Trig limits

**ritacame** · September 11th 2011, 01:45 PM

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?

**skeeter** · September 11th 2011, 01:50 PM

Originally Posted by ritacame

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 ...

$\frac{3}{6x \csc(3x)} = \frac{3 \sin(3x)}{6x} = \frac{3}{2} \cdot \frac{\sin(3x)}{3x}$

**ritacame** · September 11th 2011, 01:53 PM

That means the answer would be 3/2 if one followed those steps. Thank You!

**Siron** · September 11th 2011, 01:54 PM

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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