lim-x-0-1-tanx-1-sinx-1-sinx-

Question Number 144645 by mathdanisur last updated on 27/Jun/21

\underset{x \to 0}{l i m} {(\frac{1 + t a n x}{1 + s i n x})}^{\frac{1}{s i n x}} = ?

Answered by liberty last updated on 27/Jun/21

\lim_{x \to 0} {(\frac{1 + \sin x + \tan x - \sin x}{1 + \sin x})}^{\frac{1}{\sin x}}

= \lim_{x \to 0} {(1 + \frac{\tan x - \sin x}{1 + \sin x})}^{\frac{1}{\sin x}}

= \lim_{x \to 0} {[{(1 + \frac{\tan x - \sin x}{1 + \sin x})}^{\frac{1 + \sin x}{\tan x - \sin x}}]}^{\frac{\tan x - \sin x}{\sin x (1 + \sin x)}}

= e^{\lim_{x \to 0} (\frac{\tan x - \sin x}{\sin x (1 + \sin x)})}

= e^{\lim_{x \to 0} (\frac{\tan x (1 - \cos x)}{\sin x (1 + \sin x)})}

= e^{\frac{0}{1}} = e^{0} = 1 .

Commented by mathdanisur last updated on 27/Jun/21

a l o t c o o l t h a n k y o u S i r

Answered by mathmax by abdo last updated on 27/Jun/21

f (x) = {(\frac{1 + tanx}{1 + sinx})}^{\frac{1}{sinx}} \Rightarrow f (x) = {(\frac{1 + sinx + tanx - sinx}{1 + sinx})}^{\frac{1}{sinx}}

= {(1 + \frac{tanx - sinx}{1 + sinx})}^{\frac{1}{sinx}} = e^{\frac{1}{sinx} \log (1 + \frac{tanx - sinx}{1 + sinx})} we have

\log (1 + \frac{tanx - sinx}{1 + sinx}) \sim \frac{tanx - sinx}{1 + sinx} \Rightarrow \frac{1}{sinx} \log (\dots) \sim \frac{tanx - sinx}{sinx (1 + sinx)}

\sim \frac{x + \frac{x^{3}}{3} - (x - \frac{x^{3}}{6})}{x (1 + x)} = \frac{x^{3}}{6 x (1 + x)} \sim \frac{x^{2}}{6 (1 + x)} \to 0 (x \to 0) \Rightarrow

\lim_{x \to 0} f (x) = e^{0} = 1

Commented by mathdanisur last updated on 27/Jun/21

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