Question

${lim}_{x\to 1}\frac{1+cos\pi x}{(1-x{)}^2}$

Answer 1

${lim}_{x\to 1}\frac{1+cos\pi x}{(1-x{)}^2}$

$\Rightarrow x\to 1,x-1\to 0,$ let $x-1=y$

as $x\to 1,y\to 0$

$={lim}_{y\to 0}\frac{1+cos\pi (y+1)}{(-y{)}^2}$

$={lim}_{y\to 0}\frac{1+cos(\pi +\pi y)}{y^2}$

$={lim}_{y\to 0}\frac{1-cos\left(\pi y\right)}{y^2}$

$={lim}_{y\to 0}\frac{2si{n}^2\frac{\pi y}{2}}{y^2}$

$=2{({lim}_{y\to 0}.\frac{sin\frac{\pi y}{2}}{\frac{\pi y}{2}})}^2\times \frac{{\pi }^2}{4}$

$=2\times 1\times \frac{{\pi }^2}{4}=\frac{{\pi }^2}{2}$