A limit question

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

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You should use some paratheses to structure your formula. Does it read $\frac{\cos(\pi x)/2}{1 - \sqrt x}$ or $\frac{\cos \frac{\pi x}2}{1 - \sqrt x}$? –  martini Nov 10 '12 at 13:49
It looks like the second one. btw it is pi*x not pi*z –  nicegirl Nov 10 '12 at 13:51
Using L'Hôpital's rule: your expression when x->1 equals $\frac{-\sin(\frac{\pi x}{2})\frac{\pi}{2}}{-\frac{1}{2}*\frac{1}{\sqrt x}} = \pi$
Without De L'Hospital. First note that, $$\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-\sqrt{x}}= \lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-x}(1+\sqrt{x})$$ To calculate the limit of the fraction take $u=\frac{\pi (1-x)}{2}$. Then $\lim_{x\to 1}u=0$ and so $$\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-x}= \lim_{u\to 0}\frac{\cos(\frac{\pi}{2}-u)}{\frac{2u}{\pi}}=\frac{\pi}{2}\lim_{u\to 0}\frac{\sin u}{u}=\frac{\pi}{2}$$ Thus, $$\lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-\sqrt{x}}= \lim_{x\to 1}\frac{\cos(\frac{\pi x}{2})}{1-x}(1+\sqrt{x})$$=\frac{\pi}{2}2=\pi