Can we use limit comparison test with Riemann integrable functions?

Assume I have a function $$f(x)$$ and I want to know if the improper integral $$\intop_{a}^{b}f\left(x\right)dx$$ converge/diverge.

Can I use the limit comparison test with a riemann integrable function $$g$$ ?

For example, I want to check if the integral $$\intop_{0}^{1}\frac{\sin^{2}\left(x^{3}\right)}{x^{6}}dx$$ converge.

So can I use the limit comparison test with $$x^3$$ ? because that will lead me to:

$$\frac{\frac{\sin^{2}\left(x^{3}\right)}{x^{2}}}{x^{3}}=\frac{\sin^{2}\left(x^{3}\right)}{x^{6}}=\left(\frac{\sin\left(x^{3}\right)}{x^{3}}\right)^{2}\underset{x\to0}{\longrightarrow}1$$

Well, actually the "improper" integral that I gave as example is also riemann integrable since that limit exists at $$x=0$$. But thoretically speaking, will this work ? is it legit?

Riemann improper integrals are defined for 2 cases: when function is unbounded and when integration area is unbounded. In your case, as you have checked it, your function is continuous on $$(0,1]$$ and have limit in $$0^+$$, so it can be continued on $$[0,1]$$ continuously, therefore is even uniformly continuous and bounded. Integral from it is usual integral, not improper.
If $$0\leqslant g(x) \leqslant f(x)$$ on some interval $$[a, \infty)$$, $$f,g$$ integrable on each $$[a,b]$$, then if $$\int\limits_{a}^{\infty}f(x)\,dx$$ converges, so does $$\int\limits_{a}^{\infty}g(x)\,dx$$. If last diverges diverges, so does former.