# Solve the following

Question:

$f(x)=\left\{\begin{array}{ll}\frac{1-\cos 2 x}{x^{2}} & \text { if } x \neq 0 \\ 5, & \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$

Solution:

Checking the right hand and left hand limits of the given function, we have

$\lim _{x \rightarrow 0^{-}} f(x)=\frac{1-\cos 2 x}{x^{2}}$

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

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

$=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} h}{h^{2}} \quad\left[\because 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}\right]$

$\left.=\lim _{h \rightarrow 0} \frac{2 \sin h}{h} \cdot \frac{\sin h}{h}=2.1 .1=2 \quad \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$\lim _{x \rightarrow 0^{+}} f(x)=\frac{1-\cos 2 x}{x^{2}}$

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

$=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} h}{h^{2}}=\frac{2 \sin h}{h} \cdot \frac{\sin h}{h}=2.1 .1=2$

$\lim _{x \rightarrow 0} f(x)=5$

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

Therefore, the given function f(x) is discontinuous at x = 0.