find-lim-x-0-ln-1-sinx-sin-ln-1-x-x-2-

Question Number 33313 by abdo imad last updated on 14/Apr/18

f i n d {l i m}_{x \to 0} \frac{l n (1 + s i n x) - s i n (l n (1 + x))}{x^{2}}

Commented by abdo imad last updated on 17/Apr/18

l e t p u t u (x) = l n (1 + s i n x) - s i n (l n (1 + x)) a n d v (x) = x^{2}

l e t u s e h o s p i t a l t h e o r e m

{l i m}_{x \to 0} \frac{u (x)}{v (x)} = {l i m}_{x \to 0} \frac{u^{^{″}} (x)}{v^{^{″}} (x)} b u t w e h a v e

u^{^{'}} (x) = \frac{c o s x}{1 + s i n x} - \frac{1}{1 + x} c o s (l n (1 + x)) \Rightarrow

u^{^{″}} (x) = \frac{- s i n x (1 + s i n x) - {c o s}^{2} x}{{(1 + s i n x)}^{2}} - (\frac{- 1}{{(1 + x)}^{2}} c o s (l n (1 + x))

- \frac{1}{1 + x} \frac{1}{1 + x} s i n (l n (1 + x)) \Rightarrow {l i m}_{x \to 0} u^{^{″}} (x) = - 1 + 1 = 0

a l s o w e h a v e v^{^{'}} (x) = 2 x \Rightarrow v^{^{″}} (x) = 2 \Rightarrow

{l i m}_{x \to 0} \frac{u (x)}{v (x)} = 0