Limit of the sequence $\lim_{n\to\infty}\left(\frac{\log(n+2)}{\log (n+1)}\right)^{n\log n}$

I want to find the limit of the following sequence

$$x_n=\left(\frac{\log(n+2)}{\log (n+1)}\right)^{n\log n}$$

Try using the logarithmic equivalence criterion, that is:

$$\lim_{n\to\infty}{x_n}^{y_n}=e^L\leftrightarrow \lim_{n\to\infty}{y_n({x_n-1})}=L$$

With the above, try to find the limit of $n\log n\left(\frac{\log(n+2)}{\log (n+1)}-1\right)$

However, I was unsuccessful. Any help regarding this? thank you very much.

The logarithmic equivalence criterion which you are referring seems not necessary, let use instead

$$\left(\frac{\log(n+2)}{\log (n+1)}\right)^{n\log n}=e^{n \log n\log\left(\frac{\log(n+2)}{\log (n+1)}\right)}$$

and by

$$\frac{\log(n+2)}{\log (n+1)} =1+\frac{\log\frac{n+2}{n+1}}{\log (n+1)}$$

we have

$$n \log n\log\left(\frac{\log(n+2)}{\log (n+1)}\right)=\frac{\log\left(1+\frac{\log\frac{n+2}{n+1}}{\log (n+1)}\right)}{\frac{\log\frac{n+2}{n+1}}{\log (n+1)}}\frac{n \log n\log\frac{n+2}{n+1}}{\log (n+1)}$$

with

$$\frac{\log\left(1+\frac{\log\frac{n+2}{n+1}}{\log (n+1)}\right)}{\frac{\log\frac{n+2}{n+1}}{\log (n+1)}}\to 1$$

and

$$\frac{n \log n\log\frac{n+2}{n+1}}{\log (n+1)}=\frac{\log n}{\log (n+1)}\log\left(1+\frac 1{n+1}\right)^n \to 1\cdot \log e =1$$

then

$$\left(\frac{\log(n+2)}{\log (n+1)}\right)^{n\log n}=e^{n \log n\log\left(\frac{\log(n+2)}{\log (n+1)}\right)} \to e^1=e$$

Edit

Using the criterion you are referring to, in the same way as in the previous conclusion, we obtain

$$n\log n\left(\frac{\log(n+2)}{\log (n+1)}-1\right)=n\log n\frac{\log \frac{n+2}{n+1}}{\log (n+1)}\to 1$$