$\lim_{x\to 0}(1+\sin(x)-x)^{1/x^3}$ Evaluate$$\lim_{x\to 0}\left(1+\sin(x)-x\right)^{{1}/{x^3}}$$
i was able to solve it using a taylor exp' for $\sin(x)$ but id like to know if there is a "simpler " way. something along the lines of $e^{\log}$-ing it, or L'hopital-ing it...
 A: Since $\lim _{ x\rightarrow 0 }{ { \left( 1+x \right)  }^{ \frac { 1 }{ x }  }=e, } $ 
so
$$\lim _{ x\rightarrow 0 }{ { \left( 1+\sin { x } -x \right)  }^{ \frac { 1 }{ { x }^{ 3 } }  } } =\lim _{ x\rightarrow 0 }{ { \left[ { \left( 1+\sin { x } -x \right)  }^{ \frac { 1 }{ \sin { x } -x }  } \right]  }^{ \frac { \sin { x } -x }{ { x }^{ 3 } }  } } ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } -x }{ { x }^{ 3 } }  }  }$$
Now to compute the exponent term we need to use L'Hôpital's rule 4 times: 
$$\lim _{ x\rightarrow 0 }{ \frac { \sin { x } -x }{ { x }^{ 3 } }  } =\lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3{ x }^{ 2 } }  } =\lim _{ x\rightarrow 0 }{ \frac { -\sin { x }  }{ 6x } =\lim _{ x\rightarrow 0 }{ \frac { -\cos { x }  }{ 6 } =-\frac { 1 }{ 6 }  }  }, $$
so the final answer is

$$={ e }^{ -\frac { 1 }{ 6 }  }$$

A: L'Hopital's rule makes this fairly easy
$y = \lim_\limits{x\to 0} (1+sin x - x)^{\frac 1{x^3}}\\
\ln y = \lim_\limits{x\to 0}  \frac{\ln(1+sin x - x)}{x^3}$
At $x = 0,$ numerator and denominator both equal $0.$ Apply L'hopital's rule:
$\ln y = \lim_\limits{x\to 0}  \dfrac{\frac{\cos x - 1}{(1+\sin x - x)}}{3x^2}$
and now, rather than doing L'hopital's again, I think I will use a Taylor series.
$\ln y = \lim_\limits{x\to 0}  \dfrac{-\frac12x^2 + O(x^4)}{3x^2(1- O(x^3))}$
$\ln y = -\frac 16\\
y = e^{-\frac 16}$
A: You know that $\lim_{x\to a}f(x)^{g(x)}$ exists if and only if $\lim_{x\to a}g(x)\log f(x)=l$ exists and, in this case, the original limit is $e^l$ (with the extended rule that, if $l=-\infty$, the original limit is $0$ and, if $l=\infty$, the original limit is $\infty$).
So let's compute
$$
\lim_{x\to0}\frac{\log(1+\sin x-x)}{x^3}
$$
The Taylor expansion of $\log(1+t)$ is $t+o(t)$; moreover the Taylor expansion of $\sin x$ is $x-x^3/6+o(x^3)$. Thus
\begin{align}
\log(1+\sin x-x)
&=\log\left(1+x-\frac{x^3}{6}+o(x^3)-x\right) \\
&=\log\left(1-\frac{x^3}{6}+o(x^3)\right) \\
&=-\frac{x^3}{6}+o(x^3)
\end{align}
So we have
$$
\lim_{x\to0}\frac{\log(1+\sin x-x)}{x^3}=
\lim_{x\to0}\frac{-x^3/6+o(x^3)}{x^3}=-\frac{1}{6}
$$
Thus your limit is
$$
\lim_{x\to0}(1+\sin x-x)^{1/x^3}=e^{-1/6}
$$
