Limit of $\left(\left(\frac{1-x}{e}\right)^{1/x}\right)^{1/\sin x} $ when $x\to0$ How would I start for computing this limit:
$$ \lim_{x \to 0} \left( \left( \dfrac{1-x}{e} \right) ^ { \dfrac{1}{x} } \right) ^ { \dfrac{1}{\sin x} } $$
My father asked this question. He does not know whether it actually has a real limit or not.
 A: Don't use L'Hopital as it tells you almost nothing about the behaviour, and in fact cannot be used directly on this question as Shantty's answer attempts to, since the expression is not of a suitable indeterminate form.
As $x \to 0$:
  $\left( \dfrac{1-x}{e} \right) ^ { \frac{1}{x\sin(x)} } = \exp\left( \dfrac{\ln(1-x)-1}{x\sin(x)} \right)$
  $\in \exp\left( \dfrac{1}{x^2} \dfrac{-1-x-\frac{1}{2}x^2+O(x^3)}{1-\frac{1}{6}x^2+O(x^4)} \right)$ [by Taylor expansions]
  $\subseteq \exp\left( \dfrac{1}{x^2} (-1-x-\frac{1}{2}x^2+O(x^3))(1+\frac{1}{6}x^2+O(x^4)) \right)$
  $\subseteq \exp\left(-\dfrac{1}{x^2}-\dfrac{1}{x}-\dfrac{2}{3}+O(x)\right)$
  $\subseteq \exp\left(-\dfrac{1}{x^2}-\dfrac{1}{x}\right) e^{-\frac{2}{3}}(1+O(x))$
  $\to 0$ [because $\exp\left(-\dfrac{1}{x^2}-\dfrac{1}{x}\right) \to 0$].
This shows not only the limit but exactly how fast it goes to $0$.
Comments
The notation above is a rigorous Landau notation, unlike the slipshod abuse of notation using the equals-sign. To be precise, "$t(x) \in O(f(x))$ as $x \to 0$" is a standard short-form for:

$\exists c\ \exists δ>0\ \forall x\ ( |x|<δ \Rightarrow |t(x)| < c|f(x)| )$.

A: First take the logarithm of
$$\left(\frac{1-x}e\right)^{\frac1{x\sin(x)}}$$then take the limit.
You'll get
$$\lim_{x\to 0}e^{\frac{\ln(1-x)-1}{x\sin(x)}}=0$$
because 
$$\lim_{x\to 0}\frac{\ln(1-x)-1}{x\sin(x)}=-\infty \tag 1$$
and because the exponential function is continuously decreasing through positive numbers towards zero as its argument is tending to $-\infty.$
$(1)$ is easy to see without using the L'Hospital rule. (Just think of $\ln(1)=0$, and of the continuity of the logarithmic function.)
A: By taking log on both sides. 
$$\lim_{x\to0} \frac{\log(1-x) - 1}{x\sin x}$$
now proceed by l'Hôpital rule. 
