A limit question

Question

Find the limit \begin{equation}\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-\sqrt{x}} \end{equation}

Answer 1

Using L'Hôpital's rule: your expression when x->1 equals $\frac{-\sin(\frac{\pi x}{2})\frac{\pi}{2}}{-\frac{1}{2}*\frac{1}{\sqrt x}} = \pi$

Answer 2

Without De L'Hospital. First note that, \begin{equation}\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-\sqrt{x}}= \lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-x}(1+\sqrt{x})\end{equation} To calculate the limit of the fraction take $u=\frac{\pi (1-x)}{2}$. Then $\lim_{x\to 1}u=0$ and so \begin{equation}\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-x}= \lim_{u\to 0}\frac{\cos(\frac{\pi}{2}-u)}{\frac{2u}{\pi}}=\frac{\pi}{2}\lim_{u\to 0}\frac{\sin u}{u}=\frac{\pi}{2} \end{equation} Thus, \begin{equation}\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-\sqrt{x}}= \lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-x}(1+\sqrt{x})=\frac{\pi}{2}2=\pi\end{equation}