# A limit

• Oct 8th 2005, 03:40 AM
a4swe
A limit
$\lim{x\to\frac{\pi}{2}}$ : $\frac{cotx}{2x-\pi}$

Sorry for the bad latex, but I am trying you know.
• Oct 8th 2005, 05:23 AM
ticbol
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
• Oct 8th 2005, 05:35 AM
a4swe
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 PM
Jameson
Quote:

Originally Posted by a4swe
$\lim{x\to\frac{\pi}{2}}$ : $\frac{cotx}{2x-\pi}$

Sorry for the bad latex, but I am trying you know.

Here's the Latex for reference.

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

And I don't see a way to do this without L'Hopitals either.
• Nov 11th 2005, 06:42 PM
hpe
Using the substitution $x - \pi/2 \to x$,we get
$\lim_{x\to\frac{\pi}{2}} \frac{\cot(x)}{2x-\pi} = \lim_{x \to 0}\frac{\cot(x + \pi/2)}{2x} = \lim_{x \to 0}\frac{\cos(x+\pi/2)}{2x\sin(x+\pi/2)}$ which becomes $\lim_{x \to 0}\frac{-\sin(x)}{2x\cos(x)} = -\frac{1}{2}$.
The less L'Hopital, the better.