# Why one comparison test and not the other

I'm studying integral comparison tests and I come to this one.

$$f(x)=\int_{1}^{\infty}\frac{\sqrt{x}}{x^5+\sqrt[3]{x}}dx$$

The solution provided is to do $\displaystyle g(x)=\int_{1}^{\infty}\frac{\sqrt{x}}{x^5}dx$ and use the limit comparison test to find if $f(x)$ is convergent or not.

My question is this. Why can't I use the standard comparison test? Isn't $$0\leq f(x) \leq g(x)$$

Since $g(x)$ is convergent because $\displaystyle \frac{\sqrt{x}}{x^5} = \frac{1}{x^\frac{9}{2}}$ I believe that it's easier to use the standard comparison test and skip the limit comparison test

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You probably have somewhere on your notes that they're equivalent. You can use whatever you want. – Git Gud Jan 26 '13 at 10:50
I think,you mean the integrand is less than or equal to $g(x)$. Isn't it? I mean is $f(x)$ the integrand or $f(x)$ itself is an improper integral? – Babak S. Jan 26 '13 at 10:50
@GitGud Hi. No. I only have the LCT. – Favolas Jan 26 '13 at 10:52
@BabakSorouh Sorry. Yes. You're right. It's the integrand – Favolas Jan 26 '13 at 10:53

One remark: you should avoid writing things like $f(x)=\int...dx$ since the $x$ on the LHS means something else than the $x$ on the RHS.
Quick question. I can use the SCT in $\int_{0}^{1}\frac{1}{x^3+\sqrt{x}}dx$ making $g(x)=\int_{0}^{1}\frac{1}{\sqrt{x}}dx$ because $0\leq f(x) \leq g(x)$ Is this safe to assume? – Favolas Jan 26 '13 at 12:01
It is $g(x)=\frac{1}{\sqrt{x}}$. – André Nicolas Jan 26 '13 at 17:43