Prove $\lim_{n \to \infty}\frac{\ln (n+1)}{(n+1)(\ln^2 (n+1)-\ln^2 n)}=\frac{1}{2}$ Problem
Prove  $$\lim_{n \to \infty}\frac{\ln (n+1)}{(n+1)[\ln^2 (n+1)-\ln^2 n]}=\frac{1}{2},$$where $n=1,2,\cdots.$
My Proof
Consider the function $f(x)=\ln^2 x.$ Notice that $f'(x)=2\cdot \dfrac{\ln x}{x}.$ By Lagrange's Mean Value Theorem, we have $$\ln^2(n+1)-\ln^2 n=f(n+1)-f(n)=f'(\xi)(n+1-n)=f'(\xi)=2\cdot \frac{\ln \xi}{\xi},$$where $n<\xi<n+1.$ 
Moreover, consider another function $g(x)=\dfrac{\ln x}{x}.$ Since $g'(x)=\dfrac{1-\ln x}{x^2}<0$ holds for all $x>e,$ hence $g(n+1)<g(\xi)<g(n)$ holds for every sufficiently large $n.$ Therefore, $$\frac{1}{2} \leftarrow\frac{1}{2}\cdot\dfrac{g(n+1)}{g(n)}<\dfrac{\ln (n+1)}{(n+1)[\ln^2 (n+1)-\ln^2 n]}=\frac{1}{2}\cdot\dfrac{g(n+1)}{g(\xi)}<\frac{1}{2}\cdot\dfrac{g(n+1)}{g(n+1)}=\frac{1}{2}.$$ Thus, by Squeeze Theorem, we have that the limit we want equals $\dfrac{1}{2}.$
Am I right? The proof above is not natural to me. Any other proof ?
 A: You can use this 
$$
\lim_{n \to \infty}\dfrac{\ln (n+1)}{(n+1)[\ln^2 (n+1)-\ln^2 n]} = 
$$
$$
=\lim_{n \to \infty}\dfrac{\ln (n+1)}{(n+1)[\ln (n+1)-\ln n][\ln (n+1)+\ln n]} = 
$$
$$
=\lim_{n \to \infty}\dfrac{\ln (n+1)}{\ln\left[\left(1 +\frac{1}{n}\right)^{n+1}\right][\ln (n+1)+\ln n]} =
$$
$$
=\lim_{n \to \infty}\dfrac{\ln (n+1)}{\ln (n+1)+\ln n} = \frac{1}{2}
$$
A: There exists an appplication in both solutions of @Virtuoz's and mine. That is 

$$\lim_{n \to \infty}\frac{\ln(n+1)}{\ln n}=1.$$ 

Now, I give its proof for complement.
Proof 1
By L'Hospital's Rule, we have $$\lim_{n \to \infty}\frac{\ln(n+1)}{\ln n}=\lim_{n \to \infty}\frac{\dfrac{1}{n+1}}{\dfrac{1}{n}}=\lim_{n \to \infty}\frac{n}{n+1}=\lim_{n \to \infty}\dfrac{1}{1+\dfrac{1}{n}}=1.$$
Proof 2
Denote $f(x)=\ln x$. By Lagrange's Mean Value Theorem, we have $$\ln(n+1)-\ln n=f(n+1)-f(n)=f'(\xi)(n+1-n)=\frac{1}{\xi},$$where $n<\xi<n+1$. Let $n \to \infty$. Then $\xi \to \infty.$ Thus, $\ln(n+1)-\ln n \to 0.$ It follows that $$\lim_{n \to \infty}\frac{\ln(n+1)}{\ln n}=\lim_{n \to \infty}\left(\frac{\ln(n+1)-\ln n}{\ln n}+1\right)=0 \cdot 0+1=1.$$
