Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
 A: $${1+\cos\theta\over\tan^2\theta}=\cos^2\theta{1+\cos\theta\over1-\cos^2\theta}=\cos^2\theta{1\over1-\cos\theta}$$
so
$$\lim_{x\to1}{1+\cos\pi x\over\tan^2\pi x}=\lim_{x\to1}{\cos^2\pi x\over1-\cos\pi x}={1\over1+1}={1\over2}$$
A: \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$$\lim_{x\to1}\frac{1+\cos\pi x}{\tan^2\pi x}=\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}\times\lim_{x\to1}\cos^2\pi x=\frac12.$$
A: After factoring out a squared cosine, the limit is the same as
$$\lim_{\alpha\to\pi}\frac{1+\cos\alpha}{\sin^2\alpha}=\lim_{\alpha\to\pi}\frac1{1-\cos\alpha}=\frac12.$$
A: $\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{\sin^2 \pi x}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{1-\cos^2 \pi x}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{(1-\cos \pi x)(1+\cos \pi x)}=$
$=\lim_{x\to 1} \dfrac {\cos^2 \pi x}{1-\cos \pi x}=\dfrac{1}{2}$
A: 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?
A: 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.
