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How to find $\lim\limits_{n\to \infty} \dfrac{\log(n)}{n}$?

It is of no doubt that if we use L'Hospital's rule we will get $\lim\limits_{n\to\infty}\dfrac{ \frac{1}{n}}{1}$ which is of course equal to $0$. But how can we find the limit without using the rule?

I tried to substitute $n = x+1$ so that I could apply exponential series but that also seems to be not working. Is there any other possible method? Or do I have to do another substitution?

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  • $\begingroup$ $2 log\, (n^{1/2}) \leq 2(n^{1/2}-1)$ $\endgroup$
    – Kavi Rama Murthy
    Jan 9 at 5:33
  • $\begingroup$ $$n=e^{\log n}>\frac{1}2\log^2 n$$ $\endgroup$
    – Thomas Andrews
    Jan 9 at 5:39

2 Answers 2

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$x\leq e^ x$ for all $x\in \mathbb R$.

So $\sqrt n\le e^{\sqrt n}$. Taking log on both sides gives: $\frac 12\log n\le\sqrt n$. It follows that $0\leq \frac 12\frac{\log n}{n}\leq\frac 1{\sqrt n}$. The result follows by Squeeze principle.

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  • $\begingroup$ May I know please, how does $\frac12\log(n) ≤ \sqrt{n} $ follows that $0 ≤ \frac12 \frac{\log(n)}{n} ≤ \frac{1}{\sqrt{n}}$? $\endgroup$
    – Yooo
    Jan 9 at 5:44
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    $\begingroup$ @Utkarsh: log of a natural number is always non negative and hence the first inequality in $0\leq \frac 12\frac{\log n}{n}\color{red}{\leq}\frac 1{\sqrt n}$. The red coloured inequality follows by dividing both sides of $\frac 12\log n\le\sqrt n$ by $n$. $\endgroup$
    – Koro
    Jan 9 at 5:49
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Here is another proof, using the integral definition of the logarithm (and the power rule). For $n\geq 1$, we have the following.

\begin{align*} \log n &= \int_1^n \frac1x\ dx\\ &\leq \int_1^n \frac{1}{x^{0.9}}\ dx\\ &=\int_1^n x^{-0.9}\ dx\\[0.5em] &= \frac{x^{0.1}}{0.1}\Big|_{x=1}^{n}\\[0.5em] &= 10n^{0.1} - 10 \end{align*}

Dividing through by $n$, we obtain

$$\frac{\log n}{n} \leq \frac{10}{n^{0.9}} - \frac{10}{n}$$

and the rest follows from the squeeze theorem

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