# 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!

• Well, what about cases where we do not have differentiability? Nov 17, 2015 at 23:09
• A lot of the limits can be solved with series expansions and approximations. Nov 18, 2015 at 11:32
• Nov 18, 2015 at 21:15
• Well, limits that don't have indeterminate forms, like $\lim_{x \to 2}3$, are much easier to solve without l'Hopital's rule than with it. Nov 18, 2015 at 21:50
• It probably goes without saying that it's a poor choice for expressions that aren't in an indeterminate form ($\frac{0}{0}$, $\frac{\infty}{\infty}$, $0\cdot \infty$, $0^0$, etc.). Nov 19, 2015 at 20:31

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 $2$).

• Doesn't l'Hopital's Rule typically state that the limit $\frac{f'(x)}{g'(x)}$ must converge? In this case, l'Hopital's Rule is simply not applicable. Nov 17, 2015 at 23:29
• x @user291130: Sure -- isn't that a prime example of a case where other methods are easier (because they work at all) than L'Hospital (which doesn't work at all)? Nov 17, 2015 at 23:30
• Fair enough, thanks! Nov 17, 2015 at 23:33
• The key word in this answer was "try". You can try L'H&ocirc;pital's Rule, but you won't succeed. Nov 19, 2015 at 15:22
• @Henning, How did you know that the denominator tends to $\infty$? (Since, we don't know what sin(x) tends to when x$\to\infty$). Oct 18, 2017 at 13:49

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.

• This one is amusing. Nov 18, 2015 at 1:08
• Of course, if you do a bit of simplification after one round of L'Hospital, you get $\tfrac{(1/2) x^{-1/2} \cos \sqrt{x}}{(-1/2) x^{-1/2}} = \cos \sqrt{x}$, which approaches $1$. One could probably think of how to make a harder version that doesn't simplify like this, but at that point doing it by other means is also harder... Nov 18, 2015 at 3:01
• Wouldn't $\dfrac{2xe^{x^2}}{e^x} = 2x e^{x(x-1)}$ count as a minor algebraic cancellation? Nov 18, 2015 at 9:12
• $$\lim_{x\to \infty}\frac{2^x}{3^x}$$ is even simpler. Nov 19, 2015 at 20:25
• @MrReality You're right, that's an error. I can't edit a 2 year old comment but, for the record, that minus sign is wrong. Oct 20, 2017 at 17:54

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.

• This is the flavour of what I was looking for, thanks! Nov 17, 2015 at 23:33
• @user291130 Actually, my example is far from the best. The answer by Henning Makholm is much better. My limit can be calculated by L'Hospital, it's just hard to do it. His actuallly cannot
– 5xum
Nov 17, 2015 at 23:37
• It's also worth noting that if we use L'Hôpital's rule, then we also need to spend some time verifying the hypotheses. Nov 19, 2015 at 22:40

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.

• Is it possible to conclude that, since the limit is equal to its reciprocal by LHR, it must be 1 or -1, or did we just get lucky in this specific case? Nov 19, 2015 at 4:13
• For LHR to be applicable, one needs to show that the limit of the ratio of derivatives exists. Have we shown this, or are we flailing away? Nov 19, 2015 at 5:12

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

• +1. This doesn't answer the letter of the OPs question, but does get to the spirit. L'Hopital is a "rule". This is a method that helps you understand what is going on. Nov 18, 2015 at 19:13
• @EthanBolker: Exactly. And if any student can convince me that L'Hopital's rule is valid, that same student will probably hardly use it! Nov 19, 2015 at 6:00
• Expanding $\sqrt{x}$ into a polynomial at $x=0$ is not easy to justify pedagogically at the level of first year calculus... although it works :) Jun 22, 2020 at 22:16

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.

$$\lim_{x\to\infty}\frac{2^x}{3^x}$$

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 $|a| < 1$.

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!