$\displaystyle \lim_{x\,\to\,\infty}e^{-e^{x}}\int_0^x e^{e^{t}}dt$
Thanks in advance.
Express the limit as $\displaystyle \lim_{x\to\infty} \frac{\int_0^x e^{e^t}dt}{e^{e^x}}$. Both the numerator and denominator approach infinity, so by L'Hopital's rule and the fundamental theorem of calculus, the limit equals $\displaystyle \lim_{x\to\infty} \frac{e^{e^x}}{e^{e^x}e^x} = \lim_{x\to\infty} e^{-x} = 0$.