Recognizing Indeterminate Forms

5 min read

Reference: Stewart §6.8

Why Some Limits Are Tricky

You’ve learned that limits often work by direct substitution: plug in the value and you’re done. But sometimes substitution gives you something strange like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. What does that mean?

These expressions are called indeterminate forms because they don’t tell you what the limit actually is. The expression $\frac{0}{0}$ could equal 1 (like $\lim_{x \to 0} \frac{x}{x}$), or 0 (like $\lim_{x \to 0} \frac{x^2}{x}$), or any other value. The form itself doesn’t determine the answer, hence the name “indeterminate.”

Recognizing indeterminate forms is the first step to solving them. Once you identify the type, you can choose the right technique.

Prerequisite Map

Quick Reference

Property	Value
Concept	L’Hospital’s Rule
Chapter	6, Section 8
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Seven Indeterminate Forms

Form	Example	Why It’s Indeterminate
$\frac{0}{0}$	$\lim_{x \to 0} \frac{\sin x}{x}$	Numerator and denominator both vanish
$\frac{\infty}{\infty}$	$\lim_{x \to \infty} \frac{e^x}{x^2}$	Both grow without bound
$0 \cdot \infty$	$\lim_{x \to 0^+} x \ln x$	One factor shrinks, one explodes
$\infty - \infty$	$\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{\sin x}\right)$	Difference of large quantities
$0^0$	$\lim_{x \to 0^+} x^x$	Base and exponent both approach 0
$\infty^0$	$\lim_{x \to \infty} x^{1/x}$	Infinite base, vanishing exponent
$1^\infty$	$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$	Base approaching 1, infinite exponent

Forms That Are NOT Indeterminate

These forms have definite values:

Form	Value	Example
$\frac{c}{0}$ (where $c \neq 0$)	$\pm\infty$ or DNE	$\lim_{x \to 0^+} \frac{1}{x} = +\infty$
$\frac{0}{c}$ (where $c \neq 0$)	$0$	$\lim_{x \to 0} \frac{x^2}{1+x} = 0$
$\frac{c}{\infty}$	$0$	$\lim_{x \to \infty} \frac{5}{x} = 0$
$0 \cdot c$	$0$	Direct multiplication
$\infty + \infty$	$\infty$	Sum of large quantities
$0^\infty$	$0$	$\lim_{x \to \infty} \left(\frac{1}{2}\right)^x = 0$
$\infty^\infty$	$\infty$	Large base, large exponent

How to Identify the Form

Step 1: Evaluate the limit of the numerator (or first part) separately.

Step 2: Evaluate the limit of the denominator (or second part) separately.

Step 3: Combine to identify the form.

Example: For $\lim_{x \to 0} \frac{e^x - 1}{x}$:

Numerator: $e^0 - 1 = 0$
Denominator: $0$
Form: $\frac{0}{0}$ ✓ (indeterminate)

Example: For $\lim_{x \to 0} \frac{e^x}{x}$:

Numerator: $e^0 = 1$
Denominator: $0$
Form: $\frac{1}{0}$ (NOT indeterminate; this is $\pm\infty$ depending on direction)

Common Patterns

Quotients approaching $\frac{0}{0}$:

$\frac{\sin x}{x}$ as $x \to 0$
$\frac{\ln x}{x - 1}$ as $x \to 1$
$\frac{e^x - 1}{x}$ as $x \to 0$
Any $\frac{f(x) - f(a)}{x - a}$ as $x \to a$ (derivative definition!)

Quotients approaching $\frac{\infty}{\infty}$:

$\frac{e^x}{x^n}$ as $x \to \infty$
$\frac{\ln x}{x}$ as $x \to \infty$
$\frac{P(x)}{Q(x)}$ (polynomials of same degree) as $x \to \infty$

The Connection to Derivatives:

The form $\frac{0}{0}$ is closely related to the derivative. Notice that: $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$ is always a $\frac{0}{0}$ form (assuming $f$ is continuous at $a$).

Practice Problems

Level 1 Classifying Forms

Identify the form (indeterminate or not) for each limit:

$\lim_{x \to 0} \frac{x^2}{\sin x}$
$\lim_{x \to \infty} \frac{x^3}{e^x}$
$\lim_{x \to 0^+} \frac{1}{x^2}$
$\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$

Thought Process

For each limit, evaluate the numerator and denominator separately as the variable approaches the limit point.

If both approach 0: $\frac{0}{0}$ (indeterminate)
If both approach $\infty$: $\frac{\infty}{\infty}$ (indeterminate)
If one is nonzero and the other is 0: not indeterminate
If one is finite and the other is $\infty$: not indeterminate

Show Answer

(a) $\lim_{x \to 0} \frac{x^2}{\sin x}$

Numerator: $0^2 = 0$
Denominator: $\sin 0 = 0$
Form: $\frac{0}{0}$ (indeterminate)

(b) $\lim_{x \to \infty} \frac{x^3}{e^x}$

Numerator: $\infty$
Denominator: $\infty$
Form: $\frac{\infty}{\infty}$ (indeterminate)

(c) $\lim_{x \to 0^+} \frac{1}{x^2}$

Numerator: $1$
Denominator: $0^+$
Form: $\frac{1}{0^+} = +\infty$ (NOT indeterminate)

(d) $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$

Numerator: $1 - 1 = 0$
Denominator: $1 - 1 = 0$
Form: $\frac{0}{0}$ (indeterminate)

Level 2 Product and Power Forms

Identify the indeterminate form for each limit:

$\lim_{x \to 0^+} x^2 \ln x$
$\lim_{x \to 0^+} x^{\sin x}$
$\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^x$

Thought Process

For products: evaluate each factor separately and combine.

For powers $f(x)^{g(x)}$: evaluate the base $f(x)$ and exponent $g(x)$ separately.

$0^0$, $\infty^0$, and $1^\infty$ are the three indeterminate power forms.

Show Answer

(a) $\lim_{x \to 0^+} x^2 \ln x$

First factor: $x^2 \to 0$
Second factor: $\ln x \to -\infty$
Form: $0 \cdot (-\infty) = 0 \cdot \infty$ (indeterminate)

(b) $\lim_{x \to 0^+} x^{\sin x}$

Base: $x \to 0^+$
Exponent: $\sin x \to 0$
Form: $0^0$ (indeterminate)

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

Base: $1 + \frac{3}{x} \to 1$
Exponent: $x \to \infty$
Form: $1^\infty$ (indeterminate)

Level 3 Tricky Cases

Determine whether each limit is indeterminate. If so, identify the form.

$\lim_{x \to \infty} (x - \sqrt{x^2 + 1})$
$\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}$
$\lim_{x \to \infty} x^{1/\ln x}$

Thought Process

(a) This is a difference. As $x \to \infty$, both $x \to \infty$ and $\sqrt{x^2+1} \to \infty$.

(b) Check numerator and denominator separately. $1 - \cos 0 = 0$ and $0 \cdot \sin 0 = 0$.

Show Answer

(a) $\lim_{x \to \infty} (x - \sqrt{x^2 + 1})$

First term: $x \to +\infty$
Second term: $\sqrt{x^2 + 1} \to +\infty$
Form: $\infty - \infty$ (indeterminate)

(b) $\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}$

Numerator: $1 - \cos 0 = 1 - 1 = 0$
Denominator: $0 \cdot \sin 0 = 0 \cdot 0 = 0$
Form: $\frac{0}{0}$ (indeterminate)

(c) $\lim_{x \to \infty} x^{1/\ln x}$

Base: $x \to \infty$
Exponent: $\frac{1}{\ln x} \to 0$ (since $\ln x \to \infty$)
Form: $\infty^0$ (indeterminate)

Level 4 Distinguishing Similar-Looking Limits

For each pair, determine which limit is indeterminate and which is not. Explain the difference.

(Pair A)

$\lim_{x \to 0^+} x^x$ vs. $\lim_{x \to 0^+} x^{1/x}$

(Pair B)

$\lim_{x \to 0} \frac{\sin x}{x}$ vs. $\lim_{x \to 0} \frac{\sin x}{x^2}$

Thought Process

For each limit, carefully identify base/exponent or numerator/denominator behavior.

The key is that small changes in the expression can change whether the form is indeterminate.

Show Answer

Pair A:

$\lim_{x \to 0^+} x^x$:

Base: $0^+$
Exponent: $0^+$
Form: $0^0$ (indeterminate)

$\lim_{x \to 0^+} x^{1/x}$:

Base: $0^+$
Exponent: $\frac{1}{0^+} = +\infty$
Form: $0^\infty$ (NOT indeterminate, equals 0)

Difference: In the first, the exponent approaches 0. In the second, the exponent approaches $+\infty$, making the base (which is small and positive) raised to a huge power, giving 0.

Pair B:

$\lim_{x \to 0} \frac{\sin x}{x}$:

Form: $\frac{0}{0}$ (indeterminate, equals 1)

$\lim_{x \to 0} \frac{\sin x}{x^2}$:

Numerator: $\sin 0 = 0$
Denominator: $0^2 = 0$
Form: $\frac{0}{0}$ (also indeterminate)

Both are indeterminate! But they have different values: the first equals 1, while the second does not exist (it approaches $+\infty$ from the right and $-\infty$ from the left).

Level 5 Why $\frac{0}{0}$ Is Related to Derivatives

Show that $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ is always of the form $\frac{0}{0}$ when $f$ is continuous at $a$.
Use this observation to explain why L'Hospital's Rule might involve derivatives.
Given that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ is a $\frac{0}{0}$ form, what does this tell you about $\frac{d}{dx}[\sin x]$ at $x = 0$?

Thought Process

(a) As $h \to 0$, the numerator $f(a+h) - f(a) \to f(a) - f(a) = 0$ by continuity. The denominator $h \to 0$.

(b) The definition of derivative IS a $\frac{0}{0}$ limit. So derivatives are intimately connected to resolving this indeterminate form.

(c) The limit $\frac{\sin x}{x}$ as $x \to 0$ can be viewed as $\frac{\sin x - \sin 0}{x - 0}$, which is the derivative of $\sin x$ at $x = 0$.

Show Answer

(a) For $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$:

Numerator: As $h \to 0$, we have $a + h \to a$, so by continuity: $$f(a+h) \to f(a) \implies f(a+h) - f(a) \to 0$$
Denominator: $h \to 0$
Form: $\frac{0}{0}$ ✓

(b) The derivative is defined as exactly this $\frac{0}{0}$ limit: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

Since every derivative is a resolved $\frac{0}{0}$ form, it makes sense that derivatives would help resolve other $\frac{0}{0}$ limits. L’Hospital’s Rule formalizes this: when you have $\frac{f(x)}{g(x)}$ approaching $\frac{0}{0}$, the ratio of their rates of change (derivatives) determines the limit.

(c) We can write: $$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\sin x - \sin 0}{x - 0} = \left.\frac{d}{dx}[\sin x]\right\vert _{x=0} = \cos 0 = 1$$

This confirms that $\frac{d}{dx}[\sin x] = \cos x$, since the derivative at 0 equals 1, which matches $\cos 0 = 1$.

The limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ is essentially computing the derivative of $\sin x$ at $x = 0$ using the definition!

Common Misconceptions

Common misconception

$\frac{0}{0} = 1$ because any number divided by itself equals 1.

This is the limit-as-unreachable-barrier error. The expression $\frac{0}{0}$ is not a number with a determined value; it is an indeterminate form that signals a competition between two tendencies. The actual limit $\lim_{x \to a} \frac{f(x)}{g(x)}$ where both approach zero depends on the relative rates of $f$ and $g$ and may equal any real number or fail to exist. For example, $\lim_{x \to 0} \frac{\sin x}{x} = 1$ while $\lim_{x \to 0} \frac{x^2}{x} = 0$ and $\lim_{x \to 0} \frac{x}{x^2}$ diverges.

Mastery Checklist

I can identify $\frac{0}{0}$ and $\frac{\infty}{\infty}$ indeterminate forms
I can identify $0 \cdot \infty$ and $\infty - \infty$ indeterminate forms
I can identify $0^0$, $\infty^0$, and $1^\infty$ indeterminate power forms
I know which forms are NOT indeterminate (like $\frac{c}{0}$, $0^\infty$)
I can distinguish between similar-looking limits with different forms
I understand the connection between $\frac{0}{0}$ and derivatives

Mental Model

The Racing Analogy:

Think of indeterminate forms as a race between two quantities:

$\frac{0}{0}$: Both runners are approaching the finish line (0). Who gets there first? The ratio of their speeds determines who “wins.”
$\frac{\infty}{\infty}$: Both runners are racing away to infinity. Who’s faster? Again, the ratio of speeds (derivatives) determines the outcome.
$0 \cdot \infty$: One runner shrinks to nothing while the other explodes. Does the shrinking happen faster than the explosion, or vice versa?

Indeterminate forms are “races” where the outcome isn’t obvious from the starting positions alone. You need to look at the speeds (derivatives) to determine the winner.

Connections

Looking back:

Limit Laws explain when direct evaluation works
Limits at Infinity help identify $\frac{\infty}{\infty}$ forms

Looking ahead:

L’Hospital’s Rule shows how to resolve these forms
Indeterminate Powers and Products handles the non-quotient cases

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Inverse Trigonometric Functions	Skills Index	L’Hospital’s Rule

Last updated: 2026-01-22