Pessimist Singularitarian

#6
October 14th, 2016,
00:01
Originally Posted by

**Monoxdifly**
Won't it give $ \displaystyle \frac{0}{0}$ again in the $ \displaystyle \frac{\cos{\frac{\pi}{2}}\cdot\cos{u}}{u}$ part?

Recall the co-function identity:

$ \displaystyle \cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$

And so our limit becomes:

$ \displaystyle L=-\lim_{u\to0}\left(\frac{\sin(u)}{u}\right)$

And yes, this is an indeterminate form, but it is well-known to students and usually allowed as a result to use:

$ \displaystyle \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$