Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$

\begin{align*}\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}&=\lim_{x\to1}\frac{-\pi\sin\pi x}{2\pi\sin\pi x\cos\pi x}\\&=-\lim_{x\to1}\frac1{2\cos\pi x}\\&=\frac12\end{align*}and therefore$$\lim_{x\to1}\frac{1+\cos\pi x}{\tan^2\pi x}=\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}\times\lim_{x\to1}\cos^2\pi x=\frac12.$$
