# Why is $\ln(1-x) \approx -x$ when $x$ is small?

I saw this in a proof for the Central Limit Theorem:

$\ln(1-x) \approx -x$ when $x$ is small

It seems to be true when I plug in small values of $x$. But why does it work?

• Compute $$\lim_{x \to 0} \frac{\ln(1-x)}{-x}$$ – user384138 Jun 12 '17 at 18:02
• You can look at the Taylor expansion to obtain this result. – Severin Schraven Jun 12 '17 at 18:05
• Dang I blinked after I clicked post your answer and now there's 5 answers lol – mrnovice Jun 12 '17 at 18:07
• @mrnovice The "fastest gun in the west" problem – user223391 Jun 12 '17 at 18:11

We can write $$\ln(1-x)$$ as the infinite series:

$$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots$$

so for $x$ is small, we have that $x^2, x^3, x^4, \dots$ are even smaller, neglectible amounts so we can say:

$$\ln(1-x) \approx -x$$

• It's a good intuitive argument if one knows the sum, but.... formalizing it to argue that the sum of infinitely many negligible terms is still negligible is not that trivial. – Clement C. Jun 12 '17 at 18:09
• You are absolutely right. If the OP wants me to give further information, I will do that. – user370967 Jun 12 '17 at 18:11
• I agree with @ClementC. Taylor's theorem is extremely relevant in proving this, and at that point, you're over-killing it since it uses the MVT to prove in the first place and you may as well stop after once step and appeal to the more basic theorem. – Adam Hughes Jun 12 '17 at 18:27
• Thank you, Taylor series is a good enough argument for me. Also Open Ball's suggestion in the comments is also good. – foobar Jun 12 '17 at 22:59

Assuming $x$ is real, the Taylor series of $\ln (1-x)$ about zero is $$\ln (1-x) = \ln(1) + \frac{d}{dx}\ln(1-x)|_{(x = 0)}x + \mathcal{O}(x^2)$$ or $$\ln (1-x) = 0 - x + \mathcal{O}(x^2) = -x + \mathcal{O}(x^2)$$ For small $x$ (that is, much less than one) all terms of order $x^2$ are negligible so we have $$\ln(1-x) \approx -x.$$ Note that the statement as written cannot be true, for if $y = \ln(1-x)$ then $e^y = 1-x$, which is $< 1$ when $x$ is positive. Thus $y$ must be negative when $x$ is positive and positive when $x$ is negative in the linear limit.

The easiest way to see this is to use the Mean Value Theorem. Note that since $\log(1-x)$ is twice differentiable (in particular its derivative is continuous) that means that

$$f(x)\approx f'(0)(x-0)+f(0)={-1\over 1-0}(x-0)+0=-x$$

for $x$ near $0$. (i.e. small $x$)

You may also know this process by another name: "linearization."

Because $\ln(1)=0$ and $\left.\frac{d}{dt}\ln(1-x)\right|_{t=0}=-1$. Therefore, when $x$ is small, $\ln(1-x)$ is approximately $0+(-1)\times x=-x$.

This can also be seen using Taylor expansions. Let $f(x) = \ln(1-x)$. We have $f(0) = \ln(1) = 0$. And since $f'(x) = \frac{-1}{1-x}$, $f'(0) = -1$. This gives a linear Taylor approximation of

$$f(x) = \ln(1-x) \approx 0 + (-1)\cdot x = -x.$$

I am surprised by the earlier answers, all of which agree that the statement

$\ln(1 - x) \approx -x$ when $x$ is small

was true. Actually, it is not, and the problem lies in the notion of smallness.

### Smallness

All earlier answers seem to assume that the closer a number $x$ is to $0$, the smaller it is. However, if we assume the usual order of the real numbers, i.e., $$\ldots < -2 < -1 < 0 < 1 < 2 < \mathrm{e} < 3 < \pi < \ldots \;,$$ then a number $x$ is the smaller, the closer it is to $-\infty$.

It is true that the closer the absolute value $|x|$ is to $0$, the smaller this absolute value. But the absolute value $|x|$ is not the same as $x$ itself.

Now let us investigate two approaches suggested among the answers.

### Suggested approach 1

The first approach suggests to look at the series $$\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \ldots$$ @Math_QED claims that

"for $x$ is small, we have that $x^2, x^3, x^4, \dots$ are even smaller".

This is not true. If $x$ is small, i.e., a negative number with large absolute value, then the odd powers $x^1,x^3,x^5,\ldots$ are indeed smaller, but the even powers $x^2,x^4,x^6,\ldots$ are greater. So, this approach fails.

### Suggested approach 2

Another approach suggests to compute the limit of $$\dfrac{\ln(1-x)}{-x}$$ as $x$ becomes smaller. Actually, @Open Ball suggests to compute $$\lim\limits_{x \to 0} \dfrac{\ln(1-x)}{-x} = 1 \;.$$ However, instead of letting $x \to 0$, we have to let $x \to -\infty$, because any negative number is still smaller than $0$, and we want that $x$ becomes as small as possible. The result of the limit is $$\lim\limits_{x \to -\infty} \dfrac{\ln(1-x)}{-x} = 0 \;,$$ and this disproves the statement.

A visual depiction is given in the figure. The smaller $x$ becomes, the more do $\ln(1 - x)$ and $-x$ diverge from each other.

• lol, take x = 0,01, then x^2 = 0,0001 is not smaller? – user370967 Jun 12 '17 at 19:28
• It is, but your example is inappropriate, because $0.01$ is not small! A number $x$ is small if it is negative and its magnitude is large. But the magnitude of $0.01$ is not large! By the way, if you pick this out to claim that I was wrong, then you really miss the point. – Björn Friedrich Jun 12 '17 at 19:31
• To the downvoter: Next time, you can just hint to the error, so that I can fix it. :-) – Björn Friedrich Jun 12 '17 at 19:35
• Small means close to 0 – Flame Trap Jun 12 '17 at 19:38
• @FlameTrap: Assuming the usual order of real numbers, $-5$ is smaller than $1$ even though $1$ is closer to $0$. – Björn Friedrich Jun 12 '17 at 19:45