Find the limit of $\lim_{x\to 0} (\frac{1+\tan x}{1+\sin x})^{\csc^3x}$ I failed to find the limit of:lim(x->0) $(\frac{1+tan(x)}{1+sinx})^{\frac{1}{sin^3(x)}}$? 
as X approches 0
How do I find the answer for this? 
Thanks in advance. the answer supposed to be sqr(e). but my answer was 1.
Can anyone please help me find my mistake?
I DID:
$lim_{x \to 0} (\frac{1+tan(x)}{1+sin(x)})^{\frac{1}{sin^3(x)}}$
$lim_{x \to 0} (\frac{((1+tan(x))^{1/sin(x)}}{((1+ sin(x))^{1/sin(x)}})^{1/sin^2(x)}$ 
now I look inside:
$lim_{x \to 0} ((1+tan(x))^{1/sin(x)}$ is e
$lim_{x \to 0} ((1+sin(x))^{1/sin(x)}$ is also e
so we get:
$lim_{x \to 0} (\frac{e}{e})^{\frac{1}{sin^2(x)}}$
$lim_{x \to 0} (1)^{\frac{1}{sin^2(x)}}$   =    1
 A: $$\\ \lim _{ x\rightarrow 0 }{ { \left( \frac { 1+tan(x) }{ 1+sin\left( x \right)  }  \right)  }^{ \frac { 1 }{ sin^{ 3 }(x) }  } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { tan(x)-\sin { \left( x \right)  }  }{ 1+sin\left( x \right)  }  \right)  }^{ \frac { 1 }{ sin^{ 3 }(x) }  } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { 1 }{ \frac { 1+sin\left( x \right)  }{ tan(x)-\sin { \left( x \right)  }  }  }  \right)  }^{ \frac { 1+sin\left( x \right)  }{ tan(x)-\sin { \left( x \right)  }  } \left( \frac { tan(x)-\sin { \left( x \right)  }  }{ 1+sin\left( x \right)  } \frac { 1 }{ sin^{ 3 }(x) }  \right)  } } =\\ =\lim _{ x\rightarrow 0 }{ { \left[ { \left( 1+\frac { 1 }{ \frac { 1+sin\left( x \right)  }{ tan(x)-\sin { \left( x \right)  }  }  }  \right)  }^{ \frac { 1+sin\left( x \right)  }{ tan(x)-\sin { \left( x \right)  }  }  } \right]  }^{ \frac { tan(x)-\sin { \left( x \right)  }  }{ 1+sin\left( x \right)  } \frac { 1 }{ sin^{ 3 }(x) }  } }$$
if we simply this expression: $$\frac { tan(x)-\sin { \left( x \right)  }  }{ 1+sin\left( x \right)  } \frac { 1 }{ sin^{ 3 }(x) } $$ 
$$\frac { tan(x)-\sin { \left( x \right)  }  }{ 1+sin\left( x \right)  } \frac { 1 }{ sin^{ 3 }(x) } =\frac { \sin { \left( x \right)  } \left( \frac { 1 }{ \cos { \left( x \right)  }  } -1 \right)  }{ 1+sin\left( x \right)  } \frac { 1 }{ sin^{ 3 }(x) } =\frac { 1-\cos { \left( x \right)  }  }{ \cos { \left( x \right) \left( 1+sin\left( x \right)  \right)  }  } \frac { 1 }{ \sin ^{ 2 }{ \left( x \right)  }  } =\frac { 1-\cos { \left( x \right)  }  }{ \cos { \left( x \right) \left( 1+sin\left( x \right)  \right)  }  } \frac { 1 }{ 1-\cos ^{ 2 }{ \left( x \right)  }  } =\\ \frac { 1-\cos { \left( x \right)  }  }{ \cos { \left( x \right) \left( 1+sin\left( x \right)  \right)  }  } \frac { 1 }{ \left( 1-\cos { \left( x \right)  }  \right) \left( 1+\cos { \left( x \right)  }  \right)  } =\frac { 1 }{ \cos { \left( x \right) \left( 1+sin\left( x \right)  \right) \left( 1+\cos { \left( x \right)  }  \right)  }  } $$ so 
$$ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { tan(x)-\sin { \left( x \right)  }  }{ 1+sin\left( x \right)  } \frac { 1 }{ sin^{ 3 }(x) }  }  }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ \cos { x } \left( 1+sin\left( x \right)  \right)  } \frac { 1 }{ 1+{ \cos { x }  } }  }  }={ e }^{ \frac { 1 }{ 2 }  }$$
A: If you already know about Taylor series, the problem can easily be solved considering $$A=\left(\frac{1+\tan (x)}{1+\sin(x)}\right)^{\frac 1{\sin^3(x)}}\implies \log(A)=\frac 1{\sin^3(x)}\log\left(\frac{1+\tan (x)}{1+\sin( x)}\right)$$ Now, for small values of $x$ $$\tan( x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Replacing and performing the long division $$\frac{1+\tan (x)}{1+\sin( x)}=1+\frac{x^3}{2}+\cdots$$ Now, using that fact that, for small $y$, $\log(1+y)\approx y$ and using $y=\frac{x^3}{2}$, we then have $$\log(A)\approx \frac 1{\sin^3(x)}\frac{x^3}{2}=\frac 1 2 \left(\frac x {\sin(x)}\right)^3$$ So, $\log(A)\to \frac 12$ and $A\to \sqrt e$.
