<s>[INST] 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 ∞
 [/INST] 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 $\exp(a) = e \Rightarrow \boxed{a = 1}.$
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