Find limit of the following sequence: $\frac{\log(n+1)}{\log(n)}$?
Hint: $\log(n+1)=\log(n) + \log\left(1+\frac{1}{n}\right)$
More general result: If $f$ is differentiable on $(a,\infty)$ and $\lim_{x\to\infty} f'(x) = 0,$ then $\lim_{x\to\infty} (f(x+1) - f(x)) = 0.$ Proof: By the MVT, $\,f(x+1) - f(x)= f'(c_x)\cdot 1,$ for some $c_x \in (x,x+1).$ As $x\to \infty, c_x \to \infty,$ hence $f'(c_x)\to 0,$ giving the result.
Hint:
$$\frac{\log(n+1)}{\log(n)}=\frac{\log(n)+\log\left(1+\frac{1}{n}\right)}{\log(n)}=1+\frac{\log\left(1+\frac{1}{n}\right)}{\log(n)}.$$