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

(A) The given function is $f(x) = x^{2} - \sin x + 5$.
First,we evaluate the function at $x = \pi$:
$f(\pi) = \pi^{2} - \sin \pi + 5 = \pi^{2} - 0 + 5 = \pi^{2} + 5$.
Next,we calculate the limit of the function as $x \to \pi$:
$\lim_{x \to \pi} f(x) = \lim_{x \to \pi} (x^{2} - \sin x + 5)$.
Using the property of limits for sums and differences:
$\lim_{x \to \pi} f(x) = \lim_{x \to \pi} x^{2} - \lim_{x \to \pi} \sin x + \lim_{x \to \pi} 5$.
Substituting $x = \pi$:
$\lim_{x \to \pi} f(x) = \pi^{2} - \sin \pi + 5 = \pi^{2} - 0 + 5 = \pi^{2} + 5$.
Since $\lim_{x \to \pi} f(x) = f(\pi) = \pi^{2} + 5$,the function $f(x)$ is continuous at $x = \pi$.