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$?$
 A: When $x>0$ we know that $\ln(1+x)<x$ so $$0<\ln(x^2+x)\ln\left(1+\frac1x\right)<{\ln(x^2+x)\over x}\to0\text{ as }x\to\infty$$
A: $$\ln(x+1)=\ln x+\ln\left(1+\frac1x\right)=\ln x+O(1/x),$$
$$\ln(x+1)^2=(\ln x)^2+O\left(\frac{\ln x}x\right)=(\ln x)^2+o(1).$$
Therefore, your limit is zero.
A: Note that 
$$0\lt\ln(x+1)-\ln x=\int_x^{x+1}{dt\over t}\lt\int_x^{x+1}{dt\over x}={1\over x}$$
so
$$0\lt(\ln(x+1)-\ln x)(\ln(x+1)+\ln x)\lt{\ln(x+1)+\ln x\over x}$$
Can you take it from there?
A: Now, $$\lim_{x \to \infty} \ln \frac{x+1}{x}\ln x(x+1)=\lim_{x\rightarrow\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\lim_{x\rightarrow\infty}\frac{\ln(x^2+x)}{x}=\lim_{x\rightarrow\infty}\frac{2x+1}{x^2+x}=0$$
A: 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
$$A=2 \frac{\log(x)}x+\frac{1-\log (x)}{x^2}+O\left(\frac{1}{x^3}\right)$$
A: For continuously differentiable functions $f$ we have:
$$ f(b) - f(a)\ =\ (b-a)\cdot f'(t) $$
where $\ t\ $ exists and is such that $\ a < t < b\ $ whenever $\ a<b$.
Let $\ f(x)\ := \log^2(x).\ $ Then
$$ f'(x)\ =\ 2\cdot\frac{\log(x)}x\ $$
hence
$$ \log^2(x+1) - \log^2(x)\ =\ 2\cdot\frac{\log(t_x)}{t_x}
\ \longrightarrow\ 0\qquad \mbox{for}\,\ x\rightarrow\infty $$
where $\ \forall_{x>0}\,\ x\, <\, t_x\, <\, x+1.$
A: You may proceed as follows by substituting $x = \frac{1}{t}$ for $t\to 0^+$ and using


*

*$(\star):\frac{\ln(1+t)}{t}\stackrel{t\to 0^+}{\longrightarrow}\left.\frac{d\ln(1+t)}{dt} \right|_{t=0} = 1$ and

*$(\star\star): t\ln t \stackrel{t\to 0^+}{\longrightarrow} 0$, which is easily checked using L'Hospital, for example


You have
\begin{eqnarray*} (\ln (x+1))^{2} - (\ln x)^2
& \stackrel{x = \frac{1}{t}}{=} & \ln^2 \frac{1+t}{t} - \ln^2 \frac{1}{t} \\
& = & (\ln (1+t) - \ln t)^2 - (\ln t)^2 \\
& = & \ln^2(1+t)- 2\ln(1+t)\ln t \\
& = & \ln^2(1+t)- 2\underbrace{\frac{\ln(1+t)}{t}}_{\stackrel{(\star)}{\rightarrow}1}\cdot \underbrace{t\ln t}_{\stackrel{(\star\star)}{\rightarrow}0} \\
& \stackrel{t\to 0^+}{\longrightarrow} & 0-2\cdot 1\cdot 0 = 0
\end{eqnarray*}
A: From your penultimate step, expand to have $$\log(1+\frac{1}{x})\left(\log(1+x)+\log x\right)=\log(1+\frac{1}{x})\left(\log x(1+\frac{1}{x})+\log x\right)=\log(1+\frac{1}{x})\left(\log x+\log(1+\frac{1}{x})+\log x\right)=\log(1+\frac{1}{x})\left(2\log x+\log(1+\frac{1}{x})\right)=\log(1+\frac{1}{x})\cdot2\log x+(\log(1+\frac{1}{x}))^2.$$ The second term vanishes when $x$ is infinite, so focus on the first term and write it as $$2\frac{\log(1+\frac{1}{x})}{\frac{1}{\log x}}.$$ Then applying good old L'hopital, we get $$\frac{\frac{-1}{x(1+x)}}{\frac{-1/x}{(\log x)^2}}=\frac{(\log x)^2}{1+x},$$ and again to get $$2\frac{\log x}{x},$$ and finally again, $$\frac{2}{x},$$ which vanishes for $x=\infty.$
