Indeterminate Forms: Rules & Examples

Indeterminate forms are limit results like $0 \cdot \infty$ , $\infty - \infty$ , $1^\infty$ , $0^0$ and $\infty^0$ that don’t tell you the answer by themselves. To find these limits, rewrite the expression as a fraction of the form 0/0 or ∞/∞, then use L’Hôpital’s rule. For the three power forms, take the natural log first.

Before this lesson: L’Hôpital’s Rule, Properties of Logarithms and Adding and Subtracting Rational Expressions

Why these forms are indeterminate

In $0 \cdot \infty$ , one factor pulls toward 0 and the other toward ∞. Which one wins depends on how fast each one moves, so the limit could be 0, ∞ or any number in between. The same tug-of-war happens in the other forms: in $1^\infty$ , a base near 1 is raised to a huge power.

Some forms that look strange are not indeterminate. For example, $0^\infty = 0$ , $\infty \cdot \infty = \infty$ and $\frac{1}{\infty} = 0$ always. Only the seven forms below need special work.

Form	How to rewrite it	Then
$\frac{0}{0}$ , $\frac{\infty}{\infty}$	Already a fraction	L’Hôpital’s rule
$0 \cdot \infty$	Write $fg$ as $\frac{f}{1/g}$ or $\frac{g}{1/f}$	0/0 or ∞/∞
$\infty - \infty$	Combine into one fraction (common denominator) or factor	0/0 or ∞/∞
$1^\infty$ , $0^0$ , $\infty^0$	Let y be the expression and take ln y = (exponent)(ln base)	Find lim ln y, then y = e to that power

How to handle the power forms

Let y equal the expression, such as $y = \left(1 + \frac{3}{x}\right)^x$ .
Take ln of both sides and bring the exponent down: $\ln y = x\ln\left(1 + \frac{3}{x}\right)$ .
Find $\lim \ln y$ . It is usually a 0 · ∞ form, so rewrite it as a fraction and use L’Hôpital’s rule.
If $\lim \ln y = L$ , then $\lim y = e^L$ .

The last step works because $e^x$ is continuous: $y = e^{\ln y}$ , and the limit can move inside the exponent.

Graph of y = (1 + 1/x)^x for x from 0 to 30 (solid blue). It rises from near 1, passes 2 at x = 1, and levels off just under the dashed horizontal line y = e ≈ 2.718. — As x → ∞, (1 + 1/x)^x has the form 1^∞, but its limit is e, not 1.

Indeterminate forms examples

Example 1: A 0 · ∞ form

Find $\lim_{x \to 0^+} x\ln x$ .

0 \cdot (-\infty)

As x → 0⁺, x → 0 and ln x → −∞.

\lim_{x \to 0^+}\frac{\ln x}{\frac{1}{x}}

Rewrite as a fraction: dividing by 1/x is multiplying by x. Now it is −∞/∞.

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

L’Hôpital’s rule: differentiate the top and the bottom.

\lim_{x \to 0^+}(-x) = 0

Simplify: (1/x)(−x²) = −x.

Answer: 0. Put ln x on top, not x. The other choice, $\frac{x}{1/\ln x}$ , gets more complicated after differentiating.

Example 2: An ∞ − ∞ form

Find $\lim_{x \to 1^+}\left(\frac{1}{\ln x} - \frac{1}{x - 1}\right)$ .

\infty - \infty

As x → 1⁺, both fractions grow without bound.

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

Combine into one fraction with a common denominator. Now it is 0/0.

\lim_{x \to 1^+}\frac{1 - \frac{1}{x}}{\ln x + \frac{x - 1}{x}}

L’Hôpital’s rule; product rule in the denominator. Still 0/0.

\lim_{x \to 1^+}\frac{x - 1}{x\ln x + x - 1}

Multiply the top and the bottom by x to clear the small fractions.

\lim_{x \to 1^+}\frac{1}{\ln x + 1 + 1}

L’Hôpital’s rule again: the derivative of x ln x is ln x + 1.

\frac{1}{0 + 2} = \frac{1}{2}

Substitute x = 1.

Answer: $\frac{1}{2}$ .

Example 3: A 1^∞ form

Find $\lim_{x \to \infty}\left(1 + \frac{3}{x}\right)^x$ .

\ln y = x\ln\left(1 + \frac{3}{x}\right)

Let y be the expression; take ln and bring the exponent down.

\lim_{x \to \infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{1}{x}}

This is ∞ · 0. Rewrite as 0/0.

\lim_{x \to \infty}\frac{\frac{-3/x^2}{1 + 3/x}}{-\frac{1}{x^2}}

L’Hôpital’s rule, with the chain rule on top.

\lim_{x \to \infty}\frac{3}{1 + \frac{3}{x}} = 3

The −1/x² factors cancel, and 3/x → 0.

\lim_{x \to \infty} y = e^3

Since ln y → 3, y → e³.

Answer: $e^3 \approx 20.09$ . In general, $\lim_{x \to \infty}\left(1 + \frac{k}{x}\right)^x = e^k$ .

Example 4: A 0^0 form

Find $\lim_{x \to 0^+} x^x$ .

\ln y = x\ln x

Let y = xˣ and take ln.

\lim_{x \to 0^+}x\ln x = 0

This is Example 1.

\lim_{x \to 0^+} x^x = e^0 = 1

Undo the log.

Answer: 1. (See the graph of xˣ in logarithmic differentiation.)

Example 5: Word problem with compound interest

You invest $1,000 at 5% interest per year, compounded n times a year. After 1 year the balance is $1000\left(1 + \frac{0.05}{n}\right)^n$ . What happens as interest is compounded more and more often?

\lim_{n \to \infty}\left(1 + \frac{0.05}{n}\right)^n = e^{0.05}

This is a 1^∞ form with k = 0.05, just like Example 3.

1000e^{0.05} \approx 1000(1.051271)

Multiply by the $1,000 deposit.

1000e^{0.05} \approx 1051.27

Round to the nearest cent.

Answer: the balance approaches $1,051.27. This limit is continuous compounding, $A = Pe^{rt}$ (see compound interest). Compounding more often helps, but only up to this cap.

Common mistakes

Mistake	Example of the error	Fix
Treating 1^∞ as 1	“ $(1 + \frac{3}{x})^x \to 1$ “	The base is only near 1. Take ln: the limit is e³.
Treating 0 · ∞ as 0	“ $x\ln x \to 0$ , because 0 times anything is 0″	The answer happens to be 0 here, but $x \cdot \frac{1}{x} \to 1$ . Always rewrite and check.
Using L’Hôpital’s rule on a product	Differentiating x and ln x separately in x ln x	The rule only works on a fraction. Rewrite first.
Forgetting to undo the log	Stopping at “the limit is 3” in Example 3	That is the limit of ln y. The answer is e³.
Calling ∞ − ∞ equal to 0	“ $\frac{1}{\ln x} - \frac{1}{x - 1} \to 0$ “	Combine into one fraction. The limit in Example 2 is ½.

Related methods

Algebra instead of L’Hôpital’s rule

Some ∞ − ∞ limits are easier with algebra. For $\lim_{x \to \infty}\left(\sqrt{x^2 + x} - x\right)$ , multiply by the conjugate to get $\frac{x}{\sqrt{x^2 + x} + x} \to \frac{1}{2}$ (see evaluating limits algebraically).

The number e as a limit

Example 3 with k = 1 gives $\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^x = e$ , one of the standard definitions of the number e.

Practice problems

Find $\lim_{x \to 0^+} x^2\ln x$ .
Find $\lim_{x \to \infty} xe^{-x}$ .
Find $\lim_{x \to \infty} x\sin\frac{1}{x}$ .
Find $\lim_{x \to 0^+}\left(\frac{1}{x} - \frac{1}{e^x - 1}\right)$ .
Find $\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^{2x}$ .
Find $\lim_{x \to 0^+}(1 + x)^{1/x}$ .
Find $\lim_{x \to \infty} x^{1/x}$ .
Find $\lim_{x \to 0^+} x^{2x}$ .
Find $\lim_{x \to \infty}(x - \ln x)$ .
$500 is invested at 4% per year, compounded continuously. Find the balance after 10 years as a limit of $500\left(1 + \frac{0.04}{n}\right)^{10n}$ .

Show answers

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

0 · (−∞): rewrite as −∞/∞.

\lim_{x \to 0^+}\frac{\frac{1}{x}}{-\frac{2}{x^3}} = \lim_{x \to 0^+}\left(-\frac{x^2}{2}\right)

L’Hôpital’s rule, then simplify.

-\frac{0}{2} = 0

Substitute x = 0.

Answer: 0

\lim_{x \to \infty}\frac{x}{e^x}

∞ · 0: rewrite as ∞/∞.

\lim_{x \to \infty}\frac{1}{e^x} = 0

L’Hôpital’s rule.

Answer: 0

\lim_{x \to \infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}

∞ · 0: rewrite as 0/0.

\lim_{x \to \infty}\frac{\cos\frac{1}{x} \cdot \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}

L’Hôpital’s rule with the chain rule.

\lim_{x \to \infty}\cos\frac{1}{x} = \cos 0 = 1

Cancel, then let 1/x → 0.

Answer: 1

\frac{e^x - 1 - x}{x(e^x - 1)}

∞ − ∞: combine into one fraction. Now 0/0.

\lim_{x \to 0^+}\frac{e^x - 1}{e^x - 1 + xe^x}

L’Hôpital’s rule; product rule on the bottom. Still 0/0.

\lim_{x \to 0^+}\frac{e^x}{2e^x + xe^x} = \frac{1}{2}

L’Hôpital’s rule again, then substitute x = 0.

Answer: $\frac{1}{2}$

\ln y = 2x\ln\left(1 + \frac{1}{x}\right)

1^∞: take ln.

\lim_{x \to \infty}\ln y = 2 \cdot 1 = 2

x ln(1 + 1/x) → 1, as in Example 3 with k = 1.

Answer: $e^2 \approx 7.389$

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

1^∞: take ln. This is 0/0.

\lim_{x \to 0^+}\frac{\frac{1}{1 + x}}{1} = 1

L’Hôpital’s rule, then substitute x = 0.

Answer: $e^1 = e$

\ln y = \frac{\ln x}{x}

∞⁰: take ln. This is ∞/∞.

\lim_{x \to \infty}\frac{\frac{1}{x}}{1} = 0

L’Hôpital’s rule.

Answer: $e^0 = 1$

\ln y = 2x\ln x

0⁰: take ln.

\lim_{x \to 0^+} 2x\ln x = 2 \cdot 0 = 0

Use Example 1.

Answer: $e^0 = 1$

x - \ln x = x\left(1 - \frac{\ln x}{x}\right)

∞ − ∞: factor out x.

\lim_{x \to \infty}\frac{\ln x}{x} = 0

From L’Hôpital’s rule, so the parentheses → 1.

Answer: ∞. Here ∞ − ∞ turned out to be infinite.

10.

\lim_{n \to \infty}\left(1 + \frac{0.04}{n}\right)^{10n} = e^{0.04 \cdot 10} = e^{0.4}

1^∞ form: ln y = 10n ln(1 + 0.04/n) → 10(0.04) = 0.4.

500e^{0.4} \approx 500(1.491825) \approx 745.91

Multiply by 500.

Answer: about $745.91

Indeterminate forms FAQ

What are the seven indeterminate forms?

$\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $0 \cdot \infty$ , $\infty - \infty$ , $1^\infty$ , $0^0$ and $\infty^0$ . Each one can lead to different limits depending on the functions involved.

Why is 1 to the infinity indeterminate?

The base is only close to 1, not exactly 1. A base slightly above 1 raised to a huge power can grow to any size, so the result depends on how fast each part moves.

How do you solve a 0 times infinity limit?

Rewrite the product fg as a fraction, $\frac{f}{1/g}$ or $\frac{g}{1/f}$ , to get 0/0 or ∞/∞. Then use L’Hôpital’s rule.

Is 0 to the power infinity indeterminate?

No. A base near 0 raised to a huge power goes to 0, so $0^\infty = 0$ . Likewise, $\infty^\infty = \infty$ .

Summary

Seven indeterminate forms: 0/0, ∞/∞, 0 · ∞, ∞ − ∞, 1^∞, 0⁰ and ∞⁰.
0 · ∞: rewrite the product as a fraction.
∞ − ∞: combine into one fraction, or factor.
Power forms: take ln, find the limit L of ln y, then the answer is e^L.
$\lim_{x \to \infty}\left(1 + \frac{k}{x}\right)^x = e^k$ .

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