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I am trying to solve $$\lim \limits_{x \to 0}x\ln{x}$$ which according to WolframAlpha (and Wikipedia) equals $0$.

I managed to solve it by substituting such that $y = \dfrac{1}{x}$ and then using L'Hôpital's rule:

$$\begin{align} \lim \limits_{x \to 0}x\ln{x} & = \lim \limits_{y \to \infty}\frac{\ln{\frac{1}{y}}}{y} \\ & = \lim \limits_{y \to \infty}\frac{-\frac{1}{y^2}{\frac{1}{\frac{1}{y}}}}{1} \\ & = \lim \limits_{y \to \infty}\frac{-1}{y} \\ & = 0 \end{align}$$

but my question is when I try to solve it using L'Hôpital's rule without making the substitution, I get:

$$\begin{align} & \lim \limits_{x \to 0}x\ln{x} \\ & = \lim \limits_{x \to 0}\frac{x}{x}+\ln{x} \\ & = \lim \limits_{x \to 0}1+\ln{x} \\ & = -\infty \end{align}$$

So what went wrong here? Is it because I made $\frac{x}{x}=1$? If so how would I proceed from that point?

Or is this one of those cases where the caveat in L'Hôpital's rule that:

$$\lim \limits_{x \to c}\frac{f'(x)}{g'(x)}$$

has to exist is violated? Does equaling $-\infty$ count as not existing?

Aditya Hase
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Dan
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  • Remember what form you actually need for L'Hôpital's to apply. Differentiating $x\ln x$ isn't in that form. – mardat Dec 05 '14 at 12:45
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    You seem to have substituted $\frac{x}{x} + \ln x $ for $x \ln x$, which is incorrect at t the beginning. – hardmath Dec 05 '14 at 12:47
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    @mardat oh because the denominator is $1$ and that doesn't tend to $0$? So in that case how should I solve this limit? – Dan Dec 05 '14 at 12:48
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    And doesn't have the same limit as the numerator.

    To solve it, you could instead try, for example, $\frac{\ln x}{1\div x}$

     – mardat Dec 05 '14 at 12:48
  • @hardmath Not sure where you mean? I was differentiating but I realise that it's no good because it's not a fraction... – Dan Dec 05 '14 at 12:49
  • Indeed the problem is based on not knowing the limit of the "numerator", so we can't justify using L'Hopital's rule for that reason as well. – hardmath Dec 05 '14 at 12:50
  • So is my first method with the substitution correct? Or just a lucky coincidence? – Dan Dec 05 '14 at 12:51
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    Try again but write $ x \ln x = \frac{\ln x}{1/x}$ – Autolatry Dec 05 '14 at 12:51
  • @Dan your first method with the substitution is correct; the reason it worked where the second one didn't is essentially because it forced you to convert the equation into the form you needed. – mardat Dec 05 '14 at 12:52
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    Dan: You wrote $\lim_{x\to 0} x \ln x = \lim_{x\to 0} \frac{x}{x} + \ln x$, without justifying the step. It appears then to be merely substituting $\frac{x}{x} + \ln x$ for $x \ln x$. By the way, the limit should actually be taken from above (the right), by writing $\lim_{x\to 0^+} x \ln x$. – hardmath Dec 05 '14 at 12:57
  • @hardmath I see that now – Dan Dec 05 '14 at 13:12
  • possible duplicate of How to compute $\lim_{x\to 0^+}x^\epsilon \ln x, \epsilon>0$? – Alex Silva Dec 05 '14 at 13:42

1 Answers1

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In short if the limit of $f$ and $g$ are both zero or both $\pm\infty$, and the limit $f'/g'$ exists, then the limit $f/g$ equals it.

What's wrong is the expression $x\ln x$ as you are implicitly defining $f$ and $g$ doesn't meet the hypothesis. However if we write it as

$$\frac{\ln x}{1/x}$$

we can use l'Hopital with $f(x) = \ln x$, $g(x) = 1/x$ as

  • The limits $\lim_{x\to 0^+} \ln x = -\infty$ and $\lim_{x \to 0^+} \frac{1}{x} = \infty$; and
  • The limit $\displaystyle \lim_{x\to 0^+} \frac{f'(x)}{g'(x)}$ exists as $\displaystyle \lim_{x\to 0^+} \frac{1/x}{-1/x^2} = \lim_{x\to 0^+} -x = 0$

Hence $\lim\limits_{x\to 0^+} x\ln x = 0$.

Simon S
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  • Right that's pretty simple. Thanks! – Dan Dec 05 '14 at 12:52
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    @Dan If this answered your question, you should consider setting it as accepted. – AlexR Dec 05 '14 at 13:00
  • NB I removed an extra $=$ sign. – AlexR Dec 05 '14 at 13:05
  • I have a small doubt. When x tends to 0 lnx will be a very small number say (-infinity) and x will tend to 0. So why does multiplying infinity to zero gives us 0 ? And LHL does not exist so, will the limit exist ? – Dhiraj Barnwal Aug 16 '15 at 16:55
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    @user2414789 Yes, as the original question stands, $\lim_{x\to 0} x \ln x$ is potentially problematic as it's not necessarily clear that negative $x$ are excluded. Everywhere the limits should be the right-hand limit, $x\to 0^+$. I've ended my answer to make that explicit. As to why in the product, $x$ and $\ln x$, $x \to 0$ "dominates" $\ln x \to -\infty$, that's what the mathematics shows. Of course there are other products where the limit could be other values. But in this case, $x\to 0$ "wins out". – Simon S Aug 16 '15 at 17:00