Question

be
Evaluate :
Lt (sin x) tan x
x→/2​

Answer 1

Answer:

Let φ(x) = x - tan x and ψ(x) = sin x - x, then φ(x)/ψ(x) = (x - tan x)/(sin x - x). When x = 0, φ(x)/ψ(x) = (0 - tan 0)/(sin 0 - 0) = (0 - 0)/(0 - 0) = 0/0 is indeterminate of the form 0/0 . So we evaluate φ'(x)/ψ' (x) at x = 0 .

φ'(x)/ψ' (x) = [d/dx (x - tan x)]/ [d/dx (sin x - x)] =( 1 – sec² x)/(cos x - 1) = ( 1 – sec² x) /(cos x - 1) = [1 – (1/cos² x)]/(cos x - 1) = (cos² x - 1)/cos² x (cos x - 1) = (cos x + 1)(cos x - 1)/cos² x(cos x - 1) . Cancelling out (cos x - 1) from numerator & denominator, φ'(x)/ψ' (x) = (cos x + 1)/cos² x which tends to the limit (cos 0 + 1)/cos² 0 = (1 + 1)/1² = 2/1 = 2 .

Now by a theorem, since f ' (x)/φ' (x) exists and tends to the limit 2 as x→0, limit f(x)/φ(x) exists and is equal to 2 as x → 0 .

Hence in the limit x→0, (x - tan x)/(sin x - x) = 2 (Answer)