1. ## Trig Limit

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:

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

I want to say that $\frac {sin 4x}{sin 6x} = \frac{2}{3} sin x$, but I know this isn't correct. The answer to this is $\frac{2}{3}$, 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 $\frac {2}{3} * \frac {sin x}{x}$, but I just can't figure it out.

Thanks.

2. 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:

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

I want to say that $\frac {sin 4x}{sin 6x} = \frac{2}{3} sin x$, but I know this isn't correct. The answer to this is $\frac{2}{3}$, 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 $\frac {2}{3} * \frac {sin x}{x}$, but I just can't figure it out.

Thanks.
Can you use L'Hopitals Rule?

Never mind. I just read you haven't covered it.

Try double angle formulas.

3. Originally Posted by reastland

I want to say that $\frac {sin 4x}{sin 6x} = \frac{2}{3} sin x$, but I know this isn't correct.
It's not correct.

Have you tried to use double angle formulas?

4. $\displaystyle \frac{\sin{4x}}{\sin{6x}} = \frac{2\sin{2x}\cos{2x}}{2\sin{3x}\cos{3x}}$

$\displaystyle = \frac{\sin{2x}\cos{2x}}{\sin{3x}\cos{3x}}$

$\displaystyle = \frac{2\sin{x}\cos{x}\cos{2x}}{\sin{x}(3 - 4\sin^2{x})\cos{3x}}$

$\displaystyle = \frac{2\cos{x}\cos{2x}}{(3 - 4\sin^2{x})\cos{3x}}$.

You should now be able to evaluate the limit.

5. 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:

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

I want to say that $\frac {sin 4x}{sin 6x} = \frac{2}{3} sin x$, but I know this isn't correct. The answer to this is $\frac{2}{3}$, 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 $\frac {2}{3} * \frac {sin x}{x}$, but I just can't figure it out.

If you want to use $\lim\limits_{x\to0}\dfrac{\sin x}{x}=1$, we can think of your problem as:
$\lim\limits_{x\to0}\dfrac{\sin(4x)}{\sin(6x)}=\lim \limits_{x\to0}\dfrac{4\dfrac{\sin(4x)}{4x}}{6\dfr ac{\sin(6x)}{6x}}=\dfrac{4\lim\limits_{x\to0}\dfra c{\sin(4x)}{4x}}{6\lim\limits_{x\to0}\dfrac{\sin(6 x)}{6x}}$
Now can you see why the solution is $\dfrac{2}{3}$?