0*log(0) = 0

$x\log(ax) = x\log(a) + x\log(x)$.

$\begin{array}{lll}\lim_{x \rightarrow 0} x\log(ax) & = & \lim_{x \rightarrow 0}x\log(a) + \lim_{x \rightarrow 0} x\log(x) \\ { } & = & 0 + \lim_{x \rightarrow 0} x\log(x) \\ {} & \equiv & y \end{array}$.

$x\log(ax) = \frac{1}{a}u\log(u), u=ax$.

$\lim_{u \rightarrow 0} \frac{1}{a} u\log(u) = \lim_{x \rightarrow 0} x\log(ax) = y$.

$\lim_{u \rightarrow 0} \frac{1}{a} u\log(u) = \frac{1}{a} y$.

$\frac{1}{a} y = y$.

$y\left(1-\frac{1}{a}\right) = 0$.

$y = 0$.

$\lim_{x \rightarrow 0} x\log(x) = 0$.

4 Responses to “0*log(0) = 0”

1. Tommi Says:

x*log(x) approaching y is somewhat suspicious.

2. meichenl Says:

I guess you’re right – there’s no guarantee that y is finite. In that case we can use Hopital’s rule. I suppose this just tells us that if the limit exists, it is zero.

3. Tommi Says:

That, I think, is correct. The result is still unintuitive, as log gets very small very quickly near zero.

Goes to show that relying too much on one’s intuition is something to avoid.

4. 0*log(0) = 0 (for real) « Arcsecond Says:

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