What does an indeterminate form mean in limits?

An indeterminate form occurs when substituting a limit leads to expressions like 0/0 or ∞/∞, where the limit cannot be directly computed and requires further techniques such as L'Hôpital's Rule or algebraic simplification.

When should I use L'Hôpital's Rule?

L'Hôpital's Rule applies when a limit results in 0/0 or ∞/∞ forms. You differentiate the numerator and denominator separately and then re-evaluate the limit.

Can indeterminate forms be solved without L'Hôpital's Rule?

Yes. Sometimes algebraic simplification, trigonometric identities, or series expansion methods help resolve indeterminate forms, especially when derivatives are complicated.

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Indeterminate Forms

In calculus, while evaluating limits, we often encounter situations where direct substitution leads to undefined expressions such as 0/0 or $\frac{\infty}{\infty}$ _. These are known as indeterminate forms. They don’t directly reveal the limit’s value and require special techniques to solve, helping us analyze the behavior of functions near critical points.

1.0What Are Indeterminate Forms?

An indeterminate form occurs in limit problems when the substitution of a point leads to an expression that does not provide clear information about the limit. Instead of giving a specific value or infinity, it remains ambiguous, hence the term “indeterminate.” These forms arise frequently in calculus indeterminate forms problems and need careful handling.

2.0All Indeterminate Forms

In calculus, there are 7 types of indeterminate forms commonly encountered:

0/0
∞/∞
0 × ∞
∞ − ∞
0⁰
∞⁰
1^∞

Each type has its own approach for resolution, often involving algebraic manipulation, factoring, or applying special theorems.

3.0Indeterminate Forms in Limits

Consider the limit:

$lim_{x \to 0} \frac{s i n x}{x}$

Substituting x = 0 gives _0/0, an indeterminate form. We cannot conclude the limit directly, but using known limits or L'Hôpital's Rule helps us solve it.

Example:

$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$

Direct substitution:

$\frac{1 - c o s 0}{0 ^{2}} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form.

Using L'Hôpital's Rule (differentiate numerator and denominator):

$lim_{x \to 0} \frac{s i n x}{2 x}$

Substitute x = 0 → $\frac{s i n 0}{2 \times 0} = \frac{0}{0}$

Still indeterminate, so apply L'Hôpital’s Rule again:

$lim_{x \to 0} \frac{c o s x}{2}$

Substitute x = 0 →

$\frac{c o s 0}{2} = \frac{1}{2}$

Final Answer: _1/2

4.0Indeterminate Forms and L'Hôpital's Rule

One of the most powerful tools to resolve indeterminate forms is L'Hôpital's Rule, which applies when limits result in the forms 0/0 or ∞/∞.

The rule states:

If $lim_{x \to a} \frac{f ( x )}{g ( x )}$ results in 0/0 or ∞/∞ then

$lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ ,

provided the limit of the derivatives exists.

Example: Find $lim_{x \to \infty} \frac{x}{e ^{x}}$

Step 1: Direct substitution → ∞/∞

This is an indeterminate form.

Step 2: Apply L'Hôpital’s Rule:

$lim_{x \to \infty} \frac{1}{e ^{x}}$

Step 3: As $x \to \infty, e^{x} \to \infty$ so:

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

Final Answer: 0

5.0Indeterminate Forms Examples

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

Substitute x = 0:

$\frac{e ^{0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form.

Apply L'Hôpital’s Rule:

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

Final Answer: 1

$lim_{x \to \infty} \frac{l n x}{x}$

Substitute : $x \to \infty : \frac{\infty}{\infty}$

An indeterminate form.

Apply L'Hôpital’s Rule:

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

Final Answer: 0

These examples demonstrate how indeterminate forms occur in limits and how they are resolved using techniques like L'Hôpital's Rule.

6.0Solved Example on Indeterminate Forms

Example 1: Evaluate the limit:

$lim_{x \to 0} \frac{s i n x}{x}$

Solution:

Step 1: Substitute x = 0 → $\frac{s i n 0}{0} = \frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Apply the standard limit result:

$lim_{x \to 0} \frac{s i n x}{x} = 1$

So, the limit is 1.

Example 2: Evaluate: $lim_{x \to 0} \frac{s i n x}{x} = 1$

Solution:Step 1: Substitute x → ∞ → ∞/∞

An indeterminate form of type ∞/∞.

Step 2: Apply L'Hôpital’s Rule:

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

So, the limit is 0.

Example 3: Evaluate the limit:

$lim_{x \to 0} \frac{t a n x - x}{x ^{3}}$

Solution:

Step 1: Direct substitution gives:

$\frac{t a n 0 - 0}{0 ^{3}} = \frac{0 - 0}{0} = \frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Apply the series expansion of tan x near x = 0:

$tan x = x + \frac{x ^{3}}{3} + O (x^{5})$

Step 3: Substitute in the limit:

$lim_{x \to 0} \frac{( x + \frac{x ^{3}}{3} + \dots ) - x}{x ^{3}} = lim_{x \to 0} \frac{\frac{x ^{3}}{3}}{x ^{3}} = \frac{1}{3}$

Answer: 1/3

Example 4: Evaluate:

$lim_{x \to 0} \frac{e ^{s i n x} - 1}{x}$

Solution:

Step 1: Substitute x = 0 → $\frac{e ^{s i n 0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$

Indeterminate form of type 0/0.

Step 2: Apply L'Hôpital’s Rule:

Differentiate numerator and denominator with respect to x:

$lim_{x \to 0} \frac{e ^{s i n x} \cdot c o s x}{1}$

Step 3: Substitute x = 0 → $e^{s i n 0} \cdot cos 0 = e^{0} \cdot 1 = 1$

Answer: 1

Example 5: Evaluate:

$lim_{x \to 0} \frac{l n ( 1 + x ) - s i n x}{x ^{2}}$

Solution:

Step 1: Substitute x = 0 → $\frac{l n ( 1 + 0 ) - s i n 0}{0 ^{2}} = \frac{0 - 0}{0} = \frac{0}{0}$

Indeterminate form of type 0/0.

Step 2: Apply Taylor Series expansions near x = 0:

$ln (1 + x) = x - \frac{x ^{2}}{2} + O (x^{3})$
$sin x = x - \frac{x ^{3}}{6} + O (x^{5})$

Step 3: Substituting the expansions:

$lim_{x \to 0} \frac{( x - \frac{x ^{2}}{2} + \dots ) - ( x - \frac{x ^{3}}{6} + \dots )}{x ^{2}} = lim_{x \to 0} \frac{- \frac{x ^{2}}{2} + \frac{x ^{3}}{6}}{x ^{2}}$

Simplify numerator:

$\frac{- \frac{x ^{2}}{2} + \frac{x ^{3}}{6}}{x ^{2}} = - \frac{1}{2} + \frac{x}{6}$

Step 4: Take limit as x → 0:

$- \frac{1}{2}$

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

Example 6: Evaluate the limit:

$lim_{x \to 0} \frac{t a n x - s i n x}{x ^{3}}$

Solution:

Step 1: Substitute x = 0:

$\frac{t a n 0 - s i n 0}{0 ^{3}} = \frac{0 - 0}{0} = \frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Expand both tan x and sin x using Taylor series near x = 0:

$tan x = x + \frac{x ^{3}}{3} + O (x^{5})$
$tan x = x + \frac{x ^{3}}{6} + O (x^{5})$

Step 3: Substitute expansions:

$lim_{x \to 0} \frac{( x + \frac{x ^{3}}{3} ) - ( x - \frac{x ^{3}}{6} )}{x ^{3}}$

$= lim_{x \to 0} \frac{\frac{x ^{3}}{3} + \frac{x ^{3}}{6}}{x ^{3}}$

Final Answer: 1/2

Example 7: Evaluate:

$lim_{x \to 0} \frac{l n ( 1 + x ^{2} )}{1 - c o s x}$

Solution:

Step 1: Substitute x = 0:

$\frac{l n ( 1 + 0 ^{2} )}{1 - c o s 0} = \frac{l n 1}{1 - 1} = \frac{0}{0}$

This is an indeterminate form of type _0/0.

Step 2: Apply L'Hôpital's Rule:

Differentiate numerator and denominator:

$lim_{x \to 0} \frac{\frac{2 x}{1 + x ^{2}}}{s i n x}$

Step 3: Substitute x = 0:

$\frac{\frac{2 \cdot 0}{1 + 0 ^{2}}}{s i n 0} = \frac{0}{0}$

Still indeterminate, so apply L'Hôpital's Rule again:

Differentiate numerator and denominator again:

Numerator derivative:

$\frac{d}{d x} (\frac{2 x}{1 + x ^{2}}) = \frac{2 ( 1 + x ^{2} ) - 2 x ( 2 x )}{( 1 + x ^{2} ) ^{2}}$

$= \frac{2 ( 1 + x ^{2} ) - 4 x ^{2}}{( 1 + x ^{2} ) ^{2}}$

$= \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2}}$

Denominator derivative: cos x

Step 4: Now evaluate the limit as _{x \to 0}:

$lim_{x \to 0} \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2} \cdot c o s x} = \frac{2 - 0}{1 ^{2} \cdot 1} = 2$

Final Answer: 2

Example 8: Evaluate:

$lim_{x \to 0} \frac{x - s i n x}{x ^{3}}$

Solution:
Step 1: Substitute x = 0:

$\frac{0 - 0}{0 ^{3}} = \frac{0}{0}$

An indeterminate form of type 0/0.

Step 2: Expand sin x in Taylor series near 0:

$sin x = x - \frac{x ^{3}}{6} + O (x^{5})$

Step 3: Substitute the expansion:

$lim_{x \to 0} \frac{x - ( x - \frac{x ^{3}}{6} )}{x ^{3}} = lim_{x \to 0} \frac{\frac{x ^{3}}{6}}{x ^{3}} = \frac{1}{6}$

Final Answer: _1/6

Example 9: How many types of indeterminate forms are there?

Solution: There are 7 standard indeterminate forms in calculus:

0/0
∞/∞
0 × ∞
∞ − ∞
0⁰
∞⁰
$1^{\infty}$

7.0Practice Questions on Indeterminate Forms

Evaluate: $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$
Find: $lim_{x \to 0} \frac{e ^{x} - 1}{x}$
Compute: $lim_{x \to \infty} \frac{l n x}{x}$
Calculate: $lim_{x \to 0^{+}} x^{x}$

Also Read:

Derivative Function Calculus	Limits and Derivatives	Limits and Continuity
First Derivative Test	Application of Derivatives	Continuity in Interval
Limits of Functions	Central Limit Theorem	Partial Derivative

Join ALLEN!

(Session 2026 - 27)

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Class

Choose class

Goal

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Preferred Mode

State

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I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Indeterminate Forms

In calculus, while evaluating limits, we often encounter situations where direct substitution leads to undefined expressions such as 0/0 or $\frac{\infty}{\infty}$ _. These are known as indeterminate forms. They don’t directly reveal the limit’s value and require special techniques to solve, helping us analyze the behavior of functions near critical points.

1.0What Are Indeterminate Forms?

An indeterminate form occurs in limit problems when the substitution of a point leads to an expression that does not provide clear information about the limit. Instead of giving a specific value or infinity, it remains ambiguous, hence the term “indeterminate.” These forms arise frequently in calculus indeterminate forms problems and need careful handling.

2.0All Indeterminate Forms

In calculus, there are 7 types of indeterminate forms commonly encountered:

0/0
∞/∞
0 × ∞
∞ − ∞
0⁰
∞⁰
1^∞

Each type has its own approach for resolution, often involving algebraic manipulation, factoring, or applying special theorems.

3.0Indeterminate Forms in Limits

Consider the limit:

$lim_{x \to 0} \frac{s i n x}{x}$

Substituting x = 0 gives _0/0, an indeterminate form. We cannot conclude the limit directly, but using known limits or L'Hôpital's Rule helps us solve it.

Example:

$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$

Direct substitution:

$\frac{1 - c o s 0}{0 ^{2}} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form.

Using L'Hôpital's Rule (differentiate numerator and denominator):

$lim_{x \to 0} \frac{s i n x}{2 x}$

Substitute x = 0 → $\frac{s i n 0}{2 \times 0} = \frac{0}{0}$

Still indeterminate, so apply L'Hôpital’s Rule again:

$lim_{x \to 0} \frac{c o s x}{2}$

Substitute x = 0 →

$\frac{c o s 0}{2} = \frac{1}{2}$

Final Answer: _1/2

4.0Indeterminate Forms and L'Hôpital's Rule

One of the most powerful tools to resolve indeterminate forms is L'Hôpital's Rule, which applies when limits result in the forms 0/0 or ∞/∞.

The rule states:

If $lim_{x \to a} \frac{f ( x )}{g ( x )}$ results in 0/0 or ∞/∞ then

$lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ ,

provided the limit of the derivatives exists.

Example: Find $lim_{x \to \infty} \frac{x}{e ^{x}}$

Step 1: Direct substitution → ∞/∞

This is an indeterminate form.

Step 2: Apply L'Hôpital’s Rule:

$lim_{x \to \infty} \frac{1}{e ^{x}}$

Step 3: As $x \to \infty, e^{x} \to \infty$ so:

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

Final Answer: 0

5.0Indeterminate Forms Examples

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

Substitute x = 0:

$\frac{e ^{0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form.

Apply L'Hôpital’s Rule:

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

Final Answer: 1

$lim_{x \to \infty} \frac{l n x}{x}$

Substitute : $x \to \infty : \frac{\infty}{\infty}$

An indeterminate form.

Apply L'Hôpital’s Rule:

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

Final Answer: 0

These examples demonstrate how indeterminate forms occur in limits and how they are resolved using techniques like L'Hôpital's Rule.

6.0Solved Example on Indeterminate Forms

Example 1: Evaluate the limit:

$lim_{x \to 0} \frac{s i n x}{x}$

Solution:

Step 1: Substitute x = 0 → $\frac{s i n 0}{0} = \frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Apply the standard limit result:

$lim_{x \to 0} \frac{s i n x}{x} = 1$

So, the limit is 1.

Example 2: Evaluate: $lim_{x \to 0} \frac{s i n x}{x} = 1$

Solution:Step 1: Substitute x → ∞ → ∞/∞

An indeterminate form of type ∞/∞.

Step 2: Apply L'Hôpital’s Rule:

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

So, the limit is 0.

Example 3: Evaluate the limit:

$lim_{x \to 0} \frac{t a n x - x}{x ^{3}}$

Solution:

Step 1: Direct substitution gives:

$\frac{t a n 0 - 0}{0 ^{3}} = \frac{0 - 0}{0} = \frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Apply the series expansion of tan x near x = 0:

$tan x = x + \frac{x ^{3}}{3} + O (x^{5})$

Step 3: Substitute in the limit:

$lim_{x \to 0} \frac{( x + \frac{x ^{3}}{3} + \dots ) - x}{x ^{3}} = lim_{x \to 0} \frac{\frac{x ^{3}}{3}}{x ^{3}} = \frac{1}{3}$

Answer: 1/3

Example 4: Evaluate:

$lim_{x \to 0} \frac{e ^{s i n x} - 1}{x}$

Solution:

Step 1: Substitute x = 0 → $\frac{e ^{s i n 0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$

Indeterminate form of type 0/0.

Step 2: Apply L'Hôpital’s Rule:

Differentiate numerator and denominator with respect to x:

$lim_{x \to 0} \frac{e ^{s i n x} \cdot c o s x}{1}$

Step 3: Substitute x = 0 → $e^{s i n 0} \cdot cos 0 = e^{0} \cdot 1 = 1$

Answer: 1

Example 5: Evaluate:

$lim_{x \to 0} \frac{l n ( 1 + x ) - s i n x}{x ^{2}}$

Solution:

Step 1: Substitute x = 0 → $\frac{l n ( 1 + 0 ) - s i n 0}{0 ^{2}} = \frac{0 - 0}{0} = \frac{0}{0}$

Indeterminate form of type 0/0.

Step 2: Apply Taylor Series expansions near x = 0:

$ln (1 + x) = x - \frac{x ^{2}}{2} + O (x^{3})$
$sin x = x - \frac{x ^{3}}{6} + O (x^{5})$

Step 3: Substituting the expansions:

$lim_{x \to 0} \frac{( x - \frac{x ^{2}}{2} + \dots ) - ( x - \frac{x ^{3}}{6} + \dots )}{x ^{2}} = lim_{x \to 0} \frac{- \frac{x ^{2}}{2} + \frac{x ^{3}}{6}}{x ^{2}}$

Simplify numerator:

$\frac{- \frac{x ^{2}}{2} + \frac{x ^{3}}{6}}{x ^{2}} = - \frac{1}{2} + \frac{x}{6}$

Step 4: Take limit as x → 0:

$- \frac{1}{2}$

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

Example 6: Evaluate the limit:

$lim_{x \to 0} \frac{t a n x - s i n x}{x ^{3}}$

Solution:

Step 1: Substitute x = 0:

$\frac{t a n 0 - s i n 0}{0 ^{3}} = \frac{0 - 0}{0} = \frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Expand both tan x and sin x using Taylor series near x = 0:

$tan x = x + \frac{x ^{3}}{3} + O (x^{5})$
$tan x = x + \frac{x ^{3}}{6} + O (x^{5})$

Step 3: Substitute expansions:

$lim_{x \to 0} \frac{( x + \frac{x ^{3}}{3} ) - ( x - \frac{x ^{3}}{6} )}{x ^{3}}$

$= lim_{x \to 0} \frac{\frac{x ^{3}}{3} + \frac{x ^{3}}{6}}{x ^{3}}$

Final Answer: 1/2

Example 7: Evaluate:

$lim_{x \to 0} \frac{l n ( 1 + x ^{2} )}{1 - c o s x}$

Solution:

Step 1: Substitute x = 0:

$\frac{l n ( 1 + 0 ^{2} )}{1 - c o s 0} = \frac{l n 1}{1 - 1} = \frac{0}{0}$

This is an indeterminate form of type _0/0.

Step 2: Apply L'Hôpital's Rule:

Differentiate numerator and denominator:

$lim_{x \to 0} \frac{\frac{2 x}{1 + x ^{2}}}{s i n x}$

Step 3: Substitute x = 0:

$\frac{\frac{2 \cdot 0}{1 + 0 ^{2}}}{s i n 0} = \frac{0}{0}$

Still indeterminate, so apply L'Hôpital's Rule again:

Differentiate numerator and denominator again:

Numerator derivative:

$\frac{d}{d x} (\frac{2 x}{1 + x ^{2}}) = \frac{2 ( 1 + x ^{2} ) - 2 x ( 2 x )}{( 1 + x ^{2} ) ^{2}}$

$= \frac{2 ( 1 + x ^{2} ) - 4 x ^{2}}{( 1 + x ^{2} ) ^{2}}$

$= \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2}}$

Denominator derivative: cos x

Step 4: Now evaluate the limit as _{x \to 0}:

$lim_{x \to 0} \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2} \cdot c o s x} = \frac{2 - 0}{1 ^{2} \cdot 1} = 2$

Final Answer: 2

Example 8: Evaluate:

$lim_{x \to 0} \frac{x - s i n x}{x ^{3}}$

Solution:
Step 1: Substitute x = 0:

$\frac{0 - 0}{0 ^{3}} = \frac{0}{0}$

An indeterminate form of type 0/0.

Step 2: Expand sin x in Taylor series near 0:

$sin x = x - \frac{x ^{3}}{6} + O (x^{5})$

Step 3: Substitute the expansion:

$lim_{x \to 0} \frac{x - ( x - \frac{x ^{3}}{6} )}{x ^{3}} = lim_{x \to 0} \frac{\frac{x ^{3}}{6}}{x ^{3}} = \frac{1}{6}$

Final Answer: _1/6

Example 9: How many types of indeterminate forms are there?

Solution: There are 7 standard indeterminate forms in calculus:

0/0
∞/∞
0 × ∞
∞ − ∞
0⁰
∞⁰
$1^{\infty}$

7.0Practice Questions on Indeterminate Forms

Evaluate: $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$
Find: $lim_{x \to 0} \frac{e ^{x} - 1}{x}$
Compute: $lim_{x \to \infty} \frac{l n x}{x}$
Calculate: $lim_{x \to 0^{+}} x^{x}$

Also Read:

Derivative Function Calculus	Limits and Derivatives	Limits and Continuity
First Derivative Test	Application of Derivatives	Continuity in Interval
Limits of Functions	Central Limit Theorem	Partial Derivative

Frequently Asked Questions

What does an indeterminate form mean in limits?

When should I use L'Hôpital's Rule?

Can indeterminate forms be solved without L'Hôpital's Rule?

Frequently Asked Questions

What does an indeterminate form mean in limits?

When should I use L'Hôpital's Rule?

Can indeterminate forms be solved without L'Hôpital's Rule?

Join ALLEN!

Indeterminate Forms

1.0What Are Indeterminate Forms?

2.0All Indeterminate Forms

3.0Indeterminate Forms in Limits

4.0Indeterminate Forms and L'Hôpital's Rule

5.0Indeterminate Forms Examples

6.0Solved Example on Indeterminate Forms

7.0Practice Questions on Indeterminate Forms

Table of Contents

Join ALLEN!

Indeterminate Forms

1.0What Are Indeterminate Forms?

2.0All Indeterminate Forms

3.0Indeterminate Forms in Limits

4.0Indeterminate Forms and L'Hôpital's Rule

5.0Indeterminate Forms Examples

6.0Solved Example on Indeterminate Forms

7.0Practice Questions on Indeterminate Forms

Table of Contents