Find limit of the following sequence: $\frac{\log(n+1)}{\log(n)}$? Find the limit of $\frac{\log(n+1)}{\log(n)}$ where $n\rightarrow\infty$. Here $n$ is a natural number so I guess we can't use L'Hopital
 A: 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.
A: Hint: $\log(n+1)=\log(n) + \log\left(1+\frac{1}{n}\right)$
A: Hint:
$$\frac{\log(n+1)}{\log(n)}=\frac{\log(n)+\log\left(1+\frac{1}{n}\right)}{\log(n)}=1+\frac{\log\left(1+\frac{1}{n}\right)}{\log(n)}.$$
A: L'Hopital's rule can be used here: If the limit has a certain value if $n$ is allowed to take real values, then it still has the same value if $n$ is restricted to integers.
But L'Hopital's rule, although it can sometimes find a limit very fast, often doesn't give much insight.
A: We have, $$\lim_{n\to \infty}\frac{\log(n+1)}{\log (n)}$$ Let $n=\frac{1}{t}\implies t\to 0\ as \ n\to \infty$ $$=\lim_{t\to 0}\frac{\log\left(\frac{1}{t}+1\right)}{\log \left(\frac{1}{t}\right)}$$ $$=\lim_{t\to 0}\frac{\log\left(\frac{t+1}{t}\right)}{\log(1)-\log \left(t\right)}$$$$=\lim_{t\to 0}\frac{\log\left(t+1\right)-\log(t)}{-\log (t)}$$ $$=-\lim_{t\to 0}\frac{\log\left(t+1\right)}{\log (t)}+1$$$$=1-\lim_{t\to 0}\frac{\frac{d}{dt}(\log\left(t+1\right))}{\frac{d}{dt}\log(t)}$$ $$=1-\lim_{t\to 0}\frac{\frac{1}{t+1}}{\frac{1}{t}}=1-\lim_{t\to 0}\frac{t}{t+1}$$$$=1-\frac{0}{0+1}=1$$
