Question

Lim as x -> 0 Sin5x/Sin2x

Answer 1

\[
\lim_{x\to 0}\frac{\sin(5x)}{\sin(2x)}
\]Is this it?

Answer 2

Yes

Answer 3

Well, the main identity we're looking for is:

\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }= 1\]Now, as long as the angle of sin and the denominator match, the value as x goes to 0 is 1. So if I had sin(lnx)/lnx, its 1, even though the values may look funky. So for you, you have a sin5x, meaning we want to cram a 5x underthere somehow. And then you also have a sin2x, meaning we want to cram a 2x underneath that value as well.

Now in order to do this, we'll have to multiply in a 1/5x and a 1/2x. We have to multiply it in as a fraction like that or we won't get it in the proper spot. So what I can so is multiply top and bottom, at the same time, by (1/(5x*2x))Doing that gives us this: