Prove that for $0Prove that for $0<x<1$, $\ln(1+x) - \ln(1-x)<2x$
My direction: $\ln(x)$ is differential and continuous at $(0,∞)$, and since $0<x<1$, clearly $[1-x,1+x]\subseteq(0,∞)$ and so $\ln(x)$ is continuous in $[1-x,1+x]$ and differential in $(1-x,1+x)$.
So, by Lagrange's theorem, there is a $c$ in $(1-x, 1+x)$, so that: $${{\ln(1+x)-\ln(1-x)}\over{1+x-1+x}}=\ln'(c)$$
$${{\ln(1+x)-\ln(1-x)}\over{2x}}=\ln'(c)$$
$$\ln(1+x)-\ln(1-x)=\ln'(c)2x$$
And here I got stuck... Would appreciate any help.
 A: $\ln(1 + x) = x - x^2/2 + x^3/3 -...$, and $\ln(1 - x) = -x - x^2/2 - x^3/3 -...$, so you have
$$\ln(1 + x) - \ln(1 - x) = 2x + {2x^3 \over 3} + {2x^5 \over 5} + ...$$
Looking at the first term you see that this is greater than $2x$.
A: 
For every $0\lt x\lt1$, $\log(1+x)-\log(1-x)\gt2x$.

To show this, define $u(x)$ as the LHS and note that $u(0)=0$, $u'(0)=2$ and $u''(x)\gt0$ for every $x$ in $(0,1)$. Since the function $u$ is convex, the graph of $u$ lies above its tangent at $0$, which is the line $y=2x$. This is the result.
Edit: Alternatively, since $u''\gt0$, $u'\gt u'(0)$ on $(0,x)$ hence $u(x)-u(0)=u'(c)x\gt2x$.
A: [Proof fixed after did's comment.]
Let $g(x) = \ln (1+ x) - \ln(1-x) - 2x.$ 
We have $ g'(x) = \dfrac{1}{1+x} + \dfrac{1}{1-x} - 2 $, hence $g{'}(0)= 0 = g(0)$, also, $g^{''}(x) = -\dfrac{1}{ ( 1 + x )^2 } + \dfrac{1}{ ( 1 - x )^2 } > 0,$ for any $x\in (0,1)$ as $\dfrac{1}{1-x} > 1$ while $ 0 < \dfrac{1}{1+x} < 1$.
From Taylor expansion,
$$g(x) = g(0) + g^{'}(0)x + g''(\eta) x^2/2 = g''(\eta) x^2/2,$$ 
where $\eta \in (0,x).$ 
So $g(x) > 0$  which implies $\ln(1 + x) - \ln(1 -x) > 2x. $
A: Note that for $x\in(0,1)$:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(\vphantom{\frac{\mathrm{d}}{\mathrm{d}x}}\log(1+x)-\log(1-x)-2x\right)=\frac{2x^2}{1-x^2}\gt0
$$
Therefore, the mean value theorem says that for some $\xi\in(0,x)$
$$
\frac{(\log(1+x)-\log(1-x)-2x)-(\log(1+0)-\log(1-0)-2\cdot0)}{x-0}=\frac{2\xi^2}{1-\xi^2}\gt0
$$
Therefore,
$$
\log(1+x)-\log(1-x)>2x
$$
