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!
 A: If you try to use L'Hospital's rule to evaluate
$$ \lim_{x\to\infty} \frac{2x}{x+\sin x} $$
you end up with
$$ \lim_{x\to\infty} \frac{2}{1+\cos x} $$
which spectacularly fails to converge. But the original limit does exist (it is $2$).
A: 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
$a = b$, 
$r(x) \to \frac{F(0)}{G(0)}$.
A: It was suggested in the comment but never given explicit mention in an answer, so here is a simple limit that cannot be computed by L'Hopital rule simply because it is not differentiable.

$\lim_{t\to 0} \dfrac{1}{t \lfloor \frac{1}{t} \rfloor} = 1$.

For more details on a more intuitive technique to handle everything that L'Hopital's rule can and more, see this related post on MathEducators SE, where usually an asymptotic form of the Taylor theorem (I give a proof here) gives far more information than L'Hopital's rule.
A: L'Hospital will never clear up
$$\lim_{x \to 0} \frac{\sin \sqrt{x}}{\sqrt{x}},$$
you just get $\infty/\infty$ over and over. 
On thinking it over,
$$\lim_{x \to \infty} \frac{e^{x^2}}{e^x}$$ 
is probably a better example. It also is easy to compute by other means, and gives $\infty/\infty$ with every L'Hopital iteration, but doing minor algebraic cancellations to the L'Hopital result doesn't solve the problem.
A: $$\lim_{x\to\infty}\frac{2^x}{3^x}$$
has the form $\frac{\infty}{\infty}$, and it's very tempting to use l'Hopital's rule (at least tempting to a calculus student who has learned "limit of fraction $\Rightarrow$ l'Hopital's rule"). But applying it just gives 
$$\lim_{x\to\infty}\frac{\ln(2)2^x}{\ln(3)3^x},$$
which of course is just the same limit with a constant factor in front. The better way to compute this limit is to rewrite it as $$\lim_{x\to\infty}\left(\frac{2}{3}\right)^x$$ and use a theorem that says that $\lim_{x\to\infty}{a^x} = 0$ if $|a| < 1$.
A: If you want to evaluate
$$\lim_{x\to\infty}\frac{x^{1000000}+x^{999999}+\cdots+x+1}{x^{1000000}}$$
purely by use of L'Hopital's Rule, you will have to apply the rule a million times!
A: If you ask me, factoring is very often easier than L'Hospital.
For example, try using L'Hospital on the limit
$$\lim_{x\to 0}\frac{xe^{\cos x^2} + x\ln\left(\frac{1}{\arctan x+1}\right)}{x^2 + x\sin\left(1+\arccos(x)\right)}$$
Using L'Hospital is of course possible here, but not really recommended, since you can just factor out $\frac{x}{x}$ and then plug in $x=0$ into whatever remains.
A: There are classic examples in which successive use of L'Hospital's Rule (LHR) result in an indefinite loop.  For example, examine the limit 
$$\lim_{x\to \infty}\frac{x}{\sqrt{x^2+1}} =1$$
If we attempt to evaluate using LHR, we find
$$\lim_{x\to \infty}\frac{x}{\sqrt{x^2+1}}=\lim_{x\to \infty}\frac{1}{\frac{x}{\sqrt{x^2+1}}}=\lim_{x\to \infty}\frac{\sqrt{x^2+1}}{x}$$
Oh no!  The new limit of indeterminate form has flipped the numerator and denominator.  Therefore, a second application of LHR will simply recover the original form of this limit.  And we can never escape this loop.
Of course, we can evaluate this limit easily.  But, the "blind" use of LHR will fail in this case and similar cases in which one never escapes a dreaded loop.
