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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?

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    $\begingroup$ You seem to have substituted $\frac{x}{x} + \ln x $ for $x \ln x$, which is incorrect at t the beginning. $\endgroup$ – hardmath Dec 5 '14 at 12:47
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    $\begingroup$ @mardat oh because the denominator is $1$ and that doesn't tend to $0$? So in that case how should I solve this limit? $\endgroup$ – Dan Dec 5 '14 at 12:48
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    $\begingroup$ And doesn't have the same limit as the numerator. To solve it, you could instead try, for example, $\frac{\ln x}{1\div x}$ $\endgroup$ – mardat Dec 5 '14 at 12:48
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    $\begingroup$ Try again but write $ x \ln x = \frac{\ln x}{1/x}$ $\endgroup$ – Autolatry Dec 5 '14 at 12:51
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    $\begingroup$ 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 $\lim_{x\to 0^+} x \ln x$. $\endgroup$ – hardmath Dec 5 '14 at 12:57
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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 $\lim\limits_{x\to 0^+} x\ln x = 0$.

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  • $\begingroup$ Right that's pretty simple. Thanks! $\endgroup$ – Dan Dec 5 '14 at 12:52
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    $\begingroup$ @Dan If this answered your question, you should consider setting it as accepted. $\endgroup$ – AlexR Dec 5 '14 at 13:00
  • $\begingroup$ NB I removed an extra $=$ sign. $\endgroup$ – AlexR Dec 5 '14 at 13:05
  • $\begingroup$ 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 ? $\endgroup$ – Dhiraj Barnwal Aug 16 '15 at 16:55
  • $\begingroup$ @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". $\endgroup$ – Simon S Aug 16 '15 at 17:00

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