Limit of $\ln^2(x+1)-\ln^2(x)$ at infinity How can we show that $\ln^2(x+1)-\ln^2(x)$ converges to zero as $x\rightarrow \infty$?
Can I use $$\ln^2(x+1)-\ln^2(x)=(\ln (x+1)- \ln(x))(\ln (x+1)+ \ln(x))$$
 A: $$\lim_{x \to +\infty}\ln^2{(x+1)}-\ln^2{x}=\lim_{x \to +\infty}\ln{(x^2+x)}\ln{(1+1/x)}$$ $$=\lim_{x \to +\infty}\frac{\ln{(x^2+x)}}{\frac{1}{\ln{(1+1/x)}}}=^{L'HOSPITAL}=\lim_{x \to +\infty}(2+\frac{1}{x})\ln{(1+1/x)}=0$$
A: Using MVT, for $f(x)=\ln^2(x)$, $\exists \varepsilon \in(x,x+1)$
$$\ln^2(x+1)-\ln^2(x)=
f'(\varepsilon)\cdot(x+1-x)=
2\cdot \frac{\ln(\varepsilon )}{\varepsilon }$$
Because $x\to+\infty \Rightarrow \varepsilon \to+\infty$ and $\frac{\ln(\varepsilon )}{\varepsilon }\to 0, \varepsilon \to+\infty$.
A: We have that
$$\ln^2(x+1)-\ln^2(x)=(\ln (x+1)- \ln(x))(\ln (x+1)+ \ln(x))=$$$$=\ln\left(1+\frac1x\right)(\ln (x+1)+ \ln(x))$$
and by standard limits
$$\ln\left(1+\frac1x\right)(\ln (x+1)+ \ln(x))=\frac{\ln\left(1+\frac1x\right)}{\frac1x}\frac{\ln (x+1)+ \ln(x)}{x} \to 1 \cdot 0=0$$
A: $$\ln(x+1)-\ln x=\ln\left(1+\frac1x\right)=O(1/x)$$
as $x\to\infty$.
$$\ln(x+1)+\ln x=O(\ln x)$$
as $x\to\infty$.
Therefore
$$(\ln(x+1))^2-(\ln x)^2=O\left(\frac{\ln x}x\right)$$
and so this tends to zero as $x\to\infty$.
A: Set $x=\frac 1u$ and consider $u\to 0^+$ and use the standard limits


*

*$(\star)$: $\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1$ and $\lim_{u\to 0}(u\ln u)=0$
\begin{eqnarray} \ln^2(x+1)-\ln^2(x)
& \stackrel{x=\frac 1u}{=} & \left(\ln(1+u) -\ln u\right)^2 - \ln^2 u \\
& = & \ln^2(1+u) - 2\ln u\ln(1+u) \\
& = & \ln^2(1+u) - 2u\ln u\frac{\ln(1+u)}{u} \\
& \stackrel{(\star): u \to 0^+}{\longrightarrow} & 0-2\cdot 0\cdot 1 = 0
\end{eqnarray}
A: Yes,you can do that, it would be:
$$ln(1+\frac {1}{x})(ln(x^2+x))$$ as $x$ approaches infinity it tends to infinity 
