Question

Find the limit of sin(4x)/sin(2x) as x approaches to 0-.

Answer 1

So I got 4cos(4x)/2cos(2x)=2, is this right?

Answer 2

yes, but you can also use that sin(x) is about equal to x for small x

Answer 3

\[\frac{\sin(4x)}{4x}\times \frac{2x}{\sin(2x)}\times 2\]

Answer 4

get \(1\times 1\times 2=2\) w/o l'hopital
basically this is what @ybarrap said

Answer 5

So what's the limit of tanx/x^2 as x approaches to 0-? Isn't it sec^2 x/2x?

Answer 6

Lot of ways to approach this one :)
You could also apply Sine Double Angle and change the top to 2sin(2x)cos(2x)

Answer 7

@Idealist that is a start, but you see now that the limit is undefined since the denominator goes to zero but the numerator goes to one

Answer 8

Or can I do (sinx/cosx)/x^2 and then take the derivative?

Answer 9

Ya I like your first approach.\[\Large \lim_{x\to0^-}\frac{\tan x}{x^2} \to \frac{0}{0}\]Taking the derivative gave you,\[\Large \lim_{x\to0^-}\frac{\sec^2x}{2x}\]

Answer 10

So what is the limit giving you now? :)