# Thread: Trig limit

1. ## Trig limit

The question is:

lim t-->0 (sin^2 3x / x^2)

How do you break it up? The sin^2 is throwing me off. Do you do this?

3 (sinx . sinx / x . x). Then get the answer is 3?

I understand the sinx/x=1 rule. PLEASE guide me! I am good at calculating limits, but trig always makes me depressed

2. Take note: $\displaystyle \displaystyle\lim _{x \to 0} \frac{{\sin (3x)}} {x} = 3$

3. Yes.
So am I right?

4. Originally Posted by Marconis
The question is:

lim t-->0 (sin^2 3x / x^2) should that be x-->0 ?

How do you break it up? The sin^2 is throwing me off. Do you do this?

3 (sinx . sinx / x . x). Then get the answer is 3?

I understand the sinx/x=1 rule. PLEASE guide me! I am good at calculating limits, but trig always makes me depressed
you could write also

$\displaystyle \displaystyle\lim_{x \to 0}\frac{Sin^2y}{\left(\frac{y}{3}\right)^2}=\lim_{ y \to 0}3^2\frac{Siny}{y}\ \frac{Siny}{y}$

5. Originally Posted by Marconis
Yes.
So am I right?
No. From what is said in post #2, it should be clear that the answer will be 3^2.

6. Ah, alright. Thanks guys!