Trig limits

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September 11th 2011, 01:45 PM
ritacame

Trig limits

I know that csc is 1/sin.

Would i do 3/6(1/sin)(3x)?

Please show me the correct steps if i'm wrong. Wouldn't it just be 0/0 and become DNE?
September 11th 2011, 01:50 PM
skeeter

Re: Trig limits

Quote:

Originally Posted by ritacame

http://i76.photobucket.com/albums/j3...kets/stsji.gif

I know that csc is 1/sin.

Would i do 3/6(1/sin)(3x)?

Please show me the correct steps if i'm wrong.

note ...

$\frac{3}{6x \csc(3x)} = \frac{3 \sin(3x)}{6x} = \frac{3}{2} \cdot \frac{\sin(3x)}{3x}$
September 11th 2011, 01:53 PM
ritacame

Re: Trig limits

That means the answer would be 3/2 if one followed those steps. Thank You!
September 11th 2011, 01:54 PM
Siron

Re: Trig limits

Rewrite:
$\lim_{x\to 0} \frac{3}{6x\csc(3x)}=\lim_{x\to 0} \frac{3}{\frac{6x}{\sin(3x)}}=\lim_{x\to 0}\frac{3\sin(3x)}{6x}$

Use the fact:
$\lim_{x\to 0} \frac{\sin(x)}{x}=1$

And yes $\frac{3}{2}$ is correct.

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