Limit of $x \log x$ as $x$ tends to $0^+$

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?

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

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

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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).

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.