Question

Is the function defined by $x^2$ - sin x+5 continuous at $x=\pi$ ?

Answer 1

Here, $f\left(x\right)=x^2-sinx+5$

LHL = $\underset{x\to p{i}^{-}}{lim}f\left(x\right)=\underset{x\to p{i}^{-}}{lim}\lambda (x^2-sinx+5)$

Putting $x=\pi -h$ as $x\to {\pi }^{-}$ when $h\to 0$

$\therefore \underset{x\to 0^{-}}{lim}\left(\right(\pi -h{)}^2-sin(\pi -h)+5)$

$\underset{x\to 0^{-}}{lim}\left(\right(\pi -h^2-2\pi h-sinh+5)$

=${\pi }^2$+0+0-sin(0)+5

RHL = $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(x^2-sinx+5)$

Putting $x=\pi -h$ as $x\to a^{+}$ when $h\to 0$

$\therefore \underset{h\to 0}{lim}\left[\right(\pi +h{)}^2-sin(\pi +h)+5]=\underset{h\to 0}{lim}[{\pi }^2+h^2+2\pi h+sinh+5\left)\right]={\pi }^2+5$

Also, f$\left(\pi \right)=({\pi }^2-sin\pi +5={\pi }^2+5)$

$\therefore$ LHL=RHL=f(x), Hence, function is continuous at $x=\pi$.