Evaluate the limit $\lim \limits_{x \to 2} \frac{\cos\frac{\pi}{x}}{x-2}$

I was asked to determine the limit: $$\lim \limits_{x \to 2} \frac{\cos\frac{\pi}{x}}{x-2}$$ without using L'Hopital's rule. Also, I was given the hint: let $t = \frac{\pi}{2} - \frac{\pi}{x}$. However, I'm not sure how to begin. A little assistance would be appreciated.

• – lab bhattacharjee Feb 22 '17 at 16:17
• Did you get something from an answer below? – Did Feb 28 '17 at 16:06

We have $$L = \lim_{x \to 2} \frac{\cos \frac{\pi}{x}}{x-2} = \lim_{x \to 2} \frac{\sin(\frac{\pi}{2}-\frac{\pi}{x})}{\frac{\pi}{2}-\frac{\pi}{x}}\times \frac{\frac{\pi}{2}-\frac{\pi}{x}}{x-2}$$ We let $\displaystyle t = \frac{\pi}{2}-\frac{\pi}{x}$. Then, we get, $$L =\lim_{t \to 0} \frac{\sin t}{t}\times \frac{t}{\displaystyle \frac{t}{\frac{\pi}{2}-t}-2}$$ $$\Rightarrow L = ?$$

Hope you can take it from here.

take $\phi(x)= \cos(\frac{\pi}{x})$ then, since $\cos(\frac{\pi}{2})=0$ $$L= \lim_{x\to 2}\frac{\cos(\frac{\pi}{x})-\cos(\frac{\pi}{2})}{x-2} =\lim_{x\to 2}\frac{\phi(x)-\phi(2)}{x-2} = \phi'(2) = \frac{\pi}{2^2}\sin(\frac{\pi}{2})$$