Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers

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

share|improve this answer
add comment

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}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.