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

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

-

## closed as off-topic by Thursday, RecklessReckoner, Mike Miller, Claude Leibovici, WilliamAug 9 at 6:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thursday, RecklessReckoner, Mike Miller, Claude Leibovici, William
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

-

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.

-

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

-
The core of a workable idea is buried in this, but I’ve rarely seen such mangled notation. In particular, the parenthetical remarks between vertical bars are completely non-standard; you cannot expect them to be understood. –  Brian M. Scott Jul 15 '12 at 8:41