<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] Make the problem simpler using $x=y+1$ and later $\pi y=z$ $$\lim_{x\to 1} \dfrac {1+\cos (\pi x)}{\tan^2 (\pi x)}=\lim_{y\to 0} \dfrac {1-\cos (\pi y)}{\tan^2 (\pi y)}=\lim_{z\to 0} \dfrac {1-\cos (z)}{\tan^2 (z)}$$ and consider either equivalents $$\cos(z)\sim 1-\frac{z^2}2 \qquad \text{and}\qquad\tan(z)\sim z \implies \tan^2(z)\sim z^2$$ to conclude.
If you want more than the limit itself, use Taylor series
$$\cos(z)=1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^6\right)$$
$$\tan(z)=z+\frac{z^3}{3}+O\left(z^5\right)$$
$$\tan^2(z)=z^2+\frac{2 z^4}{3}+O\left(z^6\right)$$ which make 
$$\dfrac {1-\cos (z)}{\tan^2 (z)}=\dfrac {\frac{z^2}{2}-\frac{z^4}{24}+O\left(z^6\right) }{z^2+\frac{2 z^4}{3}+O\left(z^6\right) }=\frac{1}{2}-\frac{3 z^2}{8}+O\left(z^4\right)$$ which shows the limit and how it is approached.
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