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? 
 A: 
Is there a more straightforward way to show that the above integral converges?

We can actually evaluate the integral of interest.  Note that we have
$$\begin{align}
\int_0^\infty (x^{80}+\sin(x))e^{-x}\,dx&=\int_0^\infty x^{80}e^{-x}\,dx+ \int_0^\infty \sin(x)e^{-x}\,dx\\\\
&\Gamma(81)+\frac12\\\\
&=80!+0.5
\end{align}$$
A: 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.
A: There is no need for L'Hopital's Rule here. To show that $\int_0^{\infty} \frac {x^{80}} {e^{x}}\, dx <\infty$ note that $\frac {x^{80}} {e^{x}} <e^{-x/2}$ or $x^{80} <e^{x/2}$ for $x$ sufficiently large: $e^{x/2} > \frac {(x/2)^{100}} {100!} >x^{80}$ if $x^{20}>\frac {2^{100}} {100!}$. Since $e^{-x/2}$ is integrable we are done. 
A: As was pointed out in the comments, you could have simply use $e^{x/2}$. However even the function you have used can be proved in the limit. Start with dividing $f(x)$ by $g(x)$
$$\frac{f(x)}{g(x)} = \frac{x^{79.5}}{e^{\sqrt{x}}}$$
Now, generally we know that for any $a$, the limit $x^a/e^x$ as x tends to infinty is $0$. In this case, however, we have $e^{\sqrt{x}}$. Now if you start applying the $L'Hôpital's Rule$, you will see that this limit also degrades albeit at only half the exponential rate i.e. the numerator decreases by $\sqrt{x}$ every time.
For example, applying the rule for the first time will give you 
$$2(79.5)\frac{x^{79.5 - 1}\sqrt{x}}{e^{\sqrt{x}}} = 2(79.5)\frac{x^{79}}{e^{\sqrt{x}}}$$
Here the numerator decreased by $0.5$ power of x whereas the denominator remained the same. So if you apply the rule sufficiently enough number of times, we'll get the limit as $0$.
