<s>[INST] Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
 [/INST] Set $\pi(1-x)=2y$ 
and use $\cos(\pi-A)=-\cos A,\tan(\pi-B)=-\tan B$
and $\cos2y=1-2\sin^2y,\sin2y=2\sin y\cos y$
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}=\lim_{y\to0}\dfrac{1-\cos2y}{\tan^22y}$$
$$=\lim_{y\to0}\dfrac{2\sin^2y}{(2\sin y\cos y)^2}\cdot\lim_{y\to0}\cos^22y=?$$
Can you take it from here?
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