Solution:
The provided function is $f(x)=x^{2}-\sin x+5$
Left Hand Limit $=\lim _{x \rightarrow z^{-}}\left(x^{2}-\sin x+5\right)=\lim _{x \rightarrow n^{-}}\left[(\pi-h)^{2}-\sin (\pi-h)+5\right]=\pi^{2}+5$
Right Hand Limit $=\lim _{x \rightarrow-^{-}}\left(x^{2}-\sin x+5\right)=\lim _{x \rightarrow \pi^{-}}\left[(\pi+h)^{2}-\sin (\pi+h)+5\right]=\pi^{2}+5$
And $f(\pi)=\pi^{2}-\sin \pi+5=\pi^{2}+5$
As L.H.L. $=$ R.H.L. $=f(\pi)$
As a result, $f$ is continuous at $x=\pi$