How I can solve this limit only with algebra I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.
$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$
first all I use the logarithms properties and rewrite like this:
$$\lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$$
and tried to make to this expression goes to  $$\lim_{x\to 0} \log e^\frac{1}{2}$$
and then limit will be 1/2
I can't reach to this because the limit goes to 0 instead of ∞
 A: Let $a = \lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$. Then:
$$\exp(a) = \lim_{x\to 0} \exp \left(\log({\frac{1 + x}{1 - x}})^{\frac{1}{2x}} \right) = \lim_{x\to 0} \left(\frac{1 + x}{1 - x} \right)^{\frac{1}{2x}} = \lim_{x\to 0} \left(\frac{1/x + 1}{1/x - 1} \right)^{\frac{1}{2x}}$$
Now let $u = \frac{1}{x}$. Then we have:
$$\lim_{u \to \infty} \left(\frac{u + 1}{u - 1} \right)^{\frac{u}{2}} =\lim_{u \to \infty} \left(\frac{u-1}{u-1} + \frac{2}{u - 1} \right)^{\frac{u}{2}} = \lim_{u \to \infty} \left(1 + \frac{2}{u - 1} \right)^{\frac{u-1}{2}} \left(1 + \frac{2}{u - 1} \right)^{\frac{1}{2}}$$
$$= e \cdot 1 = e$$
Therefore $\exp(a) = e \Rightarrow \boxed{a = 1}.$
A: Your first step is half right.  Leave the $\frac 1x$ as a factor for a bit.  We have
$$\log{\sqrt\frac{1 + x}{1 - x}}=\frac 12((\log 1+x)-\log(1-x))$$
Now expand the logs as a Taylor series, which works when $x$ is small.  As we are taking the limit as $x \to 0$, we are given (or can require) that $x$ be small.  We have
$$\log(1+x) \approx x-\frac {x^2}2+\frac {x^3}3\\
\log(1-x) \approx -x-\frac {x^2}2-\frac {x^3}3\\
\log{\sqrt\frac{1 + x}{1 - x}}=\frac 12((\log 1+x)-\log(1-x))\approx x+\frac {x^3}3$$
and plugging that in the limit is $1$.
A: $f(x) :=\log (\dfrac{1+x}{1-x})^{1/2}=$
$(1/2)\log (1+x)-(1/2)\log (1-x);$
$f'(0)=$
$\lim_{x \rightarrow 0}\dfrac{\log (\dfrac {x+1} {x-1})^{1/2}-\log 1}{x}=$
$(1/2)(1)-(1/2)(-1)=1.$
Note:
$f'(x)=$
$(1/2)\dfrac{1}{1+x}-(1/2)\dfrac{1}{1-x}(-1).$
A: \begin{align*}
L &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 +(-x+x)+ x}{1 - x}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 -x +2x}{1 - x}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt{1 + \frac{2x}{1 - x}}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\left(1 + \frac{2x}{1 - x} \right)^{1/2}}  \\
    &= \lim_{x\to 0} \log{\left(1 + \frac{2x}{1 - x} \right)^{1/2x}}  \\
    &= \log \lim_{x\to 0} \left(1 + \frac{2x}{1 - x} \right)^{1/2x}  \text{.} 
\end{align*}
Let $x = \frac{1}{2u+1}$, so that $u = \frac{1-x}{2x}$.  Then as $x \rightarrow 0^+$, $u \rightarrow \infty$ and as $x \rightarrow 0^-$, $u \rightarrow -\infty$.  We investigate both.  First,
\begin{align*}
M^+ &= \log \lim_{u \rightarrow \infty} \left( 1 + \frac{1}{u}\right)^{\frac{1}{2} + u}  \\
    &= \log \lim_{u \rightarrow \infty} \left( \left(1 + \frac{1}{u}\right)^{1/2} \left(1 + \frac{1}{u}\right)^{u} \right)  \\
    &= \log \left( 1 \cdot \mathrm{e} \right)  \\
    &= 1  \text{.}
\end{align*}
Then,
\begin{align*}
M^- &= \log \lim_{u \rightarrow -\infty} \left( 1 + \frac{1}{u}\right)^{\frac{1}{2} + u}  \\
    &= \log \lim_{u \rightarrow -\infty} \left( \left(1 + \frac{1}{u}\right)^{1/2} \left(1 + \frac{1}{u}\right)^{u} \right)  \\
    &= \log \lim_{u \rightarrow -\infty} \left( 1 + \frac{1}{u}\right)^{u}  \\
    &= \log \lim_{v \rightarrow \infty} \left( 1 + \frac{1}{-v}\right)^{-v}  &  &  \text{u = -v}  \\
    &= \log \lim_{v \rightarrow \infty} \left(\left( 1 + \frac{-1}{v}\right)^{v}\right)^{-1}  \\
    &= \log \left(\lim_{v \rightarrow \infty} \left( 1 + \frac{-1}{v}\right)^{v}\right)^{-1}  \\
    &= -\log \lim_{v \rightarrow \infty} \left( 1 + \frac{-1}{v}\right)^{v}  \\
    &= -\log \mathrm{e}^{-1}  \\
    &= --1  \\
    &= 1  \text{.}
\end{align*}
Since $M^- = M^+ = 1$, $L$ exists and $L = 1$.
