0 or infinite limit in limit comparison test for improper integral

In this doc, the steps for using the Limit Comparison test are:

1. Use the LCT when trying to determine whether $\int_{a}^{\infty} f(x) \, dx$ converges and the function $f(x)$ is positive and looks complicated.
2. Find a function $g(x)$ which grows at a similar rate as $f(x)$ and is simpler.
3. Compute $\lim_{x \to \infty} \frac{f(x)}{g(x)}$
4. Make sure the limit exists and is not $0$ or $\infty$.
5. Note that if this limit does not exist, $0$ or $\infty$, then the LCT is inconclusive, and cannot be applied.

However multiple textbooks state the following

If $\frac{f}{g} \to 0$ and $\int_a^\infty g(x) dx$ converges, then $\int_a^\infty f(x) dx$ converges, and if $\frac{f}{g} \to \infty$ and $\int_a^\infty g(x) dx$ diverges, then $\int_a^\infty f(x)dx$ diverges.

Using test for limit as 0 gives incorrect answer in some cases as

For convergence of $\int_{0}^{\frac{\pi}{2}} \frac{\sin{x}}{x^{p}} dx$

we have choosing $g(x) = \frac{1}{x^p}$

$\lim_{x \to 0} \frac{\sin{x} / x^{p}}{1 / x^{p}} = \lim_{x \to 0} \sin{x} = 0$

which gives range of convergence for p as $p<1$, however if we take $g(x) = \frac{1}{x^{"p-1}}$ we will get limit as 1 and range as $p<2$ which is the correct answer.

So in inclusion can we use the limit comparison test if limit is $0$ or $\infty$?