Examine the following functions for continuity. f( X)= X^ 2 - 25 … - Vidyadip

Question

Examine the following functions for continuity.
$\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$

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Answer

$\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$
For f to be defined,
x + 5 $\neq$ 0 i. e. x $\neq$ -5
$\therefore$ D_f= Set of all real numbers except -5 = R - { -5}
Let$\text{c} \neq -5$ be any real number.
$\therefore$ f(c) = $\frac {\text{c}^2 - 25}{\text{c} + 25} = \frac{(\text{c}-5)(\text{c} + 5)}{\text{c}+5} = \text{c}-5$
Also $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{\text{x}^2-25}{\text{x}+5} =\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{({\text{x} -5)(\text{x} +5)}}{\text{x}+ 5}$
$= ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(x}- 5) = \text{c} - 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x}) = \text {f(c)}$
$\therefore$ f is continuous at x = c.
But $\text{c} \neq -5$ is any real number.
$\therefore$ f is continuous at every real number $\text{c} \in \text{D}_\text{f}.$
$\therefore$ f is continuous function.

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