$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=1$ Why is $$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=1$$ 
Can I compute the part inside the square bracket first?
Thanks for helping.                                                                                     
 A: The standard way to compute limits of this kind is to use the well-known limit:
$$\lim_{x\to\infty}\left(1+\frac1x\right)^{x}=e=\lim_{x\to0}(1+x)^{\frac1x}$$
In your case, rewrite the limit as:
$$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=\lim_{n\to\infty}\left[\left(1+\frac{\log(n+1)-\log n}{\log n}\right)^{\frac{\log n}{\log(n+1)-\log{n}}}\right]^{\frac{\log(n+1)-\log n}{\log n}n}$$
Now use the aforementioned limit and compute the $\lim_{n\to\infty}\frac{\log(n+1)-\log n}{\log n}n$ separately. 
A: Take $\log$s and use $\log (x+y) \leq \log x + y/x$ (which is true by MVT) twice. Note
\begin{align*}
\log\left(\frac{\log(n+1)}{\log n}\right)^n 
 &= n \log\log(n+1) - n \log\log n \\
 &\leq n\log(\log n + 1/n) - n \log \log n \\
 &\leq n(\log\log n + 1/(n \log n)) - n\log\log n\\
 &=1/\log n \longrightarrow 0
\end{align*}
A: Using the equivalence $$\log{\left(1+\dfrac{1}{n}\right)}\underset{n\to\infty}{\sim}  \dfrac{1}{n}$$
we have
$$\dfrac{\log(n+1)}{\log n}=\dfrac{\log \left(n\left(1+\dfrac{1}{n}\right)\right)}{\log n}=\dfrac{\log {n}+\log{\left(1+\dfrac{1}{n}\right)}}{\log n}=\\
=1+\dfrac{\log{\left(1+\dfrac{1}{n}\right)}}{\log n}\underset{n\to\infty}{\sim} 1+\dfrac{1}{n\log n}.$$
Then
$$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=\lim_{n\to\infty}\left(1+\dfrac{1}{n\log n}\right)^{n}=\lim_{n\to\infty}{e^{\frac{1}{\log{n}}}}=1.$$
A: We'll write  the function in $e^{\log}$ form$$\Big(\dfrac{\log(n+1)}{\log n}\Big)^n=\exp(n\log(\dfrac{\log(n+1)}{\log(n)}))=\exp(n \log(1+\dfrac{1}{n\log n }+o(\frac{1}{n^2})))$$$$=\exp(\frac{1}{\log(n)}+o(\frac{1}{n}))$$ Taking the limit we get 1.The 2nd step is by $\log(1+x)=x +o(x^2)$ for x close to 0.
