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?

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- Sep 11th 2011, 01:45 PMritacameTrig 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? - Sep 11th 2011, 01:50 PMskeeterRe: Trig limits
- Sep 11th 2011, 01:53 PMritacameRe: Trig limits
That means the answer would be 3/2 if one followed those steps. Thank You!

- Sep 11th 2011, 01:54 PMSironRe: Trig limits
Rewrite:

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

Use the fact:

$\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x}=1$

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