1. ## 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.

2. 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: $\displaystyle \cos x = \sin \left( \tfrac{\pi}{2} - x \right)$
Hint: Now use substitution $\displaystyle t= \tfrac{\pi}{2} - x$

3. alternatively, just let $\displaystyle t = x - \frac {\pi}2$

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

now use the addition formula for cosine. the result will be pleasantly surprising

4. 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:
$\displaystyle \lim_{x \to \pi/2} \frac{cos(x)}{x - \frac{\pi}{2}} = \lim_{x \to \pi/2} \frac{-sin(x)}{1} = -1$

-Dan

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

-Dan
well, the poster did say they couldn't use L'Hopital's

6. 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. (Yeah, right!)

Ah well. Thanks for the catch.

-Dan

7. 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. (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

8. You can also use the definition of the derivative:

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

EDIT: that is if you covered derivatives.