1. How to solve $\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos{x}}{x-\frac{\pi}{2}}$? At first I tried to convert cos x to $\displaystyle \frac{\tan{x}}{\sin{x}}$ but then realized that $\displaystyle \lim_{x\rightarrow c}\frac{\tan{x}}{x}$ only applies if c = 0. So, how?

2. Let's use the substitution:

$\displaystyle u=\frac{\pi}{2}-x$

What does the limit become then?

$\displaystyle \lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$?

4. Originally Posted by Monoxdifly
$\displaystyle \lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$?
Yes, and so I would factor the negative sign in the denominator out in front of the limit, and apply a co-function identity to the numerator...

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

6. 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:

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

Oops... Sorry sorry... Totally forgot about that...

8. Consider the following diagram of the unit circle in the first quadrant:

From this, we can see that for $0\le x\le\dfrac{\pi}{2}$ we have:

$\displaystyle \sin(x)\le x\le\tan(x)$

Dividing though by $\sin(x)$, we obtain:

$\displaystyle 1\le\frac{x}{\sin(x)}\le\frac{1}{\cos(x)}$

or:

$\displaystyle \cos(x)\le\frac{\sin(x)}{x}\le1$

And so we must then have:

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

or:

$\displaystyle 1\le\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)\le1$

Hence (by the squeeze theorem):

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