prove-that-lim-x-0-arcsin-x-1-x-2-ln-1-x-1-

Question Number 80000 by malwaan last updated on 30/Jan/20

p r o v e t h a t

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

Commented by jagoll last updated on 30/Jan/20

\sin y = \frac{x}{\sqrt{1 - x^{2}}} \Rightarrow \frac{x^{2}}{1 - x^{2}} = \sin^{2} y

\frac{1 - x^{2} - 1}{1 - x^{2}} = \sin^{2} y \Rightarrow 1 - \frac{1}{1 - x^{2}} = \sin^{2} y

\cos^{2} y = \frac{1}{1 - x^{2}} \Rightarrow 1 - x^{2} = \sec^{2} y

1 - x = \frac{\sec^{2} y}{1 + x}

Commented by abdomathmax last updated on 30/Jan/20

l e t f (x) = \frac{a r c s i n (\frac{x}{\sqrt{1 - x^{2}}})}{l n (1 - x)} c h a n g e m e n t x = s i n t

g i v e f (x) = \frac{a r c s i n (t a n t)}{l n (1 - s i n t)} = g (t)

x \to 0 \Rightarrow t \to 0 \Rightarrow g (t) \sim \frac{a r c s i n t}{- t} = - \frac{a r c s i n t}{t}

a n d {l i m}_{t \to 0} \frac{a r s i n (t)}{t} = {a r c s i n}^{^{'}} (0) w e h a v e

{a r c s i n}^{^{'}} (t) = \frac{1}{\sqrt{1 - t^{2}}} \Rightarrow {a r c s i n}^{^{'}} (0) = 1 \Rightarrow

{l i m}_{t \to 0} g (t) = - 1 = {l i m}_{x \to 0} f (x)

Answered by Kamel Kamel last updated on 30/Jan/20

Ω = \lim_{x \to 0} \frac{\arcsin (\frac{x}{\sqrt{1 - x^{2}}})}{Ln (1 - x)}

Put : x = \sin (t), x \to 0 \Rightarrow t \to 0

Ω = \lim_{x \to 0} \frac{\arcsin (\frac{x}{\sqrt{1 - x^{2}}})}{Ln (1 - x)} = \lim_{t \to 0} \frac{\arcsin (\tan (t))}{Ln (1 - \sin (t))} = - \lim \frac{\tan (t)}{\sin (t)} = - 1

Commented by malwaan last updated on 31/Jan/20

t h a n k y o u s o m u c h

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