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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

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  • $\begingroup$ The expression makes sense if $n$ is real, not necessarily an integer. L'Hopital is fine. $\endgroup$ – lulu Aug 1 '15 at 18:05
  • $\begingroup$ Actually you can use l'Hôpital's rule, because the limit exists even as $n\rightarrow\infty$ over real numbers, and so a fortiori over integers. $\endgroup$ – Noam D. Elkies Aug 1 '15 at 18:05
  • $\begingroup$ But fundamentally it isn't a correct proof right? $\endgroup$ – martianwars Aug 1 '15 at 18:06
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    $\begingroup$ @KalpeshKrishna sure it is. If the limit exists for real n (as it does) then any subsequence converges to the same limit. $\endgroup$ – lulu Aug 1 '15 at 18:07
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    $\begingroup$ Your point, I think, is that the integer limit might exist while the real limit does not. Think $sin(n\pi)$ for example. That is true. But if the real limit exists then the integer limit is subsumed by it. $\endgroup$ – lulu Aug 1 '15 at 18:11
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Hint: $\log(n+1)=\log(n) + \log\left(1+\frac{1}{n}\right)$

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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.

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Hint:

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

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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.

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We have, $$\lim_{n\to \infty}\frac{\log(n+1)}{\log (n)}$$ Let $n=\frac{1}{t}\implies t\to 0\ as \ n\to \infty$ $$=\lim_{t\to 0}\frac{\log\left(\frac{1}{t}+1\right)}{\log \left(\frac{1}{t}\right)}$$ $$=\lim_{t\to 0}\frac{\log\left(\frac{t+1}{t}\right)}{\log(1)-\log \left(t\right)}$$$$=\lim_{t\to 0}\frac{\log\left(t+1\right)-\log(t)}{-\log (t)}$$ $$=-\lim_{t\to 0}\frac{\log\left(t+1\right)}{\log (t)}+1$$$$=1-\lim_{t\to 0}\frac{\frac{d}{dt}(\log\left(t+1\right))}{\frac{d}{dt}\log(t)}$$ $$=1-\lim_{t\to 0}\frac{\frac{1}{t+1}}{\frac{1}{t}}=1-\lim_{t\to 0}\frac{t}{t+1}$$$$=1-\frac{0}{0+1}=1$$

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