Q.268. Limit using L'Hospital's Theorem and Power series.

Question 268.
Find the value of lim (x → 0) (sinx - arctanx) / [x^2 ln(1 + 2x + x^2)].

Answer 268.
f (x) = sinx - arctanx
=> f '(x) = cosx - 1/(1 + x^2)
=> f "(x) = - sinx + 2x/(1 + x^2)^2
=> f "' (x)
= - cosx + [2(1 + x^2)^2 - 8x^2 (1 + x^2)] / (1 + x^2)^4
= - cosx + (2 - 4x^2 - 6x^4) / (1 + x^2)^4
=> f '(0) = 0, f "(0) = 0 and f"'(0) = 1

 g(x) = x^2 ln(1 + 2x + x^2) = 2x^2 ln(1 + x)
=> g'(x) = 4x ln(1 + x) + 2x^2/(1 + x)
=> g"(x) = 4ln(1 + x) + 4x/(1 + x) + [4x(1 + x) - 2x^2] / (1 + x)^2
= 4ln(1 + x) + 4x/(1 + x) + 2(2x - x^2) / (1 + x)^2
=> g"'(x)
= 4/(1 + x) + [4(1 + x) - 4x] / (1 + x)^2 + 2[(2 - 2x) (1 + x)^2 - 4(2x - x^2) (1 + x)] / (1 + x)^4
=> g'(0) = 0, g"(0) = 0 and g"'(0) = 4 + 4 + 4 = 12

 => limit (x → 0) (sinx - arctanx) / [x^2 ln(1 + 2x + x^2)]
= lim (x → 0) f (x) / g(x)
As this becomes 0/0 on plugging x = 0, using L'Hospital's theorem,
limit = lim (x → 0) f '(x) / g '(x)
This also becomes 0/0 on plugging x = 0.
Hence, using L'Hospital's theorem again,
=> limit = lim (x → 0) f "(x) / g"(x)
This also becomes 0/0 on plugging x = 0.
Hence, using L'Hospital's theorem again,
=> limit = lim (x → 0) f "'(x) / g"'(x)
= f "'(0) / g"'(0)
= 1/12.

One of my contacts on YA!
Mr. Fred
suggested the following

ALTERNATE METHOD BASED ON POWER SERIES:

sinx = x - x^3/3! + x^5/5! - x^7/7! + .... ∞
arctanx = x - x^3/3 + x^5/5 - x^7/7 + .... ∞
ln (1 + x) = x - x^2/2 + x^3/3 - x^4/4 + .... ∞

=> limit
= lim (x → 0) [(x - x^3/3! + x^5/5! - .... ∞) - (x - x^3/3 + x^5/5 - .... ∞)] / [2x^2 (x - x^2/2 + x^3/3 - .... ∞]
= lim (x → 0) [x^3/6 - x^5/5 + .... ∞) / (2x^3 - x^4 + 2x^5/3 - .... ∞)
= 1/12.

Link to YA!