# Indeterminate Forms

The term “indeterminate” means an unknown value. The indeterminate form is a Mathematical expression that we cannot be able to determine the original value even after the substitution of the limits. In this article, we are going to discuss what is the indeterminate form of limits, different types of indeterminate forms in algebraic expressions with examples.

## What is Indeterminate Form?

In Mathematics, we cannot be able to find solutions for some form of Mathematical expressions. Such expressions are called indeterminate forms. In most of the cases, the indeterminate form occurs while taking the ratio of two functions, such that both of the function approaches to zero in the limit. Such cases are called “indeterminate form 0/0”. Similarly, the indeterminant form can be obtained in the addition, subtraction, multiplication, exponential operations also.

## Indeterminate Forms of Limits

Some forms of limits are called indeterminate if the limiting behaviour of individual parts of the given expression is not able to determine the overall limit.

If we have the limits like, $\lim_{x\rightarrow 0}f(x) = \lim_{x\rightarrow 0}g(x) = 0,$, then $\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}$.

If the limits are applied for the given function, then it becomes 0/0, which is known as indeterminate forms.

In Mathematics, there are seven indeterminate forms that include 0, 1 and ∞, They are

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

## Indeterminate Forms List

Some of the indeterminate forms with conditions and transformation are given below:

 Indeterminate form Conditions 0/0 $\lim_{x\rightarrow c}f(x) = 0, \lim_{x\rightarrow c}g(x) = 0$ ∞/∞ $\lim_{x\rightarrow c}f(x)= \infty , \lim_{x\rightarrow c }g(x) = \infty$ 0.∞ $\lim_{x\rightarrow c}f(x)= 0 , \lim_{x\rightarrow c }g(x) = \infty$ ∞-∞ $\lim_{x\rightarrow c}f(x)= 1 , \lim_{x\rightarrow c }g(x) = \infty$ 00 $\lim_{x\rightarrow c}f(x)=0^{+} , \lim_{x\rightarrow c }g(x) = 0$ 1∞ $\lim_{x\rightarrow c}f(x)= \infty , \lim_{x\rightarrow c }g(x) = \infty$ ∞0 $\lim_{x\rightarrow c}f(x)= \infty , \lim_{x\rightarrow c }g(x) = 0$

Now let us discuss these forms one by one.

### Indeterminate Form 1

0/0

Condition:

$\lim_{x\rightarrow c}f(x) = 0, \lim_{x\rightarrow c}g(x) = 0$

Transformation:

Transformation to ∞/∞.

Then it becomes, $\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\frac{1/g(x)}{1/f(x)}$

### Indeterminate Form 2

∞/∞

Condition:

$\lim_{x\rightarrow c}f(x)= \infty , \lim_{x\rightarrow c }g(x) = \infty$

Transformation:

Transformation to 0/0

Then it becomes, $\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\frac{1/g(x)}{1/f(x)}$

### Indeterminate Form 3

0 x ∞

Condition:

$\lim_{x\rightarrow c}f(x)= 0 , \lim_{x\rightarrow c }g(x) = \infty$

Transformation:

Transformation to 0/0

Then it becomes, $\lim_{x\rightarrow c}f(x)g(x)= \lim_{x\rightarrow c }\frac{f(x)}{1/g(x)}$

Transformation to ∞/∞.

It becomes, $\lim_{x\rightarrow c}f(x)g(x)= \lim_{x\rightarrow c }\frac{g(x)}{1/f(x)}$

### Indeterminate Form 4

$1^{\infty }$

Condition:

$\lim_{x\rightarrow c}f(x)= 1 , \lim_{x\rightarrow c }g(x) = \infty$

Transformation:

Transformation to 0/0

Then it becomes,$\lim_{x\rightarrow c}f(x)^{g(x)}=exp \lim_{x\rightarrow c }\frac{ln f(x)}{1/g(x)}$

Transformation to ∞/∞.

It becomes,$\lim_{x\rightarrow c}f(x)^{g(x)}= \lim_{x\rightarrow c }\frac{g(x)}{1/ ln f(x)}$

### Indeterminate Form 5

00

Condition:

$\lim_{x\rightarrow c}f(x)=0^{+} , \lim_{x\rightarrow c }g(x) = 0$

Transformation:

Transformation to 0/0

Then it becomes,$\lim_{x\rightarrow c}f(x)^{g(x)}=\lim_{x\rightarrow c }\frac{g(x)}{1/ ln f(x)}$

Transformation to ∞/∞.

It becomes,$\lim_{x\rightarrow c}f(x)^{g(x)}= \lim_{x\rightarrow c }\frac{ ln f(x)}{1/ g(x)}$

### Indeterminate Form 6

0

Condition:

$\lim_{x\rightarrow c}f(x)= \infty , \lim_{x\rightarrow c }g(x) = 0$

Transformation:

Transformation to 0/0

Then it becomes,$\lim_{x\rightarrow c}f(x)^{g(x)}=\lim_{x\rightarrow c }\frac{g(x)}{1/ ln f(x)}$

Transformation to ∞/∞.

It becomes,$\lim_{x\rightarrow c}f(x)^{g(x)}= \lim_{x\rightarrow c }\frac{ ln f(x)}{1/ g(x)}$

### Indeterminate Form 7

∞ – ∞

Condition:

$\lim_{x\rightarrow c}f(x)= \infty , \lim_{x\rightarrow c }g(x) = \infty$

Transformation:

Transformation to 0/0

Then it becomes,$\lim_{x\rightarrow c}(f(x)-g(x))= \lim_{x\rightarrow c }\frac{[1/g(x)]-[1/f(x)]}{1/[f(x)g(x)]}$

Transformation to ∞/∞.

It becomes,$\lim_{x\rightarrow c}(f(x)-g(x))= \lim_{x\rightarrow c }\frac{e^{f(x)}}{e^{g(x)}}$

## How to Evaluate Indeterminate Forms?

There are three methods used to evaluate indeterminate forms. They are:

Factoring Method (0/0 form)

In the factoring method, the expressions are factorized to their maximum simplest form. After that, the limit value should be substituted.

L Hospital’s Rule (0/0 or ∞/∞ form)

In this method, the derivative of each term is taken in each step successively until at least one of the terms becomes free of the variable. It means that at least one term becomes constant.

Division of Each Term by Highest Power of Variable (∞/∞ form)

In this method, each term in numerator and denominator is divided by the variable of the highest power in the expression, and then, the limit value is obtained.

## Indeterminate Forms Example

Question: Evaluate $\lim_{x\rightarrow \infty }\frac{sin 2x}{e^{x}+ x}$

Solution:

Given: $\lim_{x\rightarrow \infty }\frac{sin 2x}{e^{x}+ x}$

Let f(x) = sin 2(x) and g(x) = ex. + x

Therefore, f’(x) = 2 cos 2x , g’(x) = ex + 1

Therefore, $\lim_{x\rightarrow 0 }\frac{f'(x)}{g'(x)}= \lim_{x\rightarrow 0}\frac{2 cos 2x}{e^{x}+1}$

Now, substitute the limits, it becomes

= 2 cos(0)/ e0 + 1

= 2/ 2 = 1

Therefore, $\lim_{x\rightarrow \infty }\frac{sin 2x}{e^{x}+ x} = 1$

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