For small $x$, one has $\ln(1+x)=x$? What does it mean that for small $x$, one has $\ln(1+x)=x$? How can you explain this thing ? Thanks in advance for your reply.
 A: Since no one else said it clearly in words. One does not have that 
$$\ln(1 + x) = x$$
for small $x$. One does, however, have that for small values of $x$, $\ln(1+x)$ can be approximated by $x$. As the other answers have already pointed out, this you see from the Taylor expansion
$$
\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \dots.
$$
Now if $x$ is a small enough number, then $x^2, x^3, \dots$ are all insignificant. And so for small $x$ you can approximate $\ln(1+x)$ by $x$. 
A: Take the tangent line at of $f(x) = \ln(1+x)$ in $x = 0$. 
\begin{align*}
f(x) & \approx f(0) + f'(0) (x - 0) \\
& = \ln(1+0) + \left[\frac{d}{dx} \ln(1+x)\right]_{x = 0} (x-0) \\
& = 0 + 1 x \\
& = x
\end{align*}
A: If you write its Taylor's expansion then you have:
$\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{k}$.
For small values of $x$, the values of $x^2, x^3,...$ are small in comparing with $x$(note that positive numbers which are less than 1, will decrees as we multiply themselves), hence we can ignore the terms with degree larger than $1$, and estimate $\ln{(1+x)}$ as its first degree part which is $x$.
A: Look at the equivalent $\text{exp}(x) = 1 + x$ ($x$ teensy) and compare it to the Taylor series.
A: Hint:  One can prove [1] that $$ f(x) \approx f(a) + f'(a)(x - a).$$ for $x$ near $a$.
A: First, let me parrot Thomas' answer:  it is not true that $\log(1+x) = x$ for small $x$.  This equation holds only when $x=0$.  However, it is true that if $x$ is small, then $\log(1+x)$ is "well-approximated" by $x$.  Slightly more formally, the function $x \mapsto x$ is the best linear approximation of the function $x \mapsto \log(1+x)$ for values of $x$ which are "near zero".
Derivatives and Linear Approximation
The essential idea is that if a function $f$ is differentiable, then the best linear approximation of that function near a point $a$ is the line with slope $f'(a)$ which passes through the point $(a, f(a))$.  An equation for this line is given by
$$ f(x) \approx f(a) + f'(a)(x-a) =: \ell(x).$$
What this means is that if $x$ is "near" $a$, then the value of $f(x)$ is "near" the value of the function $\ell(x)$, where the the graph of $\ell$ is a line.  All of this can be made rigorous by careful examination of the definition of the derivative.
As others have pointed out, taking $f(x) = \log(1+x)$ and $a=0$, this becomes
\begin{align}
\log(1+x)
&\approx \log(1+0) + \left[ \frac{\mathrm{d}}{\mathrm{d}x} \log(1+x)\right]_{x=0} (x-0) \\
&=  \log(1) + \left[\frac{1}{1+x}\right]_{x=0} x \\
&= 0 + \frac{1}{1+0} x \\
&= x.
\end{align}
Hence when $x$ is "near" $0$, then $\log(1+x)$ is "near" $x$.
As others have noted, a better approximation for the value of a function "near" some point can be obtained by evaluating the Taylor series expansion of the function near that point—this is the approach of Thomas' answer.
From the Integral Definition of $\log$
The definition of $\log(x)$ which I find most compelling is the following:  for all $x \in (0,\infty)$, define
$$ \log(x) := \int_{1}^{x} \frac{1}{t}\,\mathrm{d}t. $$
Note that we can use this definition to obtain all of the other properties we might like (e.g. $\log$ is the inverse of $\exp$; $\log$ has a particular Taylor series expansion; etc).  I want to use this definition to show that when $x$ is "small", then the error $|\log(1+x)-x|$ is also "small".  For any $x \in [0,\infty)$, we have
\begin{align}
|\log(1+x) - x|
&= | x - \log(1+x) | \\
&= \left| x - \int_{1}^{1+x} \frac{1}{t} \,\mathrm{d}t \right| \\
&= \left| \int_{1}^{1+x} 1 \,\mathrm{d}t - \int_{1}^{1+x} \frac{1}{t} \,\mathrm{d}t \right| \\
&= \int_{1}^{1+x} \frac{t-1}{t}\,\mathrm{d}t  && \text{($t \ge 1 \implies \tfrac{t-1}{t} \ge 0$, thus $|\cdot|$ is redundant)}\\
&\le \int_{1}^{1+x} t-1 \,\mathrm{d}t && \text{($t \ge 1 \implies \tfrac{t-1}{t} \le t-1$)} \\
&= \left[ \frac{1}{2} t^2 \right]_{t=1}^{1+x} - x \\
&= \frac{1}{2}(1+x)^2 - \frac{1}{2} -  x \\
&= \frac{1}{2} + x + \frac{1}{2} x^2 - \frac{1}{2}  - x \\
&= \frac{1}{2} x^2.
\end{align}
This means that if $x \ge 0$, then the error between $\log(1+x)$ and the approximation $x$ will be smaller than $\frac{1}{2}x^2$.  If $x$ is close to zero (say, $x < 1$), then $\frac{1}{2}x^2$ will be smaller than $x$, so the approximation here looks pretty good.
For $x \in (-1, 0)$, the situation is a little more complicated, as the integral "blows up" as $x \to -1$.  Hence we need to constrain $x$ a little more.  Since the goal is to show that if $x$ is "near" $0$, then $\log(1+x)$ is "near" $x$, we might as well assume that $x \ge -\frac{1}{2}$.  Under this assumption, we have
\begin{align}
|\log(1+x) - x|
&= \int_{1}^{1+x} \frac{t-1}{t} \,\mathrm{d}t && \text{(as above)} \\
&\le 2 \int_{1}^{1+x} t-1\,\mathrm{d}t && \text{($t \ge 1+x \ge \tfrac{1}{2} \implies \tfrac{t-1}{t} \le 2(t-1)$)} \\
&= x^2.
\end{align}
This says that if $x \in (-\frac{1}{2}, 0)$, then the error between $\log(1+x)$ and $x$ is smaller than $x^2$.  Again, when $x$ is small, $x^2$ is even smaller.
Summarizing the above, we have the following relatively "nice" statement:

If $|x| \le \frac{1}{2}$, then the error between $\log(1+x)$ and $x$ is smaller than $x^2$.  That is,
  $$
\operatorname{error}(x)
= |\log(1+x) - x|
\le x^2.
$$
  The closer $x$ is to zero, the smaller this error will be.

