Computing limit involving exponential function

Question

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

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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?

Answer 1

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

$$\lim_{x\rightarrow \infty}\frac{-\color{red}{e^{-x-e^{-x}}}}{\color{red}{e^{-x-e^{-x}}}(-1+e^{-x})}=\lim_{x\rightarrow \infty}\frac{-1}{-1+e^{-x}}=1$$

Answer 2

Alternatively: $$\lim_{x\to\infty}\frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}}=\lim_{x\to \infty}\frac{1-\frac{1}{e^{e^{-x}}}}{e^{-(x+e^{-x})}}=\lim_{x\to \infty}\frac{e^{e^{-x}}-1}{e^{-x}}=1.$$

Answer 3

Folllowing Matthew's hint,

$$\frac{1-e^{-t}}{t\,e^{-t}}=\frac{e^t-1}t$$

and the limit is to $t\to0^+$.