Got any leads on calculating:
$\displaystyle \lim{x\to\frac{\pi}{2}}$ : $\displaystyle \frac{cotx}{2x-\pi}$
Sorry for the bad latex, but I am trying you know.
Lim(X-->pi/2)[cotX / (2x -pi)]
= cot(pi/2) / 2*pi/2 -pi
= 0/0
Indeterminate.
Use the L'Hopital's Rule: Lim(x->a)[f(x)/g(x)] = Lim(x->a)[f'(x)/g'(x)].
= Lim(X->pi/2)[-csc^2(X) / 2
= -csc^2(pi/2) / 2
= -[1/sin(pi/2)]^2 /2
= -[1/1]^2 / 2
= -1/2 ------------answer.
Using the substitution $\displaystyle x - \pi/2 \to x$,we get
$\displaystyle \lim_{x\to\frac{\pi}{2}} \frac{\cot(x)}{2x-\pi} = \lim_{x \to 0}\frac{\cot(x + \pi/2)}{2x} = \lim_{x \to 0}\frac{\cos(x+\pi/2)}{2x\sin(x+\pi/2)}$ which becomes $\displaystyle \lim_{x \to 0}\frac{-\sin(x)}{2x\cos(x)} = -\frac{1}{2}$.
The less L'Hopital, the better.