# How to find this limit $\lim_{x\to 0} (\frac{1+\tan x}{1+\sin x})^\frac{1}{\sin x}$

Help me finding this limit $$\lim\limits_{x\to 0} \left(\frac{1+\tan x}{1+\sin x}\right)^\frac{1}{\sin x}$$

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Please show the reason of downvote explicitly. I give one here, also I didn't downvote: there seems no effort shown. –  Frank Science Jul 15 '12 at 10:55

This is the answer that Madrit Zhaku was trying to give. It makes use of the well-known limit $\lim\limits_{x\to 0}(1+x)^{\frac1x}=e$.

\begin{align*} \lim_{x\to 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac1{\sin x}}&=\lim_{x\to 0}\frac{(1+\tan x)^{\frac1{\sin x}}}{(1+\sin x)^{\frac1{\sin x}}}\\ &=\frac{\lim\limits_{x\to 0}(1+\tan x)^{\frac1{\sin x}}}{\lim\limits_{x\to 0}(1+\sin x)^{\frac1{\sin x}}}\\ &=\frac{\lim\limits_{x\to 0}(1+\tan x)^{\frac1{\tan x}\cdot\frac1{\cos x}}}e\\ &=\frac1e\lim_{x\to 0}\left((1+\tan x)^{\frac1{\tan x}}\right)^{\frac1{\cos x}}\\ &=\frac1e\cdot e^1\\ &=1\;. \end{align*}

The limit can also be calculated quite easily using l’Hospital’s rule.

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Put:

1 - $$y= \left( {\frac {1+\tan \left( x \right) }{1+\sin \left( x \right) }} \right) ^{ \left( \sin \left( x \right) \right) ^{-1}}$$, (2) - take the natural log ( ln(x) ) to both sides of the above equation,

(3) - take the limit to both sides as x goes to 0 of the equation in (2),

(4) - use L'Hôpital's rule to the right hand side in (3),

(5) - exponentiate both sides (use the inverse function of ln(x) ) in (4), you get the answer.

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$\lim_{x\to 0} (\frac{1+\tan x}{1+\sin x})^\frac{1}{\sin x}$=|$\lim_{x\to 0}$$(\frac{a}{b})^n=\lim_{x\to 0}\frac{a^n}{b^n}|=$$\lim_{x\to 0}$$\frac{(1+\tan x)^\frac{1}{\sin x}}{(1+\sin x)^\frac{1}{\sin x}}$=|$\lim_{x\to 0}({1+\sin x})^\frac{1}{\sin x}$=$e$|=$\frac{1}{e}\lim_{x\to 0}({1+\tan x)}^{\frac{1}{\tan x}\cdot\frac{1}{\sin x}}$=$\frac{1}{e}\cdot e = 1$.