Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule? Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule? It seems like for every limit that results in an indeterminate form, you may as well use lHopital's Rule rather than any of the methods that are taught prior to the rule, such as factoring, rationalizing, trig limits, etc.
EDIT: These are great answers, thanks!

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
$a = b$, 
$r(x) \to \frac{F(0)}{G(0)}$.
