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

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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.

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