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Why is the limit of $x \log x$ as $x$ tends to $0^+$, $0$?

  1. The limit of $x$ as $x$ tends to $0$ is $0$.
  2. The limit of $\log x$ as $x$ tends to $0^+$ is $-\infty$.
  3. The limit of products is the product of each limit, provided each limit exists.
  4. Therefore, the limit of $x \log x$ as $x$ tends to $0^+$ should be $0 \times (-\infty)$, which is undefined and not $0$.
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    $\begingroup$ Do you know the rule of l'Hôpital? $\endgroup$
    – quid
    Aug 19, 2013 at 5:59
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    $\begingroup$ Error in your step number 3. The correct version reads: The limit of products is the product of each limit, provided each limit exists as a number. Infinity is not a number. This is the bane of many a freshman calculus student. $\endgroup$ Aug 19, 2013 at 6:03
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    $\begingroup$ Or in other words: The hammer of the rule about the limit of products didn't solve this problem, because this problem is not about driving a nail to a board. Identifying the proper tool for each job is what these problems are all about. $\endgroup$ Aug 19, 2013 at 6:11
  • $\begingroup$ $\log x$ is not defined for $x<0$. Hence the left hand limit of this function is not even defined. Hence, the answer should be limit does not exist $\endgroup$ Sep 14, 2017 at 4:47

5 Answers 5

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As you note this is a "$0 \times -\infty$", which is indeterminate, so we can use L'Hopital's Rule. But first, we should follow Babak S' suggestion, observing that

$$x \log x = \frac{\log{x}}{1/x}.$$

Taking the limit, we obtain

$$\lim \limits_{x \to 0} x \log{x} = \lim \limits_{x \to 0} \frac{\log x}{1/x} \, \stackrel{LH}{=} \, \lim \limits_{x \to 0} \frac{1/x}{-1/x^2}=\lim \limits_{x \to 0} \frac{-x^2}{x} = \lim \limits_{x \to 0} -x = 0.$$


If you need to brush up on L'Hopital's Rule, you may want to consider watching Adrian Banner's lecture on the topic

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    $\begingroup$ This video is private. :( $\endgroup$ Apr 21, 2016 at 16:11
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Hint:

  • We have the indeterminate form $0 \cdot \infty$
  • Let $t = \dfrac{1}{x}$ and now change the limit to use $t \rightarrow \infty$.

What do you get and what can you use?

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    $\begingroup$ I get it. After the substitution, we can just use L'hospital's rule to evaluate the limit. Thanks! $\endgroup$
    – hollow7
    Aug 19, 2013 at 6:04
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    $\begingroup$ Let $x=e^{-t}$ where $t$ is large. Then $\log x=-t$, and we are looking at $-te^{-t}=\frac{-t}{e^t}$ for $t$ large. You know that $e^t$ grows much faster than $t$. $\endgroup$ Aug 19, 2013 at 6:11
  • $\begingroup$ Now you have officially capped...So I'll save any further upvotes until the new Math.SE day! you may receive more upvotes, but until the new day, won't earn points for them. But there may be more "accepts" on your way. ;-) $\endgroup$
    – amWhy
    Aug 19, 2013 at 13:05
  • $\begingroup$ @Amzoti Can this be done without rule of l'Hôpital? $\endgroup$ Apr 21, 2016 at 16:12
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Hint: Assuming this point that you may know the Hopital's rule, consider the main function as follows: $$x\log(x)=\frac{\log(x)}{\frac{1}x}$$ and then take that limit.

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  • $\begingroup$ There, now we have a "nice answer" badge...very nice! +1 $\endgroup$
    – amWhy
    Aug 20, 2013 at 0:04
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One fast way to do it, only using $\log(x) \leq x$ (thus perhaps preferable if this comes up early in a freshman calculus course), is the following:

$$0\geq \lim_{x \rightarrow 0^+} x\log x=\lim_{x \rightarrow \infty} \frac{-\log x}{x}=\lim_{x \rightarrow \infty} \frac{-2\log x}{x^2} \geq \lim_{x \rightarrow \infty} \frac{-x}{x^{2}}=0, $$

as $x\log x<0$ for small $x>0$, and then using the substitutions $x\mapsto \frac{1}{x}$, $x\mapsto x^2$. The squeeze theorem then implies that the limit indeed is 0.

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By $x=e^{-y}$ with $y \to \infty$ we have

$$x \log x =-\frac y {e^y} \to 0$$

which can be easily proved by the definition of $e^y$ or by induction and extended to reals.

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