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.

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- Oct 8th 2005, 03:40 AMa4sweA limit
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. - Oct 8th 2005, 05:23 AMticbol
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. - Oct 8th 2005, 05:35 AMa4swe
That's true, thanks.

But is there a (there is) way to solve it without using L'Hopital's Rule? - Nov 6th 2005, 05:15 PMJamesonQuote:

Originally Posted by**a4swe**

$\displaystyle \lim_{x\to\frac{\pi}{2}} \frac{\cot(x)}{2x-\pi}$

And I don't see a way to do this without L'Hopitals either. - Nov 11th 2005, 06:42 PMhpe
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.