Evaluate Limit x tends to 0 x minus sinx divided by tanx minus x?

lim (x→0) [(x - sin x)/(tan x - x)]

Since both the numerator and the denominator have limit zero as x tends to 0, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply the l'Hopital's Rule and the limit equals

lim (x→0) [(x - sin x)'/(tan x - x)']
= lim (x→0) [(1 - cos x)/(sec2 x - 1)] (form 0/0, use again the l'Hopital's Rule)
= lim (x→0) [(1 - cos x)'/(sec2 x - 1)']
= lim (x→0) [(0 - (-sin x)/(2sec x sec x tan x - 0)]
= lim (x→0) [(sin x)/(2sec2 x tan x)] (substitute 1/cos2 x for sec2 x and sin x/cos x for tan x)
= lim (x→0) [(sin x)/(2sin x/cos3 x)]
= lim (x→0) [(sin x cos3 x)/2sin x]
= lim (x→0) (cos3 x/2)
= 1/2

Thus, (x - sin x)/(tan x - x) tends to 0.5 as x tends to 0.