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Is the function defined by $f(x)=x^2-\sin x+5$ continuous at $x=\pi$ ?
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Verified Answer
Let $f(x)=x^2-\sin x+5$
At $x=\pi$, L.H.L. $=\lim _{x \rightarrow \pi^{-}}\left(x^2-\sin x+5\right)$, Put $x=\pi-h$,
$\therefore \quad \text { L.H.L. }=\lim _{\mathrm{h} \rightarrow 0}\left[(\pi-\mathrm{h})^2-\sin (\pi-\mathrm{h})+5\right]$
$=\lim _{h \rightarrow 0}\left[\pi^2-2 \pi h+h^2-\sin h+5\right]$
$=\pi^2+5$
R.H.L. $=\lim _{x \rightarrow \pi^{+}}\left(x^2-\sin x+5\right)$, Put $x=\pi+h$,
$\therefore$ L.H.L. $=\lim _{h \rightarrow 0}\left[(\pi+h)^2-\sin (\pi+h)+5\right]$
$=\lim _{\mathrm{h} \rightarrow 0}\left[\pi^2+2 \pi \mathrm{h}+\mathrm{h}^2+\sin \mathrm{h}+5\right]$
$=\pi^2+5$
$f(\pi)=\pi^2+5, \quad \therefore$ L.H.L. $=$ R.H.L. $=f(\pi)$
Hence, $f$ is continuous at $x=\pi$
At $x=\pi$, L.H.L. $=\lim _{x \rightarrow \pi^{-}}\left(x^2-\sin x+5\right)$, Put $x=\pi-h$,
$\therefore \quad \text { L.H.L. }=\lim _{\mathrm{h} \rightarrow 0}\left[(\pi-\mathrm{h})^2-\sin (\pi-\mathrm{h})+5\right]$
$=\lim _{h \rightarrow 0}\left[\pi^2-2 \pi h+h^2-\sin h+5\right]$
$=\pi^2+5$
R.H.L. $=\lim _{x \rightarrow \pi^{+}}\left(x^2-\sin x+5\right)$, Put $x=\pi+h$,
$\therefore$ L.H.L. $=\lim _{h \rightarrow 0}\left[(\pi+h)^2-\sin (\pi+h)+5\right]$
$=\lim _{\mathrm{h} \rightarrow 0}\left[\pi^2+2 \pi \mathrm{h}+\mathrm{h}^2+\sin \mathrm{h}+5\right]$
$=\pi^2+5$
$f(\pi)=\pi^2+5, \quad \therefore$ L.H.L. $=$ R.H.L. $=f(\pi)$
Hence, $f$ is continuous at $x=\pi$
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