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# Indeterminate Forms

Indeterminate form is an algebraic expression in which even after substitution of limit value does not helps in determining the original value of the expression. The term "Indeterminate" literally means an unknown value.

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## Indeterminate Forms of Limits

In mathematics, some times the given mathematical expression is not determined or not definitively then such type of expression is called indeterminate. Some forms of limits are called indeterminate if the limiting behavior of individual parts of the given expression is not able to determine the over all limit.

Like if we have $\lim_{x \rightarrow 0}f(x) = \lim_{x\rightarrow 0}g(x) = 0$ then $\lim_{x \rightarrow 0}$$\frac{f(x)}{g(x)} becomes the limit of the form of \frac{0}{0} is called the indeterminate form. There are seven Indeterminate forms involving 0, 1, and ∞. They are, \frac{0}{0}$$, 0 \times \infty, $$\frac{\infty}{\infty}$$, \infty, - \infty, 0^{0}, 1^{\infty}$

## List of Indeterminate Forms

Given below are some of the Indeterminate forms

### Indeterminate Form 1

$\frac{0}{0}$

Conditions:

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

Transformation to $\frac{\infty}{\infty}$:

$\lim_{x \to c} $$\frac{f(x)}{g(x)} = \lim_{x \to c}$$ \frac{1/g(x)}{1/f(x)}$

### Indeterminate Form 2

$\frac{\infty}{\infty}$

Conditions:

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

Transformation to $\frac{0}{0}$:

$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} $$\frac{1/g(x)}{1/f(x)} ### Indeterminate Form 3 0 \times \infty Conditions: \lim_{x \to c} f(x) = 0, \lim_{x \to c} g(x) = \infty Transformation to \frac{0}{0}: \lim_{x \to c} f(x)g(x) = \lim_{x \to c}$$\frac{f(x)}{1/g(x)}$

Transformation to $\frac{\infty}{\infty}$:

$\lim_{x \to c} f(x)g(x) = \lim_{x \to c} $$\frac{g(x)}{1/f(x)} ### Indeterminate Form 4 1^{\infty} Conditions: \lim_{x \to c} f(x) = 1, \lim_{x \to c} g(x) = \infty Transformation to \frac{0}{0}: \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c}$$\frac{\ln f(x)}{1/g(x)}$

Transformation to $\frac{\infty}{\infty}$:

$\lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c} $$\frac{g(x)}{1/\ln f(x)} ### Indeterminate Form 5 0^{0} Conditions: \lim_{x \to c} f(x) = 0^+, \lim_{x \to c} g(x) = 0 Transformation to \frac{0}{0}: \lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c}$$\frac{g(x)}{1/\ln f(x)}$

Transformation to $\frac{\infty}{\infty}$:

$\lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c} $$\frac{\ln f(x)}{1/g(x)} ### Indeterminate Form 6 \infty^{0} Conditions: \lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = 0 Transformation to \frac{0}{0}: \lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c}$$\frac{g(x)}{1/\ln f(x)}$

Transformation to $\frac{\infty}{\infty}$:

$\lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c} $$\frac{\ln f(x)}{1/g(x)} ### Indeterminate Form 7 \infty - \infty Conditions: \lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = \infty Transformation to \frac{0}{0}: \lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c}$$\frac{1/g(x) - 1/f(x)}{1/(f(x)g(x))}$

Transformation to $\frac{\infty}{\infty}$:

$\lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} $$\frac{e^{f(x)}}{e^{g(x)}} ## Evaluating Intermediate Forms Back to Top The three main methods of evaluating Indeterminate forms are: • Factoring Method(\frac{0}{0} form) • Division of each term by highest power of variable ( \frac{ \infty }{ \infty } form ) • L' Hospital Rule ( \frac{0}{0} or \frac{ \infty }{ \infty }) ### Factoring Method: In this method, the expressions are factorized to their maximum simplest form and then, the limit value is substituted. ### Division of Each Term by Highest Power of Variable: In this method, each term in numerator and denominator is divided by the variable of highest power in the expression and then, the limit value is obtained. ### L' Hospital Rule: In this method, the derivative of each term is taken in each step successively until atleast one of term becomes free of variable that is atleast one term becomes a constant. ### Solved Examples Question 1: Evaluate \lim_{x \to \infty}$$\frac{\sin (2x)}{e^{x} + x}$
Solution:
$\lim_{x \to \infty} $$\frac{\sin (2x)}{e^{x} + x} Let f(x) = \sin (2x) and g(x) = e^{x} + x f'(x) = \cos (2x) and g'(x) = e^{x} + 1 So, \lim_{x \to 0}$$\frac{f'(x)}{g'(x)}$ = $\lim_{x \to 0} $$\frac{2 \cos (2x)}{e^{x} + 1} = \frac{2 \cos (0)}{e^{x} + 1} = \frac{2}{2} = 1 Question 2: Evaluate \lim_{x \to 2}$$\frac{x^{3} - 8}{x - 2}$
Solution:
$\lim_{x \to 2} $$\frac{x^{3} - 8}{x - 2} Formula: a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) So, x^{3} - 8 = x^{3} - 2^{3} = (x - 2)(x^{2} + 2x + 4) = \lim_{x \to 2}$$\frac{(x - 2)(x^{2} + 2x + 4)}{x - 2}$
= $\lim_{x \to 2} (x^{2} + 2x + 4)$
= $2^{2} + 2(2) + 4$
= $12$

Question 3: Evaluate $\lim_{x \to \infty} $$\frac{4x^{3} + 3x^{2} + 5x - 2}{2x^{3} + 7x + 1} Solution: \lim_{x \to \infty}$$\frac{4x^{3} + 3x^{2} + 5x - 2}{2x^{3} + 7x + 1}$

Here, the variable with highest power is x3. So, each term is divided by x3.
$\lim_{x \to \infty} $$\frac{4x^{3} + 3x^{2} + 5x - 2}{2x^{3} + 7x + 1} = \lim_{x \to \infty}$$\frac{4 + \frac{3}{x} + \frac{5}{x^{2}} - \frac{2}{x^{3}}}{2 + \frac{7}{x^{2}} + \frac{1}{x^{3}}}$
= $\lim_{x \to \infty}$$(\frac{4 + 0 + 0}{2 + 0 + 0})$
= $\frac{4}{2}$
= $2$

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