If the functions f(x), defined below is continuous at x=0, find the value of k.

Solution:

Given: $f(x)=\left\{\begin{array}{c}\frac{1-\cos 2 x}{2 x^{2}}, \quad \mathrm{x}<0 \\ k, \quad x=0 \\ \frac{x}{|x|}, \quad x>0\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{c}\frac{1-\cos 2 x}{2 x^{2}}, \quad \mathrm{x}<0 \\ k, \quad x=0 \\ 1, \quad x>0\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)$

$=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2(-h)}{2(-h)^{2}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2 h}{2 h^{2}}\right)$

$=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{2 \sin ^{2} h}{h^{2}}\right)$

$=\frac{2}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{2} h}{h^{2}}\right)$

$=\frac{2}{2} \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^{2}$

$=1 \times 1$

$=1$

(RHL at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(1)=1$

Also, $f(0)=k$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow 1=1=k$

Hence, the required value of $k$ is $1 .$