# Evaluate the limit

• Mar 8th 2010, 10:44 PM
Dr Zoidburg
Evaluate the limit
Evaluate the limit of
$\displaystyle \lim_{x \to \pi} \frac {sin^2{x}} {1+cos{3x}}$

I assume I use L'Hopital's rule here?
This gives me:
$\displaystyle \frac {sin{2x}} {-3sin{3x}}$
which comes to 0/0.
wanting some reassurance here that I'm on the right track! (Itwasntme)
• Mar 9th 2010, 01:07 AM
earboth
Quote:

Originally Posted by Dr Zoidburg
Evaluate the limit of
$\displaystyle \lim_{x \to \pi} \frac {sin^2{x}} {1+cos{3x}}$

I assume I use L'Hopital's rule here? <<<<<<< OK
This gives me:
$\displaystyle \frac {sin{2x}} {-3sin{3x}}$
which comes to 0/0.
wanting some reassurance here that I'm on the right track! (Itwasntme)

I would use:

$\displaystyle \lim_{x \to \pi} \frac {sin^2{x}} {1+cos{3x}} = \lim_{x\to \pi}\dfrac{2 \sin(x) \cdot \cos(x)}{-3\sin(3x)}$

As long as the quotient becomes $\displaystyle \frac00$ you can apply l'Hopitals's rule.

In the end you should get $\displaystyle \frac29$
• Mar 9th 2010, 02:20 AM
HallsofIvy
In other words, apply L'Hopital's rule again!
• Mar 9th 2010, 03:56 PM
Dr Zoidburg
cool. Got it now.
Thanks all for your help. I'm doing this paper by correspondence so this is the only place (other than the textbooks) I can turn for advice/help when I get stuck or just when I want confirmation I'm on the right track.