find-lim-x-0-1-x-1-x-1-sinx-

Question Number 66318 by mathmax by abdo last updated on 12/Aug/19

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

Commented by mathmax by abdo last updated on 24/Aug/19

l e t f (x) = {(\frac{1 + x}{1 - x})}^{\frac{1}{s i n x}} \Rightarrow f (x) = e^{\frac{1}{s i n x} l n (\frac{1 + x}{1 - x})} b u t

l n (\frac{1 + x}{1 - x}) = l n (1 + x) - l n (1 - x) \sim x - (- x) = 2 x a n d s i n x \sim x \Rightarrow

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

Commented by kaivan.ahmadi last updated on 12/Aug/19

= e^{{l i m}_{x \to 0} (\frac{1 + x}{1 - x} - 1) \times \frac{1}{x}} = e^{{l i m}_{x \to 0} (\frac{2 x}{1 - x}) \times \frac{1}{x}} = e^{2}

Commented by Mikael last updated on 12/Aug/19

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

\underset{x \to 0}{l i m} [{(1 + \frac{2 x}{1 - x})}^{\frac{1 - x}{2 x}}] = e, t h e n \underset{x \to 0}{l i m} \frac{2 x}{1 - x} . \frac{1}{s i n x} = \underset{x \to 0}{l i m} \frac{x}{s i n x} . \frac{2}{1 - x} = 1 \times 2 = 2

= e^{2}

Commented by mathmax by abdo last updated on 12/Aug/19

t h a n k s s i r .