Determining whether an improper integral converges or diverges. $$\int_{1}^{\infty}\dfrac{\sqrt{x^7+2}}{x^4}\text{dx}$$
I was told to let $f(x)=\dfrac{\sqrt{x}}{x^4}$ and $g(x)=\dfrac{\sqrt{x^7+2}}{x^4}$ then find the limit as $x$ approaches $\infty$ of $\dfrac{f(x)}{g(x)}$ and $\int_{1}^{\infty}f(x)\text{dx}$.
I found that $\lim_{x \to \infty} \dfrac{f(x)}{g(x)} =0$. According to the limit comparison test, $0<L<\infty$, since the limit is $0$, this integral will diverge. Is this correct?
Also how do you determine $f(x)$ and $g(x)$ to use the limit comparison test? 
 A: $\displaystyle{x \gg 1\,,\quad\mbox{integrand}\ \sim x^{-1/2}\,,\quad\mbox{integral}\ \sim x^{1/2}:\
\mbox{Diverges !!!.}}$
A: By the test you have mentioned, the integral will diverge.
Think about it for a second and it will make sense. With that limit you showed that $f$ "grows slower" than g. Finding out that the integral of $f$ diverges means that $f$ grows "too fast" to converge.
Then, also g grows too fast and the integral will diverge. That's the meaning of the limit comparison test.
Well, $g$ will probably always be the function of interest to integrate and $f$ will usually be a simpler function to integrate and that grows faster or slower than $g$.
Looking at $g$, you can see that as $x \to \infty$, the function behaves like $x^{-1/2}$. So your intuition should tell that $\int g$ diverges, and you must choose $f$ such that $\int f$ also diverges (easily provable) and grows slower than $g$, so you can use limit comparison test.
A: What you were told doesn't make sense.  What I would do if I were you is to find a simpler function that is bounded by the numerator, say $f(x)=x^{3.5}$, as $x \rightarrow \infty$ and another that bounds it, say $f(x)=x^{3.5}+2$.  Now, with $g(x) = x^4$ you ought to be able to perform the resulting integral for both of these $f(x)$ and show that it either converges or diverges.
