Is the function defined by

The given function is $f(x)=x^{2}-\sin x+5$

It is evident that $f$ is defined at $x=\pi$.

At $x=\pi, f(x)=f(\pi)=\pi^{2}-\sin \pi+5=\pi^{2}-0+5=\pi^{2}+5$

Consider $\lim f(x)=\lim \left(x^{2}-\sin x+5\right)$

Put $x=\pi+h$

If $x \rightarrow \pi$, then it is evident that $h \rightarrow 0$

$\begin{aligned} \therefore \lim _{x \rightarrow x} f(x) &=\lim _{x \rightarrow x}\left(x^{2}-\sin x+5\right) \\ &=\lim _{h \rightarrow 0}\left[(\pi+h)^{2}-\sin (\pi+h)+5\right] \\ &=\lim _{h \rightarrow 0}(\pi+h)^{2}-\lim _{h \rightarrow 0} \sin (\pi+h)+\lim _{h \rightarrow 0} 5 \\ &=(\pi+0)^{2}-\lim _{h \rightarrow 0}[\sin \pi \cosh +\cos \pi \sinh ]+5 \\ &=\pi^{2}-\lim _{h \rightarrow 0} \sin \pi \cosh -\lim _{h \rightarrow 0} \cos \pi \sinh +5 \\ &=\pi^{2}-\sin \pi \cos 0-\cos \pi \sin 0+5 \\ &=\pi^{2}-0 \times 1-(-1) \times 0+5 \\ &=\pi^{2}+5 \end{aligned}$

$\therefore \lim _{x \rightarrow \pi} f(x)=f(\pi)$

Therefore, the given function $f$ is continuous at $x=\pi$