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 ∞

$f(x) :=\log (\dfrac{1+x}{1-x})^{1/2}=$
$(1/2)\log (1+x)-(1/2)\log (1-x);$
$f'(0)=$
$\lim_{x \rightarrow 0}\dfrac{\log (\dfrac {x+1} {x-1})^{1/2}-\log 1}{x}=$
$(1/2)(1)-(1/2)(-1)=1.$
Note:
$f'(x)=$
$(1/2)\dfrac{1}{1+x}-(1/2)\dfrac{1}{1-x}(-1).$
