Calculate Trigonometry limits I need help solving this task, if anyone had a similar problem it would help me.
The task is:
Calculate using the rule $\lim\limits_{x\to \infty}\left(1+\frac{1}{x}\right)^x=\large e  $:
$\lim_{x\to0}\left(\frac{1+\mathrm{tg}\: x}{1+\sin x}\right)\Large^{\frac{1}{\sin x}}
 $
I tried this:
$  \lim_{x\to0}\left(\frac{1+\mathrm{tg}\: x}{1+\sin x}\right)^{\Large\frac{1}{\sin x}}=\lim_{x\to0}\left(\frac{1+\frac{\sin x}{\cos x}}{1+\sin x}\right)^{\Large\frac{1}{\sin x}}=\lim_{x\to0}\left(\frac{\sin x+\cos x}{\cos x\cdot(1+\sin x)}\right)^{\Large\frac{1}{\sin x}}
 $
But I do not know, how to solve this task.
Thanks in advance !
 A: You have the right idea, but to get a better match with the form given in the rule you should separate an add-end of $1$ from the base.  Thus
$\left(\dfrac{1+\tan x}{1+\sin x}\right)^{\dfrac1{\sin x}}=\left(1+\dfrac{\tan x-\sin x}{1+\sin x}\right)^{\dfrac1{\sin x}}$
Now introduce the second term of the base as a factor into the exponent.  The reciprocal of the factor is then multiplied by the original exponent $1/\sin x$:
$=\left(1+\dfrac{\tan x-\sin x}{1+\sin x}\right)^{\dfrac{1+\sin x}{\tan x-\sin x}\cdot\dfrac{\tan x-\sin x}{(\sin x)(1+\sin x)}}$
$=\color{blue}{\left(\left(1+\dfrac{\tan x-\sin x}{1+\sin x}\right)^{\dfrac{1+\sin x}{\tan x-\sin x}}\right)}^{\color{brown}{\dfrac{\tan x-\sin x}{(\sin x)(1+\sin x)}}}$
The blue expression inside the largest set of parentheses follows your rule and thus has a limiting value of $e$.  The remaining terms, colored brown, are an exponent on this expression that tends to $e$, which has some limit $L$ that you are to find.  The overall limit would then be $e^L$.
A: Write
$$\left(\frac{1+\tan x}{1+\sin x}\right)^{1/\sin x}=\frac{\left(1+\tan x\right)^{1/\sin x}}{\left(1+\sin x\right)^{1/\sin x}}$$
and the denominator tends to $e$ (because $\sin x$ tends to $0$).
Then
$$\left(1+\tan x\right)^{1/\sin x}=\left(\left(1+\tan x\right)^{1/\tan x}\right)^{1/\cos x},$$
which tends to $e^1$.
A: We note that by l'Hopital rule
$$
\lim_{x\to0} \frac{\ln(1+\sin x)}{\sin x}=\lim_{x\to0} \frac{\ln(1+\tan x)}{\sin x}=1
$$
Therefore:
$$
\lim_{x\to0}\bigg(\frac{1+\tan x}{1+\sin x}\bigg)^{1/\sin x}=\lim_{x\to0}\exp\bigg[\frac{\ln(1+\tan x)}{\sin x}- \frac{\ln(1+\sin x)}{\sin x}\bigg]=e^{1-1}=1
$$
We can also show it using the required limit:
$$
\lim_{x\to0}\bigg(\frac{1+\tan x}{1+\sin x}\bigg)^{1/\sin x}= \frac{\lim_{x\to0}(1+\tan x)^{1/\sin x}}{\lim_{x\to0}(1+\sin x)^{1/\sin x}} \\
=\frac{\lim_{y\to\infty}[1+\frac{1}{y}(1-\frac{1}{y^2 })^{1/2}]^y} {\lim_{y\to\infty}[1+\frac{1}{y}]^y}=\frac{\lim_{y\to\infty}[1+\frac{1}{y} ]^y} {\lim_{y\to\infty}[1+\frac{1}{y}]^y}=\frac{e}{e}=1
$$
