find-lim-x-0-1-x-e-arcsinx-x-2-

Question Number 42785 by maxmathsup by imad last updated on 02/Sep/18

f i n d {l i m}_{x \to 0} \frac{1 + x - e^{a r c s i n x}}{x^{2}}

Commented by maxmathsup by imad last updated on 30/Sep/18

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 - e^{a r c s i n x} a n d

v (x) = x^{2} \Rightarrow u^{(1)} (x) = 1 - \frac{1}{\sqrt{1 - x^{2}}} e^{a r c s i n x} \Rightarrow

u^{(2)} (x) = - {{(1 - x^{2})}^{- \frac{1}{2}} e^{a r c s i n x}}^{(1)}

= - {- \frac{1}{2} (- 2 x) {(1 - x^{2})}^{- \frac{3}{2}} e^{a r c s i n x} + \frac{{(1 - x^{2})}^{- \frac{1}{2}}}{\sqrt{1 - x^{2}}} e^{a r c s i n x}}

= - x {(1 - x^{2})}^{- \frac{3}{2}} e^{a r c s i n x} - \frac{1}{1 - x^{2}} e^{a r c s i n x} \Rightarrow {l i m}_{x \to 0} u^{(2)} (x) = - 1

a l s o w e h a v e v^{(1)} (x) = 2 x a n d v^{(2)} (x) = 2 \Rightarrow {l i m}_{x \to 0} v^{(2)} (x) = 2 \Rightarrow

{l i m}_{x \to 0} \frac{1 + x - e^{a r c s i n x}}{x^{2}} = - \frac{1}{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

t = {s i n}^{- 1} x x \to 0 t \to 0

\lim_{t \to 0} \frac{1 + s i n t - e^{t}}{{s i n}^{2} t} (\frac{0}{0})

\lim_{t \to 0} \frac{c o s t - e^{t}}{s i n 2 t} (\frac{0}{0})

\lim_{t \to 0} \frac{- s i n t - e^{t}}{2 c o s 2 t}

= \frac{- 0 - 1}{2} = \frac{- 1}{2}