Evaluate the limit

**Dr Zoidburg** · March 8th 2010, 10:44 PM

Evaluate the limit of
$\lim_{x \to \pi} \frac {sin^2{x}} {1+cos{3x}}$

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

**earboth** · March 9th 2010, 01:07 AM

Originally Posted by Dr Zoidburg

Evaluate the limit of
$\lim_{x \to \pi} \frac {sin^2{x}} {1+cos{3x}}$

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

I would use:

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

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

In the end you should get $\frac29$

**HallsofIvy** · March 9th 2010, 02:20 AM

In other words, apply L'Hopital's rule again!

**Dr Zoidburg** · March 9th 2010, 03:56 PM

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.

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