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’.
Best Answer
$$ n\cdot \frac{\sqrt[n]{n}-1}{\log n}=\frac{\mathrm{e^{\frac{\log n}{n}}}-1}{\frac{\log n}{n}}=\frac{\mathrm{e}^{a_n}-1}{a_n} $$ 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 $\lim_{n\to\infty}\frac{\mathrm{e}^{a_n}-1}{a_n}=1$.