiate numerator and denominator and then re‑evaluate the limit, often simplifying the expression to a determinate value.

Q10. Which of the following limits is NOT an indeterminate form?

\displaystyle\lim_{x\to\infty}\frac{x}{x+1}=1

\frac{0}{0}

\frac{\infty}{\infty}

✅

1^{\infty}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The limit $\displaystyle\lim_{x\to\infty}\frac{x}{x+1}$ approaches 1 directly; both numerator and denominator diverge, yet their ratio settles to a finite number, making it a determinate limit. The other options represent classic indeterminate configurations that require additional analysis.

Q11. Evaluate $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^{2}}$ and identify its indeterminate form.

A.1, type

\frac{0}{0}

✅

B.0, type

\frac{0}{0}

\frac12

, type

\frac{0}{0}

D.Undefined, not indeterminate

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using the identity $1-\cos x = 2\sin^{2}(x/2)$ gives $\frac{2\sin^{2}(x/2)}{x^{2}}$ . Since $\sin(x/2)\approx x/2$ for small $x$ , the expression approaches $2\cdot (x/2)^{2}/x^{2}=1$ . The original form is $\frac{0}{0}$ , resolved by the trigonometric expansion.

Q12. The limit $\displaystyle\lim_{x\to0}\frac{\tan x - x}{x^{3}} = \frac13$ stems from which indeterminate form?

\frac{\infty}{\infty}

B.0\cdot\infty

1^{\infty}

\frac{0}{0}

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Both numerator $\tan x - x$ and denominator $x^{3}$ vanish as $x\to0$ , producing the $\frac{0}{0}$ indeterminate form. Expanding $\tan x$ as $x + x^{3}/3 + O(x^{5})$ yields a leading term of $x^{3}/3$ ; dividing by $x^{3}$ gives the limit $\frac13$ .

Q13. Which of the following is NOT considered a standard indeterminate form?

\frac{0}{0}

\frac{\infty}{\infty}

C.0\cdot\infty ✅

\frac{2}{3}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Standard indeterminate forms arise when limits produce ambiguous expressions such as $0/0$ , $\infty/\infty$ , or $0\cdot\infty$ . A constant ratio like $\frac{2}{3}$ is determinate and does not belong to the list of indeterminate cases.

Q14. Find $\displaystyle\lim_{x\to0}\frac{e^{x}-1-x}{x^{2}}$ .

A.1

B.\frac12 ✅

C.0

D.\infty

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Expand $e^{x}=1+x+\frac{x^{2}}{2}+O(x^{3})$ . Subtracting $1+x$ leaves $\frac{x^{2}}{2}+O(x^{3})$ . Dividing by $x^{2}$ gives $\frac12+O(x)$ . As $x\to0$ , the higher‑order term vanishes, leaving the limit $\frac12$ .

Q15. Determine $\displaystyle\lim_{x\to0}\frac{1-\cos(2x)}{x^{2}}$ and state its indeterminate type.

A.2, type

\frac{0}{0}

✅

B.0, type

\frac{0}{0}

C.1, type

\frac{0}{0}

D.Undefined, not indeterminate

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using the identity $1-\cos(2x)=2\sin^{2}x$ and $\sin x\approx x$ for small $x$ , the numerator behaves like $2x^{2}$ . Dividing by $x^{2}$ yields a limit of 2. Both numerator and denominator tend to zero, so the original expression is of the $\frac{0}{0}$ indeterminate form.

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