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?
 A: 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:
$$\ln(1-x) \approx -x$$
A: 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. 
A: 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."
A: 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.
A: 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 $0+(-1)\times x=-x$.
A: 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 
$$ f(x) = \ln(1-x) \approx 0 + (-1)\cdot x = -x.$$
