Find limit $\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$ Find limit $\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$.
My attempt 
$\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$.
$= \lim_{x \to \infty} (\ln(x+1) - \ln (x))(\ln (x+1) + ln(x))$
$=\lim_{x \to \infty} (\ln (\frac{x+1}{x})(\ln x(x+1))$
It's $0 \times \infty$ indeterminate form.
Then I'm stuck.
How to proceed$?$

Write
$$\log(x+1)=\log \left(x\left(1+\frac{1}{x}\right) \right)=\log(x)+\log \left(1+\frac{1}{x}\right) $$ and, by Taylor,
$$\log \left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$
So,
$$A=\log ^2(x+1)-\log ^2(x)=\left(\log(x)+\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right) \right)^2-\log^2(x)$$ Expand to get
$$A=2 \frac{\log(x)}x+\frac{1-\log (x)}{x^2}+O\left(\frac{1}{x^3}\right)$$
