# Trig limit

• Sep 20th 2008, 05:29 PM
uberbandgeek6
Trig limit
Unfortunately, I can't use any advanced techniques to find this limit because our class hasn't "learned" l'hopital's rule or anything. The most I think I can use are the lim(sinx)/x and lim(1-cosx)/x rules. Here is the problem:

lim as x approaches pi/2 of: (cosx)/(x - pi/2)

Thanks.
• Sep 20th 2008, 05:37 PM
ThePerfectHacker
Quote:

Originally Posted by uberbandgeek6
Unfortunately, I can't use any advanced techniques to find this limit because our class hasn't "learned" l'hopital's rule or anything. The most I think I can use are the lim(sinx)/x and lim(1-cosx)/x rules. Here is the problem:

lim as x approaches pi/2 of: (cosx)/(x - pi/2)

Thanks.

Hint: $\cos x = \sin \left( \tfrac{\pi}{2} - x \right)$
Hint: Now use substitution $t= \tfrac{\pi}{2} - x$
• Sep 20th 2008, 05:45 PM
Jhevon
alternatively, just let $t = x - \frac {\pi}2$

your limit becomes $\lim_{t \to 0} \frac {\cos (t + \pi / 2)}t$

now use the addition formula for cosine. the result will be pleasantly surprising (Wink)
• Sep 20th 2008, 05:54 PM
topsquark
Quote:

Originally Posted by uberbandgeek6
Unfortunately, I can't use any advanced techniques to find this limit because our class hasn't "learned" l'hopital's rule or anything. The most I think I can use are the lim(sinx)/x and lim(1-cosx)/x rules. Here is the problem:

lim as x approaches pi/2 of: (cosx)/(x - pi/2)

Thanks.

Or just use L'Hopital's rule:
$\lim_{x \to \pi/2} \frac{cos(x)}{x - \frac{\pi}{2}} = \lim_{x \to \pi/2} \frac{-sin(x)}{1} = -1$

-Dan
• Sep 20th 2008, 06:28 PM
Jhevon
Quote:

Originally Posted by topsquark
Or just use L'Hopital's rule:
$\lim_{x \to \pi/2} \frac{cos(x)}{x - \frac{\pi}{2}} = \lim_{x \to \pi/2} \frac{-sin(x)}{1} = -1$

-Dan

well, the poster did say they couldn't use L'Hopital's
• Sep 20th 2008, 06:50 PM
topsquark
Quote:

Originally Posted by Jhevon
well, the poster did say they couldn't use L'Hopital's

I mentioned it for the sake of simplicity. Yeah, that's why I did it. I certainly read the OP's first post and caught that. (Doh) (Yeah, right!)

Ah well. Thanks for the catch.

-Dan
• Sep 20th 2008, 06:55 PM
Jhevon
Quote:

Originally Posted by topsquark
I mentioned it for the sake of simplicity. Yeah, that's why I did it. I certainly read the OP's first post and caught that. (Doh) (Yeah, right!)

Ah well. Thanks for the catch.

-Dan

don't worry about it. i always read the OP's first post and catch stuff like that (Yeah, right!) too
• Sep 20th 2008, 07:39 PM
Chop Suey
You can also use the definition of the derivative:

$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}-0}{x-\frac{\pi}{2}}=(\cos{x})'~\text{at}~x = \frac{\pi}{2}$

EDIT: that is if you covered derivatives.