evaluate without using L'Hopital theorem the following limit

lim x-->0 [(sin(x)-x)/(x-tan(x))]

the answer is 0.5 but I want to know the steps to calculate such a problem

using Taylor series,

sinx-x = -x^3/3! + x^5/5! - ...
x-tanx = -x^3/3 - 2x^5/15 - ...

divide both by x^3 and you have

-1/3! + x^2/5! - ...
-1/3 - 2x^3/15 - ...

now divide and let x->0 and

(-1/3!)/(-1/3) = 1/2