Limit of $x \log x$ as $x$ tends to $0^+$
Why is the limit of $x \log x$ as $x$ tends to $0^+$, $0$?
- The limit of $x$ as $x$ tends to $0$ is $0$.
- The limit of $\log x$ as $x$ tends to $0^+$ is $-\infty$.
- The limit of products is the product of each limit, provided each limit exists.
- Therefore, the limit of $x \log x$ as $x$ tends to $0^+$ should be $0 \times (-\infty)$, which is undefined and not $0$.
Solution 1:
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
- Calculus I - Optimization and L'Hôpital's Rule - Lecture 10 (Start at about 1:15:00).
Solution 2:
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?
Solution 3:
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.
Solution 4:
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.