limit of indeterminate of type $$\lim_{x\to0}(1-x)^\frac{1}{x}$$
I know we have to use logarithms to evaluate the limit, but isn't it the type of $1^{\infty }$? so the answer should be prety straightforward as $1^{\infty }$ which is 1, but actually the answer is $e^{-7}$
 A: I think I know what you mean. To use the following proposition of limits:
$$\lim f=a,\lim g=b,then  \; \lim f^g=a^b $$
you have to make sure that $0\ne a,b$ should be real number, not $\infty$
So your claim:$1^{\infty}=1$ is not right 
A: For sufficiently small nonzero $x$, the quantity $(1 - x)^{1/x}$ is always positive, hence it may be rewritten as
\begin{align}
\exp(\log (1 - x)^{1/x}) = \exp\left(\frac{1}{x}\log(1 - x)\right)
\end{align} 
Now you can evaluate the exponent part by L'Hospital rule (it is of type $0/0$):
$$\lim_{x \to 0} \frac{\log(1 - x)}{x} = -\lim_{x \to 0}\frac{1}{1 - x} = -1.$$
Hence the original limit is $e^{-1}$, by the continuity of function $x \mapsto e^x$. 
A: You also could have used Taylor series $$y=(1-x)^\frac{1}{x}\implies \log(y)=\frac{1}{x}\,\log(1-x)$$ Now, Taylor series $$\log(y)=\frac{1}{x}\,\left( -x-\frac{x^2}{2}+O\left(x^3\right)\right)=-1-\frac{x}{2}+O\left(x^2\right)$$ When $x \to 0$, then $\log(y)\to -1$ and $y\to \frac 1e$.
Notice that you obtain the limit and also how it is approached since, using $y=e^{\log(y)}$ and Taylor again $$y=\frac{1}{e}-\frac{x}{2 e}+O\left(x^2\right)$$
