Find the limit as x approaches 0 of x-sinx/x-tanx?

Question

You can use l'Hospital's Rule, because 0/0. Then I put x-cosx/x-sec^2x, but not sure if that is right. Can someone help. Thanks!

Answer 1

Well, you're on the right track, but

d/dx (x - sinx) = 1 - cosx, and d/dx (x - tanx) = 1 - sec^2x.

From there you will need to apply l'Hopital's rule again, and then do some of the usual reorganizing/rewriting of trig functions to find the answer

limit as x approaches 0 of (x-sinx) / (x-tanx) = - (1/2)