Question
For what value of $k,$ the function given below is continuous at $x=0$ ? $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{4+x}-2}{x} & , x \neq 0 \\ k & , x=0\end{array}\right.$
✓
Answer
As, $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{4+x-2}}{x}, & x \neq 0 \\ k, & x=0\end{array}\right.$ is continuous at $x=0$
$ \Rightarrow \ce{LHL=RHL}=f(0) $ or $ \lim _{x \rightarrow 0} f(x)=f(0)$
$ \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}=k$
$ \Rightarrow \lim _{x \rightarrow 0} \frac{4+x-4}{x(\sqrt{4+x}+2)}=k $
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{1}{(\sqrt{4+x}+2)}$
$ \therefore k=\frac{1}{4}$
$ \Rightarrow \ce{LHL=RHL}=f(0) $ or $ \lim _{x \rightarrow 0} f(x)=f(0)$
$ \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}=k$
$ \Rightarrow \lim _{x \rightarrow 0} \frac{4+x-4}{x(\sqrt{4+x}+2)}=k $
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{1}{(\sqrt{4+x}+2)}$
$ \therefore k=\frac{1}{4}$
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