<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] $${1+\cos\theta\over\tan^2\theta}=\cos^2\theta{1+\cos\theta\over1-\cos^2\theta}=\cos^2\theta{1\over1-\cos\theta}$$
so
$$\lim_{x\to1}{1+\cos\pi x\over\tan^2\pi x}=\lim_{x\to1}{\cos^2\pi x\over1-\cos\pi x}={1\over1+1}={1\over2}$$
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