Given information
We are given with the limit \[\mathop {\lim }\limits_{x \to \pi } \sin \left( {x + \sin x} \right)\]We have to evaluate the function using continuity

Step 1:
Let \[\begin{array}{l}f\left( x \right) = \sin \left( {x + \sin x} \right)\\h\left( x \right) = \sin x\\g\left( x \right) = x + \sin x\end{array}\]

Then using composite formula
\[\begin{array}{l}f\left( x \right) = \sin \left( {g\left( x \right)} \right)\\f\left( x \right) = h\left( {g\left( x \right)} \right)\\f\left( x \right) = \left( {h{\rm{o}}g} \right)x\end{array}\]Therefore, \[f\left( x \right) = \left( {h{\rm{o}}g} \right)x\]

Step 2:
If a function $g$ is continuous at $a$ and the function $h$ is continuous at $g(a)$ then the composite function $\left( {h{\rm{o}}g} \right)x$ is continuous at $x=a$.

Continuity of g(x) at $\pi$\[\begin{array}{l}\mathop {\lim }\limits_{x \to \pi } g\left( x \right) = \mathop {\lim }\limits_{x \to \pi } \left( {x + \sin x} \right)\\ = \pi + \sin \pi \\ = \pi \\ = g\left( \pi \right)\end{array}\]Therefore, g(x) is continuous at $x=\pi$

Step 3:
Continuity of h(x) at $g(\pi)$\[\begin{array}{l}\mathop {\lim }\limits_{x \to \pi } h\left( x \right) = \mathop {\lim }\limits_{x \to \pi } \left( {\sin x} \right)\\ = \sin \pi \\ = 0\end{array}\]Therefore, h(x) is continuous at $x=\pi$

Step 4:
So by continuity theorem, the composite function $f\left( x \right) = \left( {hog} \right)x$ is continuous at $x=\pi$

\[\begin{array}{l}\mathop {\lim }\limits_{x \to \pi } f\left( x \right) = f\left( \pi \right)\\ = \sin \left( {\pi + \sin \pi } \right)\\ = \sin \left( {\pi + 0} \right)\\ = 0\end{array}\]Therefore, the limit is: \[\mathop {\lim }\limits_{x \to \pi } \sin \left( {x + \sin x} \right) = 0\]