\( \lim\limits_{x\to\pi} \dfrac{\cos{x}+\sin{(2x)}+1}{x^2-\pi^2} \)

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Substituting \(x \to \pi\) results in the indeterminate form \(\dfrac{0}{0}\).

$$ \frac{\cos(\pi)+\sin(2\pi)+1}{\pi^2-\pi^2} $$ $$ =\frac{-1+0+1}{0} $$ $$ =\frac{0}{0} $$

Use L'Hospital's Rule.

$$ \lim\limits_{x\to\pi} \dfrac{\cos{x}+\sin{(2x)}+1}{x^2-\pi^2} = \lim\limits_{x\to\pi} \dfrac{\frac{d}{dx}[\cos{x}+\sin{(2x)}+1]}{\frac{d}{dx}[x^2-\pi^2]} $$ $$ = \lim\limits_{x\to\pi} \frac{-\sin{x}+2\cos{(2x)}+1}{2x} $$ $$ = \frac{-\sin(\pi)+2\cos{2\pi}}{2\pi} $$ $$ = \frac{0+2}{2\pi} $$ $$ = \boxed{\frac{1}{\pi}} $$