How to compute $f(x)=x\ln{x}$ at $x=0$?

Question

Given

$f(x)=x\ln{x}$,

is it correct to say that $f(0)= \lim{}_{x\to{0}}f(x)=\lim{}_{x\to{0}}x\ln{x}=0$?

If this is true, why is this so? Is it because $0*n$, where $n$ is any value, is $0$?

My intuition tells me that is should be $f(0)=0ln(0)=0(-\infty)=\mathbf{undefined}$?

What is the correct reasoning?

This is in reference to this question on Cross Validated, where the first way to compute $f(0)$ was suggested in a comment. That solved the issue, but it doesn't make sense to me mathematically for the above concern when computing $f(0)$.

Answer 1

Try viewing it as $\dfrac{\ln(x)}{1/x}$, and then using L'Hopital's rule.

As you observed, just plugging in $x=0$ gives you the non-sensical quantity $$0\cdot\ln(0)=0\cdot(-\infty)=???.$$

This is an example of a limit where you cannot plug in the point being approached. Remember, the limit $\displaystyle\lim_{x\to a}f(x)$ only equals $f(a)$ if the function $f$ is continuous at $a$. The function $f(x)=x\ln(x)$ is certainly not continuous at $0$.

Answer 2

As written, the function is not defined at $0$ because $\ln(0)$ is undefined. However, if desired one could complete the function continuously by defining $f(0)=0$. This is no longer the same function, however.

Answer 3

$f(0)$ doesn't exist, because $\ln(0)$ doesn't exist. But it is a fairly common abuse of notation to write $f(x)=x \ln(x)$ when what is really meant is its continuous extension to zero. This continuous extension is indeed zero at zero, since

$$\lim_{x \to 0^+} x \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}}=\lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x \to 0^+} -x = 0$$

by L'Hospital's rule. Another proof substitutes $x=e^{y}$ as $y \to -\infty$, so that $\ln(x)=y$ and

$$\lim_{x \to 0^+} x \ln(x) = \lim_{y \to -\infty} y e^y$$

which can be shown to be zero in various ways, including L'Hospital's rule.

Answer 4

$0 \ln 0\;$ is an indefinite form, but $\;\lim\limits_{x\to 0} x\ln x = 0\;$.

So $f(0)$ is undefined although the limit to that point exists.

Note:   We do often take the convention that $0\ln 0 = 0$ when, for example, dealing with it in entropy equations or such.   While it is not strictly true, the convention makes those equations useable.