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

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