using lim (radius -0) (sin (radius))/(radius) = 1, find the limit of lim (y-0) (sin 3y)/4y. I have the solutions manual, but I can't figure how they're doing the algebra/trig to work it out so there isn't a 0 in the denominator.

Question

using lim (radius ->0) (sin (radius))/(radius) = 1, find the limit of lim (y->0) (sin 3y)/4y. I have the solutions manual, but I can't figure how they're doing the algebra/trig to work it out so there isn't a 0 in the denominator.

dumbcow · Answer

$$\frac{\sin(3y)}{4y}*\frac{3y}{3y} = \frac{3y}{4y}*\frac{\sin(3y)}{3y} = \frac{3}{4}\frac{\sin(3y)}{3y}$$
then you use fact that
$$\lim_{x \rightarrow 0}\frac{\sin(ax)}{ax} = 1$$
i don't remember the proof for that though

anonymous · Answer

okay I think I see... you're multiplying by 3/3 so you can move the 4 out of the way? Any particular reason you're using y/y as well? And.... oooooh, it's because you're trying to define the denominator to be equal to the sin on the top, I see. Oh that's going to be tricky on my tired brain, but I see how it should makes sense.

dumbcow · Answer

haha nevermind...you can just say that radius = 3y so you know that lim sin(3y) /3y  = 1

dumbcow · Answer

yeah your getting it :)

anonymous · Answer

oh yeah, I see that! I just could not figure what was up with the rearranging of 3s and 4s XD I see it now.