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!

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    Well, what about cases where we do not have differentiability? – Hagen von Eitzen Nov 17 '15 at 23:09
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    A lot of the limits can be solved with series expansions and approximations. – Gabor Bakos Nov 18 '15 at 11:32
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    Closely related: [Are there any situations in which L'Hopital's Rule WILL NOT work?](http://math.stackexchange.com/q/1521718/72968) – Silverfish Nov 18 '15 at 21:15
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    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. – user2357112 Nov 18 '15 at 21:50
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    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.). – asmeurer Nov 19 '15 at 20:31

8 Answers8


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 $2$).

hmakholm left over Monica
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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.

David E Speyer
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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.

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    This is the flavour of what I was looking for, thanks! – user291130 Nov 17 '15 at 23:33
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    @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 '15 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. – Irregular User Nov 19 '15 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.

Mark Viola
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    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? – Kevin Nov 19 '15 at 04:13
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    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? – Mark Viola Nov 19 '15 at 05:12

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

marty cohen
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    +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. – Ethan Bolker Nov 18 '15 at 19:13
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    @EthanBolker: Exactly. And if any student can **convince** me that L'Hopital's rule is valid, that same student will probably hardly use it! – user21820 Nov 19 '15 at 06:00
  • 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 :) – Miguel Jun 22 '20 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.

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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


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 $|a| < 1$.

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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!

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