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I’m currently studying a course in real analysis and I’m having some trouble with calculating limits like the following one:

$$ \lim _{n \rightarrow \infty} n \cdot \frac{\sqrt[n]{n}-1}{\operatorname{log} n} $$

The answer would be as follows:

\begin{array}{l} =\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n}-1}{\frac{\log n}{n}}=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n}-1}{\log \sqrt[n]{n}} \\ \approx \lim _{n \rightarrow \infty} \frac{\sqrt [n]{n}-1}{\sqrt [n]{n}-1}=1 \end{array}

The problem I have with this possible resolution of the limit is that we have to use an approximation, which is valid since n to the 1/n goes to 1. I suppose that this ‘approximation’ has something to do with Taylor series, but that’s something we haven’t covered yet in class. So I was wondering whether there was another way of computing this limit, which does not require the use of such ‘approximation’. There are a bunch of other limits like this which can be solved by this kind of ‘approximation’.

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2 Answers 2

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$$ n\cdot \frac{\sqrt[n]{n}-1}{\log n}=\frac{\mathrm{e^{\frac{\log n}{n}}}-1}{\frac{\log n}{n}}=\frac{\mathrm{e}^{a_n}-1}{a_n} $$ where $a_n>0$ and $a_n\to 0.$

Next, use the fact that $\lim_{h\to 0}\frac{\mathrm{e}^{h}-1}{h}=1$, to obtain that $\lim_{n\to\infty}\frac{\mathrm{e}^{a_n}-1}{a_n}=1$.

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Set $y := n^{1/n} >1$, and consider $y \rightarrow 1$.

$\log y= \log (1+(y-1))=$

$(y-1)+O((y-1)^2);$

$\lim_{y \rightarrow 1}\dfrac{y-1}{(y-1)+O((y-1)^2)}=1.$

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  • $\begingroup$ Although OP points out that $y=n^{1/n}\to1$ is something that they've not yet covered. $\endgroup$
    – Teepeemm
    Commented Nov 15, 2020 at 17:19
  • $\begingroup$ Yes, you are right. On the other hand assuming $n^{1/n}$ converges, say to $a>1$, the limit were $\dfrac{a-1}{\log a}$. $\endgroup$ Commented Nov 15, 2020 at 18:16

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