how do you prove that the limit as x approaches 0, of tan3x/tan5x is 3/5?

Question

I did the next step by differentiating and got 3sec^2(3x)/5sec^2(5x). How do i go further to prove it? i wish i could just say the answer was 3/5.......

Answer 1

You're almost there! Using L'Hopital's rule, you differentiated and got 3sec^2(3x)/5sec^2(5x), right? Now sec(0) = 1/cos(0) = 1, so just plug in x=0 and you get 3/5.

Answer 2

Your'e actually on the right track.
Just substitute the value of x, which is zero, to the differentiated formula:

lim f(x) as x approaches 0 = 3sec²(3)(0) / 5sec²(5)(0)
the reciprocal of sec²x = 1/cos²x as we know it ...
and cos²(0) = 1. Therefore,

lim f(x) as x approaches 0 = 3/5

Answer 3

Secant(y) is continuous at 0, so lim(x->0)sec^2(3x) = sec^2(0) = 1. Same for sec^2(5x).