This is a follow up to a question posted yesterday here.

For my question, suppose you wish to compute the following limit

$$ \lim _{x\to 0^+}\left(\frac{2x^7+x^2}{x^7+5x^4}\right) $$

My first thought was to apply L'Hôpital's rule 7 times because the numerator and denominator both tend to zero. Without doing any calculations, it should be obvious that if you take 7 derivatives of the top and bottom (i.e. apply L'Hôpital's rule) you'll end up with a ratio of constants which I don't believe is the correct answer.

After performing polynomial long division I reduced the limit to the following

$$ \lim _{x\to 0^+}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) =\lim _{x\to 0^+} 2 + \lim _{x\to 0^+}\left(\frac{x^2-10x^4}{x^7+5x^4}\right) $$

Taking the derivative 4 times of the top and bottom of the second limit on the right yields the following

$$ \lim _{x\to 0^+}\left(\frac{-240}{840x^3+120}\right) $$

Hence we have

$$ \lim _{x\to 0^+}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) = 2 - 2 = 0 $$

Which I still don't think is the correct answer. What's going on here!

Question: Why does the repeated naive application of L'Hôpital's rule appear not to be working

PS: According to Wolfram Alpha the answer should be $+\infty$.

  • 1
    $\begingroup$ Since after applying L'Hopital twice, the numerator no longer goes to $0$. $\endgroup$ – Hetebrij Jun 11 '16 at 12:42
  • $\begingroup$ The $\lim \limits_{x\to 0}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) =\lim \limits_{x\to 0} 2 + \lim _{x\to 0}\left(\frac{x^2-10x^4}{x^7+5x^4}\right)$ step is incorrect. The limit of sum only equals the sum of the limits if the limits actually exist. $\endgroup$ – Git Gud Jun 11 '16 at 12:43
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    $\begingroup$ You can only apply L'Hopital's rule when the limit of the numerator and denominator are both zero (or infinity). The limit at zero of the second derivative of the numerator is 2, so you cannot apply L'Hopital's a third time. $\endgroup$ – D. Mansfield Jun 11 '16 at 12:43
  • $\begingroup$ @GitGud : ​ One of the right-side-of-the-= limits exists and is finite, which together is enough. ​ ​ ​ ​ $\endgroup$ – user57159 Jun 11 '16 at 19:18
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    $\begingroup$ It appears that L'Hospital's Rule has become the panacea for any limit problem. Perhaps the day is not far when someone uses it for evaluating $\lim_{x \to 0}x/x$. The limit in question does not need any powerful/complicated tools like L'Hospital. Rather cancelling $x^{2}$ from numerator and denominator gives the answer in one step. $\endgroup$ – Paramanand Singh Jun 11 '16 at 19:34

You have to check hopital hypothesis after every differentiation. In particular after two differentiation the numerator won't be tending to $0$ anymore (there will be a constant term) so you have to stop there.

There also other problems, namely:

$$\lim _{x\to 0^+}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) =\lim _{x\to 0^+} 2 + \lim _{x\to 0^+}\left(\frac{x^2-10x^4}{x^7+5x^4}\right)$$

This is only valid if both limits on the RHS exist, something which you have to justify.

Also, there really is no reason to use long polynomial division instead of just applying hopital directly

Finally, a cool way to do it is using infinetesimals; in fact since $$2x^7 + x^2 \sim_0 x^2$$ and $$x^7 + 5x^4 \sim_0 5x^4$$, we find that

$$\frac{2x^7 + x^2 }{x^7 + 5x^4} \sim \frac{x^2}{5x^4} = \frac 1{5x^2} \to \infty$$

  • $\begingroup$ Ahhh I see. Thanks! This makes sense. I'll accept this when the time limit is up $\endgroup$ – ApprenticeOfMathematics Jun 11 '16 at 12:44
  • $\begingroup$ Don't single-sided limits of ratios of polynomials always exist? $\endgroup$ – Joshua Jun 11 '16 at 13:48
  • $\begingroup$ @Joshua Depends on what you mean by exists. If you mean is finite, then clearly not. If you mean finite or $\infty$, then I think you are right. $\endgroup$ – Ant Jun 11 '16 at 13:57
  • $\begingroup$ Yeah I meant finite or ∞, which is why his not checking before applying L'Hôpital didn't bother me. $\endgroup$ – Joshua Jun 11 '16 at 14:16
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    $\begingroup$ @Joshua But the error was precisely there. The point is that even if the ratio of polynomials admits a limit, hopital requires the numerator and the denominator to both be tending to $0$ or $\infty$. In this case after two applications of hopital the numerator was not tending to $0$ anymore $\endgroup$ – Ant Jun 11 '16 at 14:40

L'Hospital is proven and cannot "not work". Never. Provided you stick to the theorem hypothesis, among which: the numerator and denominator are going to zero. This is where your reasoning fails.

Actually, this limit doesn't require L'Hospital, as

$$ \lim _{x\to 0^+}\frac{2x^7+x^2}{x^7+5x^4} =\lim _{x\to 0^+}\frac{2x^5+1}{x^5+5x^2}$$

which is not an indeterminate form.

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    $\begingroup$ +1 for the answer that avoid's L'Hopital. I'm sorry that "rule" is taught, since even when it works correctly it can obscure what is really going on - as you point out. $\endgroup$ – Ethan Bolker Jun 11 '16 at 13:09

No, you don't apply L'Hôpital seven times in the original form of the function, that would be very wrong, just like $$ \lim_{x\to0^+}\frac{x}{x^2}=\lim_{x\to0^+}\frac{1}{x}=\lim_{x\to0^+}\frac{0}{1}=0 $$ Can you spot where the wrong $=$ is? Because you certainly know the limit is $\infty$.

When you differentiate twice the numerator you get $$ 84x^5+2 $$ and this function has limit $2$ as $x\to0$, so it is not one for which L'Hôpital applies.

If you had the limit at $\infty$, then applying L'Hôpital seven times would be possible and would also give the correct result, because up to the sixth step you still have functions that have infinite limit at $\infty$. Possible, but boring and inefficient.

Be careful with L'Hôpital: a good tool, but dangerous if not handled with care.


It is in fact $+\infty$ by the presence of $x^4$ in the denominator and $x^2$ in the numerator. After you apply twice the Rule you have $84x^5+2$ as numerator.


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