$f(x)=\left\{\begin{array}{rr}\frac{1+3 x^2-\cos 2 x}{x^2}, & \text { for } x \neq 0 \\ k, & \text { for } x=0\end{array}\right.$
RHL
$f(0+h)=\lim _{h \rightarrow 0} \frac{1+3(0+h)^2-\cos 2(0+h)}{(0+h)^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-\cos 2 h}{h^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-\left(1-2 \sin ^2 h\right)}{h^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-1+2 \sin ^2 h}{h^2}$
$=\lim _{h \rightarrow 0}\left\{3+2\left(\frac{\sin ^2 h}{h^2}\right)\right\}$
$=3+2 \cdot \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2$
$=3+2 \cdot(1)^2 \quad\left\{\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right\}$
$=3+2=5$
LHL
$f(0-h)=\lim _{h \rightarrow 0} \frac{1+3(0-h)^2-\cos 2(0-h)}{(0-h)^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-\cos 2 h}{h^2}$
$=5$
Since, the function is continuous at $x=0$, then LHL $=$ RHL $=f(0) \because f(0)=k \Rightarrow k=5$