# Thread: easy old limit I forgot how to do

1. ## easy old limit I forgot how to do

The limit of sin3x/sin6x

as x approaches 0?

There was some easy trick or way of writing it but I forgot. I know seems bad but I haven't been reviewing enough.

2. Originally Posted by BugzLooney
The limit of sin3x/sin6x

as x approaches 0?

There was some easy trick or way of writing it but I forgot. I know seems bad but I haven't been reviewing enough.
Note that $\displaystyle \frac{\sin (3x)}{\sin (6x)} = \frac{1}{2} \cdot \frac{\sin (3x)}{3x} \cdot \frac{6x}{\sin (6x)}$.

3. Originally Posted by BugzLooney
The limit of sin3x/sin6x

as x approaches 0?

There was some easy trick or way of writing it but I forgot. I know seems bad but I haven't been reviewing enough.
Same thing, but a different way of recognizing it.

$\displaystyle \frac{\sin(3x)}{\sin(6x)}=\frac{\frac{\sin(3x)-\sin(3\cdot 0)}{x-0}}{\frac{\sin(6x)-\sin(6\cdot 0)}{x-0}}$

4. I'm still confused.

EDIT: Wait I see now the first one, but how would I have known? Just something I should see I guess? And so the limit of sin3x/3x is that the same as the limit of sin(x)/x? Cuz then it would be the lim of 1/2 times 1 times the limit of the last one 6x/sin(6x) but I don't know what the limit of 6x/sin6x is.

5. He is trying to utilize the fact that the limit of sin(x)/x as x approaches 0 is equal to 1

So, substitute 3x into x and you'll get sin(3x)/3x... this is what you have, so that also equals 1 when its limit is taken as x approaches 0 (same thing with 6x).

The manipulation started by multiplying what you began with, with 6x/6x = 6x/3x * 1/2.

Thus, the sin(3x)/3x and sin(6x)/6x both evaluate to 1, leaving just 1/2 (and that's the answer!)

6. Yes but in his answer he has that its 6x/sin6x not sin6x/6x.

Where are you getting the sin6x/6x from?

Wait so x/sinx is also equal to 1 when x approaches zero then?? In that case then I would get it.

Drexel I thank you for the response but your post is a bit confusing to me. However I do understand the problem now from Mr.F's explanation. If I ever see something like this I guess I should try and rearrange it in a way similar to that.

7. Originally Posted by BugzLooney
Yes but in his answer he has that its 6x/sin6x not sin6x/6x.

Where are you getting the sin6x/6x from?
$\displaystyle \lim_{x\to0}\frac{6x}{\sin(6x)}=\frac{1}{\lim_{x\t o0}\frac{\sin(6x)}{6x}}=\frac{1}{1}=1$

8. It makes no difference. 6x/sin(6x) also equals 1 as x--> 0 since it is merely the reciprocal of sin(6x)/6x. As you know, the reciprocal of 1 is itself, so the technique applies correctly.