Q.169. Limit.

Question 169.
Find lim (x → 0) (tan x -x)/x^3 without using L'Hospital's theorem.

Answer 169.
lim (x → 0) (tanx - x)/x^3

= lim (x → 0) (sinx - x cosx) / x^3 cosx
= lim (x → 0) [(sinx - x + x - xcosx) / x^3] * lim (x → 0) (1/cosx)
= lim (x → 0) [(sinx - x)/x^3] + lim (x → 0) (1 - cosx)/x^2 ... [lim (x → 0) (1/cosx) = 1]
= - 1/6 + 1/2 = 1/3.
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lim (x → 0) [(sinx - x)/x^3] = - 1/6 is proved as under.
Assuming that this limit exists, let lim (x → 0) [(sinx - x)/x^3] = L
Let x = 3t. Then, x → 0 => t → 0.
=> L
= lim (x → 0) [(sinx - x)/x^3]
= - lim (t → 0) (3t - sin3t) / (3t)^3
= - lim (t → 0) [3t - (3sint - 4sin^3 t)] / (27t^3)
= lim (t → 0) [3 (sint - t)/(27t^3)] - (4/27) lim (t → 0) sin^3 t/t^3
= L/9 - 4/27
=> 8L/9 = - 4/27
=> L = - 1/6.

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