Limit of $\ln^2(x+1)-\ln^2(x)$ at infinity

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 $\frac{\ln(\varepsilon )}{\varepsilon }\to 0, \varepsilon \to+\infty$.

$$\lim_{x \to +\infty}\ln^2{(x+1)}-\ln^2{x}=\lim_{x \to +\infty}\ln{(x^2+x)}\ln{(1+1/x)}$$ $$=\lim_{x \to +\infty}\frac{\ln{(x^2+x)}}{\frac{1}{\ln{(1+1/x)}}}=^{L'HOSPITAL}=\lim_{x \to +\infty}(2+\frac{1}{x})\ln{(1+1/x)}=0$$

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

$$\ln\left(1+\frac1x\right)(\ln (x+1)+ \ln(x))=\frac{\ln\left(1+\frac1x\right)}{\frac1x}\frac{\ln (x+1)+ \ln(x)}{x} \to 1 \cdot 0=0$$