Is the function defined by $f(x)=x^{2}-\sin x+5$ continuous at $x=\pi ?$
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 $x=\pi$
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