Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule?

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 $2$).

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.

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.

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.

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