Evaluate the limit $\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1){{\log }^2}(n + 1) - n{{\log }^2}n}}{{{{\log }^2}n}}$ Evaluate:
$$\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1){{\log }^2}(n + 1) - n{{\log }^2}n}}{{{{\log }^2}n}}$$

Intuitively, I feel that for large $n$, ${\log}(n+1) \approx \ {\log}(n) 
$. So, the above limit should reduce to:

$$=\mathop {\lim }\limits_{n \to \infty } \frac{{\{ (n + 1) - n\} {{\log }^2}n}}{{{{\log }^2}n}} \ = 1$$

However, can someone please suggest how can one mathematically show this.

Thanks!
If you want to be strict, write $$\log(1+n)=\log(n)+\log(1+\frac 1n)$$ So, the numerator is $$A=(n + 1) \log^2(n + 1) - n\log^2(n)=(n+1)\left(\log(n)+\log(1+\frac 1n)\right)^2-n \log^2(n)$$ Expanding the square and grouping $$A=a^2 n+a^2+2 a n \log (n)+2 a \log (n)+\log ^2(n)$$ where $a=\log(1+\frac 1n)\approx \frac 1n $ is small when $n$ is large.

Is this what you are looking for ?

Edit

Alternatively, you could set $n=\frac 1x$ and use Taylor series around $x=0$. Doing so, the entire expression will be $$\left(1-\frac{2}{\log (x)}\right)+x \left(\frac{1}{\log ^2(x)}-\frac{1}{\log
   (x)}\right)+\frac{x^2}{3 \log (x)}+O\left(x^3\right)$$
It should be $$\lim_{n\to \infty} \left((n+1)(\frac{\ln(n+1)}{\ln n})^2 -n\right) = \lim_{n\to \infty} (n+1 - n) = 1.$$