Originally Posted by

**reastland** I'm really struggling with these trig limits. I have three of them that I can't get straight, which I'll post separately. Sorry in advance for dropping all of these. I've stared at them endlessly, and don't know what else to do. Here's the first one:

$\displaystyle \displaystyle \lim_{x\to 0 } \frac {sin 4x}{sin 6x}$

I want to say that $\displaystyle \frac {sin 4x}{sin 6x} = \frac{2}{3} sin x$, but I know this isn't correct. The answer to this is $\displaystyle \frac{2}{3}$, but I have no idea how to get there. We haven't covered L'Hospitals rule yet, so somehow I think that I have to get this down to $\displaystyle \frac {2}{3} * \frac {sin x}{x}$, but I just can't figure it out.

Can somebody please help?

Thanks.