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

At $x = \pi :$

$\mathop {\lim }\limits_{x \to {\pi ^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to \pi + h\atop\scriptstyle h \to 0} {(\pi + h)^2} - \sin (\pi + h) + 5$

$= \mathop {\lim }\limits_{\scriptstyle x \to \pi + \atop\scriptstyle h \to 0} [({\pi ^2} + {h^2} + 2\pi h) + \sinh + 5] = {\pi ^2} + 5$

$\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to \pi - h\atop\scriptstyle h \to 0} [{(\pi - h)^2} - \sin (\pi - h) + 5]$

$= \mathop {\lim }\limits_{\scriptstyle x \to \pi - h\atop\scriptstyle h \to 0} ({\pi ^2} + {h^2} - 2\pi h) - \sin + 5 = {\pi ^2} + 5$

Also,$f(\pi ) = {\pi ^2} + 5$

Thus, R.H.L. $=$ L.H.L. $=$ $f(\pi ).$
Hence we can say that function is continuous at x $=$ $\pi$.