Testing whether the following integral converges 
Determine whether the following integral converges or diverges:
  $$I=\int_0^\infty\frac{x^{80}+\sin(x)}{\exp(x)}\,dx.$$

Since 
$$0\overset{?}{<}\frac{x^{80}+\sin(x)}{e^x}\leq\frac{x^{80}+1}{e^x}\sim\frac{x^{80}}{e^x}$$
and if we let $f(x)=x^{80}/e^x$ and $g(x)= \sqrt{x}\,e^{-\sqrt{x}}\,$ we have that
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$$ therefore if $\int_0^\infty g(x)\,dx$ converges so does $I$. But
$$\int_0^\infty\sqrt{x}\,e^{-\sqrt{x}}\,dx\to {\small{\begin{bmatrix}&u=\sqrt{x}&\\&du=\frac{1}{2\sqrt{x}}dx&\end{bmatrix}}}
\to\int_0^\infty e^{-u}\,du=1.$$
Thus $I$ does converge. 
Now, the limit used above is not too trivial to show that it is $0$. Intuitively, it is easy to see why, but on a more rigorous aspect, L'Hôpital's Rule here would get messy. Thus, my question, is there a more straightforward way to show that the above integral converges? 

The term $\sin x\,e^{-x}$ obviously converges.
Now by the Taylor development,
$$e^x>\frac{x^{82}}{82!}$$ and
$$x^{80}e^{-x}<82!\,x^{-2},$$ which has a convergent integral.
