# Unsolvable limit of trigonometric function

1. Dec 8, 2008

### cuthecheese

I need to find the limit of sin(3t)^2/t^2 as t approaches 0. We have not yet learned L'Hopital's rule, so how do I find the limit here?

I tried to take the derivative of sin(3t)^2/t^2, but it is nowhere near cancelling out 't' from the bottom.

Thanks

2. Dec 8, 2008

### gabbagabbahey

Do you know the limit of sin(x)/x as x->0? Do you know the rules for a limit of a product?

If, so use those along with the fact that $$\frac{\sin^2(3t)}{t^2}=\frac{\sin(3t)}{t}\cdot\frac{\sin(3t)}{t}$$

3. Dec 9, 2008

### HallsofIvy

Staff Emeritus
And, if that isn't enough, the obvious
$$\frac{sin^2(3t)}{t^2}= 9\frac{sin(3t)}{3t}\cdot\frac{sin(3t)}{3t}$$

4. Dec 9, 2008

### Дьявол

$$\lim_{t \rightarrow 0}(\frac{sin(3t)^2}{t^2})$$

Now remember $$\lim_{t \rightarrow 0}\frac{sin(t)}{t}=1$$.

So $$\lim_{t \rightarrow 0}(\frac{sin(3t)^2}{t^2})=\lim_{t \rightarrow 0}(\frac{sin(3t)}{t})^2$$.

It is pretty easy from now on. In future, please consider using LaTeX code.

Regards.