calculate-lim-x-0-1-x-sinx-1-x-2-

Question Number 62435 by mathsolverby Abdo last updated on 21/Jun/19

c a l c u l a t e {l i m}_{x \to 0} \frac{{(1 + x)}^{s i n x} - 1}{x^{2}}

Commented by Smail last updated on 22/Jun/19

{(1 + x)}^{s i n x} = e^{l n ({(1 + x)}^{s i n x})} = e^{s i n x \times l n (1 + x)}

l n (1 + x) \sim x w h e n x i s n e a r 0

s i n x \sim x w h e n x i s n e a r 0

s o {(1 + x)}^{s i n x} \sim e^{x \times x} \sim e^{x^{2}}

e^{x} \sim 1 + x w h e n x n e a r 0

s o {(1 + x)}^{s i n x} \sim 1 + x^{2}

\underset{x \to 0}{l i m} \frac{{(1 + x)}^{s i n x} - 1}{x^{2}} = \underset{x \to 0}{l i m} \frac{1 + x^{2} - 1}{x^{2}}

= 1

Commented by mathmax by abdo last updated on 23/Jun/19

l e t u s e h o s p i t a l t h e o r e m u (x) = {(1 + x)}^{s i n x} - 1 \Rightarrow u (x) = e^{s i n x l n (1 + x)} - 1 \Rightarrow

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

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

+ {(c o s x l n (1 + x) + \frac{s i n x}{1 + x})}^{2} e^{s i n x l n (1 + x)} \Rightarrow {l i m}_{x \to 0} u^{(2)} (x) = 2

v (x) = x^{2} \Rightarrow g^{^{'}} (x) = 2 x a n d v^{(2)} (x) = 2 = {l i m}_{x \to 0} g^{(2)} (x) \Rightarrow

{l i m}_{x \to 0} \frac{{(1 + x)}^{s i n x} - 1}{x^{2}} = {l i m}_{x \to 0} \frac{u^{(2)} (x)}{v^{(2)} (x)} = \frac{2}{2} = 1 .