Question

iscuss the continuty of when f(x) = {X ^2-25/x-5 when xis not equal to 5 and 9 when x=5
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{ \dfrac{ {x}^{2} - 25}{x - 5} \: when \: x \ne \: 5} \\ \\ &\sf{9 \: \: when \: x \: = 5} \end{cases}\end{gathered}\end{gathered}$

We know,

By definition of Continuity of a function

A function f(x) is said to be continuous at x = a iff

$\rm :\longmapsto\:\boxed{\tt{ f(5) = \displaystyle\lim_{x \to 5}\rm f(x) \: }}$

So, from the given function we have

$\rm :\longmapsto\:\boxed{\tt{ f(5) = 9}} - - - - (1)$

Now, Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 5}\rm f(x)$

$\rm \:  =  \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - 25}{x - 5}$

$\rm \:  =  \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - {5}^{2} }{x - 5}$

$\rm \:  =  \: \displaystyle\lim_{x \to 5}\rm \frac{ \cancel{(x - 5)} \: (x + 5)}{ \cancel{x - 5}}$

$\rm \:  =  \:\displaystyle\lim_{x \to 5}\rm (x + 5)$

$\rm \:  =  \: 5 + 5$

$\rm \:  =  \: 10$

$\rm\implies \:\boxed{\tt{ \displaystyle\lim_{x \to 5}\rm f(x) = 10}} - - - - (2)$

So, from equation (1) and (2), we concluded that

$\rm :\longmapsto\:\boxed{\tt{ f(5) \: \ne \: \displaystyle\lim_{x \to 5}\rm f(5) \: }}$

$\rm\implies \:f(x) \: is \: not \: continuous \: at \: x = 5$

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MORE TO LEARN

If f and g are two continuous function then

• f + g is also continuous.

• f - g is also continuous

• f × g is also continuous

• f[g(x)] is also continuous

• g[f(x)] is also continuous