I don't understand something about L'Hôpital's rule. In this case:

$$ \begin{align} & {{}\phantom{=}}\lim_{x\to0}\frac{e^x-1-x^2}{x^4+x^3+x^2} \\[8pt] & =\lim_{x\to0}\frac{(e^x-1-x^2)'}{(x^4+x^3+x^2)'} \\[8pt] & =\lim_{x\to0}\frac{(e^x-2x)'}{(4x^3+3x^2+2x)'} \\[8pt] & =\lim_{x\to0}\frac{(e^x-2)'}{(12x^2+6x+2)'} \\[8pt] & = \lim_{x\to0}\frac{(e^x)'}{(24x+6)'} \\[8pt] & = \lim_{x\to0}\frac{e^x}{24} \\[8pt] & = \frac{e^0}{24} \\[8pt] & = \frac{1}{24} \end{align} $$

Why do we have to keep on solving after this step:


Can't I just plug in $x=0$ and compute the limit at this step giving me:


I'm very confused, because I get different probable answers for the limit, depending on when do I stop to differentiate, as clearly $-\frac1{2}\neq \frac 1{24}$.

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    $\begingroup$ You ought to stop already at $\frac{e^x-2x}{4x^3+3x^2+2x}$, which gets you to $\frac{1}{0}$. The limit doesn't exist. $\endgroup$ – Daniel Fischer Aug 6 '13 at 13:59
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    $\begingroup$ it has to be 0/0 or infinity/infinity to use L'hopital per iteration $\endgroup$ – WhizKid Aug 6 '13 at 14:00
  • $\begingroup$ Please make your question self-contained. At present it is not since to get crucial information requires going to an offsite (and possibly transient) page. $\endgroup$ – Did Aug 6 '13 at 14:10
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    $\begingroup$ You're missing "$\lim\limits_{x\to0}$" in front of your third fraction on the first line. Without that, the "$=$" just before that is not true. $\endgroup$ – Michael Hardy Aug 6 '13 at 14:20
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    $\begingroup$ You should stop using L'Hôpital's rule if you feel dizziness, if it begins to interfere with your relationships with other mathematicians, or if you find yourself obsessively thinking about how to apply L'Hôpital's rule to objects in your environment, such as your cat. $\endgroup$ – LarsH Aug 7 '13 at 15:17

Once your answer is no longer in the form 0/0 or $\frac{\infty}{\infty}$ you must stop applying the rule. You only apply the rule to attempt to get rid of the indeterminate forms. If you apply L'Hopital's rule when it is not applicable (i.e., when your function no longer yields an indeterminate value of 0/0 or $\frac{\infty}{\infty}$) you will most likely get the wrong answer.

You should have stopped differentiating the top and bottom once you got to this:

$\dfrac{e^x-2x}{4x^2+3x^2+2x}$. Taking the limit at that gives you $1/0$. The limit is nonexistent.

Also, don't be tempted to say "infinity" when you see a 0 in the denominator and a non-zero number in the top. It may not be the case. For example, the function $\frac{1}{x}$ approaches infinity and negative infinity from both sides of the limit as x approaches 0. Its not necessarily infinite; its best just to leave it as "nonexistent".

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    $\begingroup$ Maybe I should rephrase: the two-sided limit does not exist. We can say that the one sided limit 0+ is infinite, and the other is negative infinite. $\endgroup$ – ra1nmaster Aug 6 '13 at 14:38
  • $\begingroup$ What I meant to say is this: what is your logic in saying that positive infinity equals negative infinity? It doesn't make sense to me. $\endgroup$ – ra1nmaster Aug 6 '13 at 14:48
  • $\begingroup$ Sorry, I have reread the Johnson-Freyd comment, and he says that +inf is identical to -inf indeed, mathoverflow.net/questions/10124/… So, I do not take your argument that you cannot write answer =∞ neither the premice that left limit ≠ right limit. $\endgroup$ – Val Aug 6 '13 at 15:01
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    $\begingroup$ @Val, it depends. If you are adding two infinities, $+\infty$ and $-\infty$, to the reals, which is the usual approach in analysis, then those two are utterly different. If you're only adding one infinity, $\infty$, as is sometimes done in, say, topology, then it is what it is. The qestion is whether you want something that goes up forever to have the same limite as something that goes down forever. $\endgroup$ – dfeuer Aug 6 '13 at 15:59
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    $\begingroup$ @Val Saying a limit infinity is qualitative information, in this case of the behaviour of the function near the point. The difference is between saying "its values go indefinitely far from the origin", $\lim=\infty$, to saying "its values go indefinitely far from the origin in the negative direction", $-\infty$, (or "in the positive direction", $+\infty$)". It is less information. In this case may not be important at all. On the other hand, having more information allows you to use the result if you need to do other stuff with it. $\endgroup$ – OR. Aug 6 '13 at 17:22

After differentiating just once, you get $$\lim_{x \to 0} \dfrac{e^x-2x}{4x^3+3x^2+2x}$$ which "evaluates" to $\dfrac 10$, i.e., the numerator approaches $1$, and the denominator approaches $0$. Hence, L'Hopital no longer applies and we have $$\lim_{x \to 0} \dfrac{e^x-2x}{4x^3+3x^2+2x}\quad\text{does not exist}.$$

L'Hopital's rule applies provided and only while a limit evaluates to an "indeterminate" form: e.g., $\dfrac 00, \;\text{or}\;\dfrac {\pm\infty}{\pm\infty}$.

  • $\begingroup$ Lets make it an even 20 TU's! :-) +1 $\endgroup$ – Amzoti Aug 7 '13 at 0:40
  • $\begingroup$ What if the numerator is positive infinity and the denominator is negative infinity? $\endgroup$ – The 29th Saltshaker Aug 27 '17 at 0:37
  • $\begingroup$ Isn't 1/0 also indeterminate form? $\endgroup$ – PGupta Aug 8 '18 at 15:30
  • $\begingroup$ No, it is not an indeterminate form, if you mean something like $\lim_{x\to 0} \frac 1{0}$; then we say the limit approaches infinity. $\endgroup$ – amWhy Aug 8 '18 at 15:33

A quick addition to Ra1nMaster's otherwise excellent answer: you can only apply L'Hopital's rule if you have an indeterminate form and if the limit, after applying L'Hopital's rule, exists.

This second condition is equally important; for instance a classic stumper is $$\lim_{x\to\infty} \frac{x}{x+\sin x}.$$ Since this limit has the form $\frac{\infty}{\infty}$, one might naively apply L'hopital's rule, getting $$\lim_{x\to\infty} \frac{1}{1+\cos x}$$ and concluding the original limit does not exist. This is wrong; $$\lim_{x\to\infty} \frac{x}{x+\sin x} = \lim_{x\to\infty} \frac{1}{1+\frac{\sin x}{x}}=1.$$

  • $\begingroup$ Inspired me +1 ! $\endgroup$ – mick Aug 26 '13 at 0:24
  • $\begingroup$ This explains subtle details. +1! $\endgroup$ – R004 Aug 2 '17 at 6:03
  • $\begingroup$ The method used by the top answers seem to be wrong, based on this. $\endgroup$ – Abraham Zhang Dec 4 '18 at 21:06

One must be very careful about using l'Hospital's rule. It applies only when the numerator and denominator both tend to $0$ or $\infty$ and the denominator is never $0$ in a punctured neighborhood of the point. A denominator of the form $x\sin \frac1x$ is ineligable for the rule.

But, students should try to avoid the rule anyway. Here is a parable. A student is assigned the task of finding

$$ \lim_{x\to 0} \frac{\sin^6 x}{x^6}. $$

A bad student cancels the $6$ and the $x$ giving $\sin$.

A naive student applies l'Hospital's rule 6 times and eventually gets $\frac{720}{720} = 1$.

A mediocre student applies the rule once, and gets

$$ \lim_{x\to 0} \frac{6\sin^5 x \cos x}{6x^5}. $$

He cancels the $6$, and removes the $\cos x$ term since it tends to $1$. He repeats the process 5 more times.

A good student writes the expression as:

$$ \lim_{x\to 0} \left[\frac{\sin x}{x}\right]^6. $$

He uses continuity of $t^6$ to move the limit inside the brackets and gets $1^6 = 1$.

An engineer says "$\sin x = x$, so the expression is $1$." And, he'd be right!

Using l'Hospital's rule can also hurt the students understanding of a problem. Here's a classic example.

The $p$-power mean of two positive numbers, $x$ and $y$ is defined as:

$$ M_p(x,y) = \left[\frac{x^p + y^p}{p}\right]^{\frac1p}. $$

Can one provide suggestive evidence without using l'Hospital's rule that

$$ \lim_{p\to0^{+}} M_p(x,y) = \sqrt{xy}? $$

One way to do this is to use the approximation $x^p \approx 1 + p \log x$ for $p$ small, taken from the Taylor expansion of $x^p$ at $p=0$. Substitute into the power mean formula, and one gets:

$$ \left[1 + \frac p2 (\log x + \log y)\right]^{\frac1p} = \left[1 + p \log \sqrt{xy}\right]^{\frac1p}. $$

Let $s = \frac1p$, and this is

$$ \left[1 + \frac{\log \sqrt{xy}}{s}\right]^s $$

The limit of that as $s\to{+\infty}$ is

$$ e^{\log\sqrt{xy}} = \sqrt{xy}. $$

Making this rigorous is not trivial.

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    $\begingroup$ Engineers are always right, for physically plausible values of "right". $\endgroup$ – Thomas Aug 7 '13 at 7:13

You can't use L'hopital in the third equality sign because you don't have a $"0/0"$ expression.

You get $$ \lim_{x\to 0}\frac{e^x - 2x}{4x^3 + 3x^2 + 2x}. $$ Here the numerator approaches $1$ and the denominator approaches $0$, so in all the limit doesn't exist.

Also note that $$ 4x^3 + 3x^2 + 2x $$ is negative when $x <0$ and it is positive when $x>0$. Therefore one can't say that the limit is $\infty$ or $-\infty$. The limit just doesn't exist.


From the second equality we find $$\lim_{x\to0^+}\frac{e^x-2x}{4x^3+3x^2+2x}=\infty$$

thus the other equalities are false.

  • $\begingroup$ I don't think that the limit is $+\infty$. $\endgroup$ – Thomas Aug 6 '13 at 14:10
  • $\begingroup$ The limit is $+\infty$ at $0^+$ and $-\infty$ at $0^-$. $\endgroup$ – user63181 Aug 6 '13 at 14:12
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    $\begingroup$ Its not necessarily infinite; approaching from the right and left yield different values. $\endgroup$ – ra1nmaster Aug 6 '13 at 14:25
  • $\begingroup$ @Ra1nMaster How itsn't necessary infinite? It's $\pm \infty$. $\endgroup$ – user63181 Aug 6 '13 at 14:28
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    $\begingroup$ It is not infinite because the right and left limit are different. "infinity" implies positive infinity. If the limit's left and right values are different, the limit is nonexistent. $\endgroup$ – ra1nmaster Aug 6 '13 at 14:32

L'Hôpital's rule is ment to applied only when you have to limit of the form 0/0. Once your limit is not in 0/0 form, you are not supposed to apply L'Hôpital's rule.


In layman's terms:

L'Hôpital's rule should only be applied when we have the limit of the form:


Read: http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx and http://tutorial.math.lamar.edu/Classes/CalcI/TypesOfInfinity.aspx

Another tip:

If you are using the rule and you find yourself differentiating many times then there is another way to solve these limits. Like simple reduction or change of base.


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