I like expanding
the terms in
$\frac{f(x)}{g(x)}
$
into polynomials
and seeing what happens
as $x \to 0$.

I also freely use
the "big-oh" and
(less often)
the "little-oh" notation.

For example,
one of the answers used
$\lim_{x \to 0} \frac{\sin \sqrt{x}}{\sqrt{x}}
$.

Since
$\sin(x)
=x+O(x^3)
$,
$\sin(\sqrt{x})
=\sqrt{x}+O(x^{3/2})
$
so
$\frac{\sin \sqrt{x}}{\sqrt{x}}
=\frac{\sqrt{x}+O(x^{3/2})}{\sqrt{x}}
=1+O(x)
\to 1
$.

Similarly,
for the example
$\lim_{x\to \infty}\frac{x}{\sqrt{x^2+1}}
$,
since
$\sqrt{x^2+1}
=x\sqrt{1+\frac{1}{x^2}}
=x(1+\frac{1}{2x^2}+O(1/x^4))
=x(1+O(1/x^2))
$
so
$\frac{x}{\sqrt{x^2+1}}
=\frac{x}{x(1+O(1/x^2))}
=\frac{1}{1+O(1/x^2)}
\to 1
\text{ as } x \to 0
$.

Since both $f$ and $g
\to 0$
as $x \to 0$,
we must have
$f(x)
=x^aF(x)
$
and
$g(x)
=x^bG(x)
$
where
$a>0$, $b>0$,
$F(0) \ne 0$,
and
$G(0) \ne 0$.

Therefore
$r(x)
=\frac{f(x)}{g(x)}
=\frac{x^aF(x)}{x^bG(x)}
=x^{a-b}\frac{F(x)}{G(x)}
$.

If
$a > b$,
$r(x) \to 0$;
if
$a < b$,
$r(x) \to \infty$;
and
if
$a = b$,
$r(x) \to \frac{F(0)}{G(0)}$.