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:

1. 0/0
2. ∞/∞
3. 0 × ∞
4. ∞ − ∞
5. 0⁰
6. ∞⁰
7. 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:

${\lim }_{x\to 0}\frac{\sin x}{x}$

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:

${\lim }_{x\to 0}\frac{1-\cos x}{x^2}$

Direct substitution:

$\frac{1-\cos 0}{0^2}=\frac{1-1}{0}=\frac{0}{0}$

This is an indeterminate form.

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

${\lim }_{x\to 0}\frac{\sin x}{2x}$

Substitute x = 0 → $\frac{\sin 0}{2\times 0}=\frac{0}{0}$

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

${\lim }_{x\to 0}\frac{\cos x}{2}$

Substitute x = 0 →

$\frac{\cos 0}{2}=\frac{1}{2}$

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 

${\lim }_{x\to a}\frac{f(x)}{g(x)}={\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$,

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

1. ${\lim }_{x\to 0}\frac{e^x-1}{x}$

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

2. ${\lim }_{x\to \infty }\frac{\ln x}{x}$

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:

${\lim }_{x\to 0}\frac{\sin x}{x}$

Solution:

Step 1: Substitute x = 0 →$\frac{\sin 0}{0}=\frac{0}{0}$

This is an indeterminate form of type 0/0.

Step 2: Apply the standard limit result:

${\lim }_{x\to 0}\frac{\sin x}{x}=1$

So, the limit is 1.

Example 2: Evaluate: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$

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:

${\lim }_{x\to 0}\frac{\tan x-x}{x^3}$

Solution:

Step 1: Direct substitution gives:

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

This is an indeterminate form of type 0/0.

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

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

Step 3: Substitute in the limit:

${\lim }_{x\to 0}\frac{\left(x+\frac{x^3}{3}+\dots \right)-x}{x^3}={\lim }_{x\to 0}\frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$

Answer: 1/3

 Example 4: Evaluate:

${\lim }_{x\to 0}\frac{e^{\sin x}-1}{x}$

Solution:

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

Indeterminate form of type 0/0.

Step 2: Apply L'Hôpital’s Rule: 

Differentiate numerator and denominator with respect to x:

${\lim }_{x\to 0}\frac{e^{\sin x}\cdot \cos x}{1}$

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

Answer: 1

Example 5: Evaluate:

${\lim }_{x\to 0}\frac{\ln (1+x)-\sin x}{x^2}$

Solution:

Step 1: Substitute x = 0 → $\frac{\ln (1+0)-\sin 0}{0^2}=\frac{0-0}{0}=\frac{0}{0}$

Indeterminate form of type 0/0.

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

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

Step 3: Substituting the expansions:

${\lim }_{x\to 0}\frac{\left(x-\frac{x^2}{2}+\dots \right)-\left(x-\frac{x^3}{6}+\dots \right)}{x^2}={\lim }_{x\to 0}\frac{-\frac{x^2}{2}+\frac{x^3}{6}}{x^2}$

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:

${\lim }_{x\to 0}\frac{\tan x-\sin x}{x^3}$

Solution:

Step 1: Substitute x = 0:

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

This is an indeterminate form of type 0/0.

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

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

Step 3: Substitute expansions:

${\lim }_{x\to 0}\frac{\left(x+\frac{x^3}{3}\right)-\left(x-\frac{x^3}{6}\right)}{x^3}$

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

Final Answer: 1/2

Example 7: Evaluate:

${\lim }_{x\to 0}\frac{\ln (1+x^2)}{1-\cos x}$

Solution:

Step 1: Substitute x = 0:

$\frac{\ln (1+0^2)}{1-\cos 0}=\frac{\ln 1}{1-1}=\frac{0}{0}$

This is an indeterminate form of type _0/0.

Step 2: Apply L'Hôpital's Rule:

Differentiate numerator and denominator:

${\lim }_{x\to 0}\frac{\frac{2x}{1+x^2}}{\sin x}$

Step 3: Substitute x = 0:

$\frac{\frac{2\cdot 0}{1+0^2}}{\sin 0}=\frac{0}{0}$

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

Differentiate numerator and denominator again:

• Numerator derivative:

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

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

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

• Denominator derivative: cos x

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

${\lim }_{x\to 0}\frac{2-2x^2}{(1+x^2{)}^2\cdot \cos x}=\frac{2-0}{1^2\cdot 1}=2$

Final Answer: 2

Example 8: Evaluate: 

${\lim }_{x\to 0}\frac{x-\sin x}{x^3}$

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:

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

Step 3: Substitute the expansion:

${\lim }_{x\to 0}\frac{x-\left(x-\frac{x^3}{6}\right)}{x^3}={\lim }_{x\to 0}\frac{\frac{x^3}{6}}{x^3}=\frac{1}{6}$

Final Answer: _1/6

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

Solution: There are 7 standard indeterminate forms in calculus:

1. 0/0
2. ∞/∞
3. 0 × ∞
4. ∞ − ∞
5. 0⁰
6. ∞⁰
7. $1^{\infty }$

7.0Practice Questions on Indeterminate Forms

1. Evaluate: ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}$
2. Find: ${\lim }_{x\to 0}\frac{e^x-1}{x}$
3. Compute: ${\lim }_{x\to \infty }\frac{\ln x}{x}$
4. Calculate: ${\lim }_{x\to 0^{+}}{x}^x$

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