📝 0^0 infinity^0 1^infinity indeterminate forms (16 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 16 questions available

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What is 0^0 infinity^0 1^infinity indeterminate forms?

Definition:
Forms like $0^0$ , $\infty^0$ , and $1^\infty$ are indeterminate because the base and exponent compete. To solve them, take the natural logarithm of the expression $y = f(x)^{g(x)}$ , yielding $\ln(y) = g(x)\ln(f(x))$ . This transforms the problem into a $0 \cdot \infty$ form, which can then be converted to a quotient for L'Hôpital's Rule. Finally, exponentiate the result to find $y$ .

Example:
Evaluate $\lim_{x\to 0^+} x^x$ . Let $y=x^x$ , so $\ln(y) = x\ln(x)$ . We know $\lim_{x\to 0^+} x\ln(x) = 0$ . Thus, $\ln(y) \to 0$ , so $y \to e^0 = 1$ . Therefore, $\lim_{x\to 0^+} x^x = 1$ .

Reason:
These forms appear in advanced calculus and combinatorics. Using logarithms linearizes the exponent, making the limit tractable. This technique is essential for defining continuous extensions of discrete functions and analyzing the behavior of power towers and complex exponential expressions.

Easy

Medium

Hard

📝 All 0^0 infinity^0 1^infinity indeterminate forms MCQs

Q1. What is the definition of the indeterminate form $0^{0}$ in the context of limits?

A.An expression where the base approaches zero while the exponent approaches zero, making the limit undefined without further analysis. ✅

B.A product of zero and zero.

C.A constant equal to one.

D.A situation where the exponent is zero regardless of the base.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: When both the base and exponent tend to zero, the limit cannot be assigned a single value because different rates of approach can produce different results. Therefore the form $0^{0}$ is labeled indeterminate, meaning additional analysis is required to determine the actual limit, unlike the discrete convention that sets $0^{0}=1$ .

Q2. Which of the following limits correctly evaluates to 1?

A.0

B.1 ✅

C.Infinity

D.-1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The limit $\lim_{x\to0^{+}} x^{x}$ equals 1 because taking logarithms gives $\ln L = \lim_{x\to0^{+}} x\ln x = 0$ . Hence the original expression approaches $e^{0}=1$ . The other options either diverge or are undefined, making choice B the only correct evaluation.

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

A.The limit always equals 1.

B.The limit is always 0.

C.The limit can be any non‑negative number depending on the rates of approach. ✅

D.The limit does not exist.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Because both the base and exponent vanish, the resulting limit depends on how quickly each approaches zero. By writing $\ln\big(f^{g}\big)=g\ln f$ and examining the product $g\cdot\ln f$ , one sees that the limit may range from 0 to 1 or even larger, so no single value is guaranteed.

Q4. When evaluating a limit that yields the form $0^{0}$ , which of the following strategies is generally more reliable than direct substitution?

A.Replace the expression with 1.

B.Apply the logarithm and convert to a quotient of two limits. ✅

C.Differentiate numerator and denominator directly.

D.Assume the limit equals the base.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Taking logarithms transforms the indeterminate power into a product, which can often be rewritten as a quotient of two limits. This allows the use of algebraic techniques such as L'Hôpital's rule or series expansions, providing a systematic way to resolve the ambiguity, unlike direct substitution which is invalid.

Q5. Let $f(x)=x$ and $g(x)=\sin x$ . What is $\displaystyle \lim_{x\to0} f(x)^{\,g(x)}$ ?

A.0

B.1 ✅

e^{0}

D.The limit does not exist.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Write $\ln L = g(x)\ln f(x)=\sin x\cdot\ln x$ . Near zero, $\sin x\approx x$ , so $\ln L\approx x\ln x\to0$ . Hence $L=e^{0}=1$ . Direct substitution would give the ambiguous form $0^{0}$ , but the logarithmic approach reveals the limit equals 1.

Q6. Which method correctly evaluates $\displaystyle \lim_{x\to0} (\cos x)^{\tan x}$ ?

A.Take natural log:

\displaystyle \lim_{x\to0} \tan x \ln(\cos x)

and evaluate. ✅

B.Replace

\cos x

with 1 and

\tan x

with 0 directly.

C.Apply L'Hôpital to

\frac{\ln(\cos x)}{1/\tan x}

D.Differentiate numerator and denominator separately.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Set $L=(\cos x)^{\tan x}$ and consider $\ln L=\tan x\ln(\cos x)$ . Using series $\cos x\approx1-\frac{x^{2}}{2}$ and $\tan x\approx x$ , we get $\ln L\approx x\big(-\frac{x^{2}}{2}\big)=-\frac{x^{3}}{2}\to0$ . Thus $L=e^{0}=1$ . The logarithmic conversion is essential for handling the $0^{0}$ form.

Q7. Consider $h(x)=\left(e^{-1/x}\right)^{x}$ for $x>0$ . What is $\displaystyle \lim_{x\to0^{+}} h(x)$ ?

A.0

B.1

e^{-1}

✅

D.Infinity

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Rewrite the expression as $h(x)=e^{-1/x\cdot x}=e^{-1}$ . The exponent simplifies to the constant $-1$ for all positive $x$ , so the limit as $x$ approaches zero from the right is simply $e^{-1}$ . No indeterminate behavior occurs here.

Q8. If $p(x)=x^{2}$ and $q(x)=\sqrt{x}$ both tend to 0 as $x\to0^{+}$ , which of the following best describes $\displaystyle \lim_{x\to0^{+}} p(x)^{\,q(x)}$ ?

A.The limit is 0 because the base vanishes faster.

B.The limit is 1 because the exponent tends to 0 faster.

C.The limit equals

e^{0}=1

. ✅

D.The limit is indeterminate and requires further analysis.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Take logs: $\ln L = q(x)\ln p(x)=\sqrt{x}\cdot\ln(x^{2})=2\sqrt{x}\,\ln x$ . As $x\to0^{+}$ , $\sqrt{x}\to0$ while $\ln x\to-\infty$ ; their product tends to 0. Hence $\ln L\to0$ and $L=e^{0}=1$ .

Q9. To evaluate $\displaystyle \lim_{x\to0} (1 - x)^{1/x}$ , which transformation is most appropriate?

A.Set

y = (1 - x)^{1/x}

and take

\ln y = \frac{\ln(1 - x)}{x}

. ✅

B.Directly substitute

x = 0

C.Rewrite as

\exp\big(\frac{1}{x}\big)

D.Apply the binomial theorem.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Let $y=(1-x)^{1/x}$ . Then $\ln y=\frac{\ln(1-x)}{x}$ . Using the series $\ln(1-x)\approx -x$ for small $x$ , the quotient approaches $-1$ . Hence $y\to e^{-1}$ . The logarithmic transformation turns the $0^{0}$ form into a manageable quotient.

Q10. Why is the expression $0^{0}$ classified as an indeterminate form when evaluating limits?

A.Because

0^{0}

is always equal to 1 by definition.

B.Because the limit can approach any non‑negative value depending on how the base and exponent approach zero. ✅

C.Because the exponent never reaches zero.

D.Because the base cannot be zero in limit processes.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When both base and exponent tend to zero, the resulting limit depends on their relative rates. For example, $(x)^{x}\to1$ while $(0)^{x}\to0$ . Thus the form $0^{0}$ does not have a unique limit and is labeled indeterminate, requiring further analysis to determine the actual value.

Q11. Let $a_n = \frac{1}{n}$ and $b_n = \frac{1}{\ln (n+1)}$ . What is $\displaystyle \lim_{n\to\infty} a_n^{\,b_n}$ ?

A.0

B.1

e^{-1}

✅

D.Infinity

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Compute $\ln L = b_n\ln a_n = \frac{1}{\ln (n+1)}\cdot(-\ln n)$ . As $n\to\infty$ , $\ln n$ and $\ln (n+1)$ are asymptotically equal, so the ratio approaches $-1$ . Hence $\ln L\to-1$ and $L=e^{-1}$ .

Q12. Evaluate $\displaystyle \lim_{x\to0^{+}} \left( \frac{\sin x}{x} \right)^{1/x}$ .

A.0

B.1 ✅

e^{-1/6}

e^{ -\frac{1}{6} }

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Set $L=\left(\frac{\sin x}{x}\right)^{1/x}$ . Then $\ln L = \frac{1}{x}\ln\!\left(1-\frac{x^{2}}{6}+O(x^{4})\right)\approx\frac{1}{x}\left(-\frac{x^{2}}{6}\right) = -\frac{x}{6}\to0$ . Therefore $L=e^{0}=1$ .

Q13. Find $\displaystyle \lim_{x\to0^{+}} \left( x^{x} \right)^{\ln x}$ .

A.0

B.1 ✅

e^{-1}

D.Infinity

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Write the expression as $e^{x(\ln x)^{2}}$ . Since $\ln x\to-\infty$ but $x\to0$ faster, the product $x(\ln x)^{2}\to0$ . Hence the exponent tends to 0 and the whole expression approaches $e^{0}=1$ .

Q14. Why is applying L'Hôpital's rule directly to a limit that yields the form $0^{0}$ generally unsuitable, and which alternative technique should be used?

A.Because L'Hôpital only works on quotients; instead, take logarithms to convert to a quotient of limits. ✅

B.Because the derivative of 0 is undefined; instead, use series expansion.

C.Because L'Hôpital gives the same answer; no alternative needed.

D.Because the rule requires both numerator and denominator to be non‑zero.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The power form $0^{0}$ is not a quotient, so L'Hôpital's rule cannot be applied directly. By taking the natural logarithm, the expression becomes $\ln L = g(x)\ln f(x)$ , which can often be rewritten as a ratio $\frac{\ln f(x)}{1/g(x)}$ . This quotient can then be tackled with L'Hôpital or other limit techniques.

Q15. The function $f(x)=x^{x}$ for $x>0$ approaches which value as $x\to0^{+}$ , and how does this relate to the indeterminate form $0^{0}$ ?

A.Approaches 0, showing that

0^{0}=0

B.Approaches 1, illustrating that

0^{0}

can be resolved to 1 in this context. ✅

C.Diverges to infinity, indicating

0^{0}

is undefined.

D.Oscillates, making the limit ambiguous.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using $\ln f = x\ln x$ and noting that $x\ln x\to0$ as $x\to0^{+}$ , we find $f(x)=e^{x\ln x}\to e^{0}=1$ . This concrete example shows that when both base and exponent tend to zero, the limit may be 1, demonstrating why $0^{0}$ is treated as indeterminate rather than assigned a fixed value.

Q16. If a function $g(x)$ approaches 0 while its exponent $h(x)$ approaches a positive constant $c>0$ , what is $\displaystyle \lim_{x\to a} g(x)^{\,h(x)}$ ?

A.0 ✅

B.1

c

D.Infinity

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: When the base tends to zero and the exponent settles at a positive constant, the power behaves like $0^{c}$ . Since any positive power of zero is zero, the limit is 0. The exponent’s constancy prevents the ambiguous behavior seen in the $0^{0}$ case, leading to a straightforward result.

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