Computing limit involving exponential function Could you help me to show that
$$
\lim_{x\rightarrow \infty}\frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}}=1
$$

My attemp (incomplete)
$$
\lim_{x\rightarrow \infty}\frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}}=\frac{1-e^{0}}{0}=\frac{0}{0}
$$
I apply L'Hopital 
$$
\lim_{x\rightarrow \infty}\frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}}=\lim_{x\rightarrow \infty}\frac{-e^{-e^{-x}}(-e^{-x})(-1)}{e^{-x-e^{-x}}(-1-e^{-x}(-1))}=\frac{0}{0}
$$
I apply again Hopital and get again $\frac{0}{0}$ ...
What am I doing wrong?
 A: Alternatively:
$$\lim_{x\to\infty}\frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}}=\lim_{x\to \infty}\frac{1-\frac{1}{e^{e^{-x}}}}{e^{-(x+e^{-x})}}=\lim_{x\to \infty}\frac{e^{e^{-x}}-1}{e^{-x}}=1.$$
A: Folllowing Matthew's hint,
$$\frac{1-e^{-t}}{t\,e^{-t}}=\frac{e^t-1}t$$
and the limit is to $t\to0^+$.
A: Note that since $e^{-x}\to 0$ by standard limits for $t\to 0 \quad \frac{e^t-1}{t}\to 1$
$$\frac{1-e^{ -e^{-x} } }{e^{-(x+e^{-x})}}=\frac{1-e^{ -e^{-x}} }{-e^{-x}}\frac{-e^{-x}}{e^{-(x+e^{-x})}}=\frac{1-e^{ -e^{-x}} }{-e^{-x}}(-e^{-x+x+e^{-x}})=\frac{e^{ -e^{-x}} -1}{-e^{-x}}(e^{e^{-x}})\\=\frac{e^{ t} -1}{t}(e^{-t})\to1\cdot 1 =1$$
Note that by your method from here
$$\lim_{x\rightarrow \infty}\frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}}=\lim_{x\rightarrow \infty}\frac{-e^{-e^{-x}}(-e^{-x})(-1)}{e^{-x-e^{-x}}(-1-e^{-x}(-1))}=\lim_{x\rightarrow \infty}\frac{-e^{-x-e^{-x}}}{e^{-x-e^{-x}}(-1+e^{-x})}$$
we can cancel out the $e^{-x-e^{-x}}$ term and obtain
$$\lim_{x\rightarrow \infty}\frac{-\color{red}{e^{-x-e^{-x}}}}{\color{red}{e^{-x-e^{-x}}}(-1+e^{-x})}=\lim_{x\rightarrow \infty}\frac{-1}{-1+e^{-x}}=1$$
