48

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 $0$.
user2984602
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hollow7
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    Do you know the rule of l'Hôpital? – quid Aug 19 '13 at 05:59
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    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. – Jyrki Lahtonen Aug 19 '13 at 06:03
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    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. – Jyrki Lahtonen Aug 19 '13 at 06:11
  • $\log x$ 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 – Dr. N.Padmanabhan Sep 14 '17 at 04:47
  • @Dr.N.Padmanabhan Whether the function is defined for $x<0$ or not is not relevant for determining the limit as $x$ tends to $0^+$. – pfincent Sep 24 '22 at 13:25

5 Answers5

39

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

Gamma Function
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29

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?

Amzoti
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    I get it. After the substitution, we can just use L'hospital's rule to evaluate the limit. Thanks! – hollow7 Aug 19 '13 at 06:04
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    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 $t$. – André Nicolas Aug 19 '13 at 06:11
  • 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. ;-) – amWhy Aug 19 '13 at 13:05
  • @Amzoti Can this be done without rule of l'Hôpital? – Archisman Panigrahi Apr 21 '16 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.

Mikasa
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5

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.

user712939
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1

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.

user
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