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
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
1 answer