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 ∞

\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 $L = 1$.
