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.
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.
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!)
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.