## 0*log(0) = 0 (for real)

Tommi pointed out in the comments to the previous post that I had made a mistake in my purported proof that

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

because while finding the limit, I assumed the limit exists. It made me sad, because I thought the proof was really cute. But actually it was just incomplete. It feels just like that time when I was a kid and we brought home new puppy, only to discover it was missing a leg. My sister and I made a new leg out of Tinker Toys and sewed it on there, despite the puppy’s vociferous protests. I figure the same basic strategy is in order now, so let’s re-examine that expression.

We can verify the limit in a straightforward way using Hopital’s Rule

For more on Hopital see the book by Gilbert Strang, (section 3.8).

Crunching away,

$\begin{array}{lll}\lim_{x \rightarrow 0} x*\log(x) & = & \lim_{x\rightarrow 0} \frac{\log(x)}{x^{-1}} \\ { } & = & \lim_{x \rightarrow 0} \frac{x^{-1}}{-x^{-2}} \\ {} & = & \lim_{x\rightarrow 0}-x \\ { } & = & 0\end{array}$

Statisfying?

One way to build some intuition would be to make some plots. Here’s $1/x$ (red) and $-\log(x)$ (blue) plotted together.

Red line is 1/x. Blue line is -log(x). Red wins!

We have

$-x\log(x) = \frac{-\log(x)}{1/x}$.

So if $\log(x)$ is getting bigger and bigger faster than $1/x$, the numerator of the fraction is growing faster than the denominator, and the limit should increase up to infinity. But it looks instead like $1/x$ actually grows faster. Then the denominator is growing faster than the numerator, and we expect the limit to be zero.

Here’s a plot of the actual thing, $x*\log(x)$.

x*log(x) Plotted from 0 to 1, thanks to Stephen Wolfram and the Caltech Library computers

Definitely going towards zero.

Finally, one more proof. I want to take a look at the series expansions and see if the limit of that is obvious as $x$ goes to zero. The best place to take a series of logarithm is around $x=1$.

Let $x = 1-u$.

Then $u = 1-x$.

$\begin{array}{lll} x\log(x) & = &(1-u)\log(1-u) \\ { } & = & (1-u)\sum_{k=1}^\infty \frac{-u^k}{k} \\ {} & = & \sum_{k=2}^\infty \frac{u^k}{k-1} - \sum_{k=1}^\infty \frac{u^k}{k} \\ {} & = & -u + \sum_{k=2}^\infty \frac{u^k}{k-1} - \frac{u^k}{k} \\ {} & = & -u + \sum_{k=2}^\infty \frac{u^k}{k(k-1)} \end{array}$.

We want the limit as $x \rightarrow 0$, which is the same as the limit as $u \rightarrow 1$. Now there are no zeroes or infinities any more, so we can just plug $u=1$ straight into the last line of the equalities above.

$\lim_{x \rightarrow 0}x\log(x) = -1 + \sum_{k=2}^\infty \frac{1}{k(k-1)}$.

Let’s do the summation. The first few terms to add are

$\{\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}\}$

and the first few partial sums are

$\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}\}$.

From this we conjecture that

$\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}$.

It’s been shown for $n=2,3,4,5$ already. Use induction, assuming it’s true for $n$, and showing it’s true for $n+1$.

If $\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}$,

then

$\begin{array}{lll} \sum_{k=2}^{n+1} \frac{1}{k(k-1)} & = & \frac{n-1}{n} + \frac{1}{(n+1)n} \\ {} & = & \frac{(n-1)(n+1) + 1}{n(n+1)} \\ {} & = & \frac{n^2 -1 + 1}{n(n+1)} \\ {} & = & \frac{n^2}{(n+1)n} \\ {} & = & \frac{n}{n+1}\end{array}$.

Which proves the conjectured summation. The limit of the sum as $n \rightarrow \infty$ is

$\lim_{n \rightarrow \infty} \frac{n-1}{n} = 1$.

With this result for the summation, we just get

$\lim_{x \rightarrow 0} x\log(x) = 1-1 = 0$.

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### 7 Responses to “0*log(0) = 0 (for real)”

1. 0*log(0) = 0 « Arcsecond Says:

[…] Rough Pebbles Welcome « Algebraic Numbers Form a Field, and Are Countable 0*log(0) = 0 (for real) […]

2. Shailendra Says:

well, if u take limit x->0 (xlogx) then it is 0. But if u take definite integral of 1/x with limit 0 to 1, then you will have logx with limit 0 to 1 and then log1- log0. Now we need the value of log0, which is undefined so i want to know the solution with explanation.

Thanks

3. Mark Eichenlaub Says:

Hi Shailendra,

You appear to be trying to cause trouble where none exists. If you want to define the natural logarithm as

$\ln(x) = \int_1^x \frac{1}{t} dt$

that’s fine. Then to find a reasonable value for $0*\ln(0)$, we should take the limit

$\lim_{x\to0} x \int_1^x \frac{1}{t}dt$.

That limit exists and is equal to zero. I’m not sure what else you want. Yes, log(0) is undefined, but that problem is not unique to the way you’ve set things up. That problem is already there when trying to evaluate 0*log(0), no matter how you slice it.

4. Seetharaman Says:

Hello , I am a school student only but i feel log 0 can be defined and log 0 =0 :) see my proof and leave ur comments :
1/ infinity =0 ;
infinity *0=1 …(1)
log 0
log 0 * (1/0)/ (1/0) multiplying nd dividing by (1/0 )
dat means uponsimplification;
log (0* infi *1 )/infi
log 1 =log 0 =0 :) how is it ? please temme weda it is valid :D it is logical :)
Thanks

5. Mark Eichenlaub Says:

Hi Seetharaman,

Infinity is not a number. Calculations involving algebraic manipulation of infinity do not make sense.

Please try to use better grammar. Your post is almost unreadable.

6. Anonymous Says:

why not define log(0) as a new thing like square root of -1 if i ?

7. PA Says:

0*Log (0) is the same as Log (0^0) which is the same as Log(1)

all of which equal 0 at the end of the day. (or any other time of day)

right? Or am I off?