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.

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 $x$.
