I try to calculate the following limit: $$\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}$$ I think it should equal 1, because: $$\exp(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}$$ (Already proven)

Solving for $x$ gives: $$ \log x = \lim_{n \to \infty} n \left(\sqrt[n]{x}-1\right) \implies \lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}=\lim_{n\to\infty}\frac{\log n}{\log n}=1 $$ But I like to calculate the limit with just standard things like L’Hôpital’s rule, because the previous way is maybe wrong and contains too much magic.

For example, I tried this: $$ \begin{align*} \lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n} &= \lim_{n\to\infty}\frac{\frac{d}{dn}\!\!\left(n\left(\sqrt[n]{n}-1\right)\right)}{\frac{d}{dn}\log n} \\[6pt] &=\lim_{n\to\infty}\frac{n^{1/n-1}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n}-1}{1/n} \\[6pt] &=\lim_{n\to\infty}\left(n\left(n^{1/n-1}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n}-1\right)\right) \\[6pt] &=\lim_{n\to\infty}\left(n^{1/n}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n+1}-n\right) \\[6pt] &=??? \end{align*} $$ But then it becomes really ugly and looks wrong. Have I done something wrong? Is there another way to find the limit?

Thank you for any ideas.

  • $\begingroup$ Your two lines after "Solve for x" are completely incomprehensible to me. $\endgroup$
    – Timbuc
    Nov 4 '14 at 17:27
  • $\begingroup$ They are probably wrong, but i just started playing around with $\lim_{n\to\infty}\left(\exp(x)=\left(1+\frac{x}{n}\right)^{n}\right)$ and produced $\lim_{n\to\infty}\left(\log(x)=n\left(\sqrt[n]{x}-1\right)\right)$ and so on. I have no idea idea if the used operations are valid. I was just playing around. $\endgroup$ Nov 4 '14 at 17:49
  • $\begingroup$ Could you please explain how you passed from the first expression to the second one? $\endgroup$
    – Timbuc
    Nov 4 '14 at 17:54
  • $\begingroup$ Of course: $\lim_{n\to\infty}\left(\exp(x)=\left(1+\frac{x}{n}\right)^{n}\right)$. Replace $\exp\left(x\right)$ with $y$, this creates $\lim_{n\to\infty}\left(y=\left(1+\frac{x}{n}\right)^{n}\right)$. Solve this now for x to get the $\log\left(y\right)$ function: $\Rightarrow\lim_{n\to\infty}\left(\sqrt[n]{y}=1+\frac{x}{n}\right)$ $\Rightarrow\lim_{n\to\infty}\left(\sqrt[n]{y}-1=\frac{x}{n}\right)$ $\Rightarrow\lim_{n\to\infty}\left(n\left(\sqrt[n]{y}-1\right)=x\right)$ $=lim_{n\to\infty}\left(n\left(\sqrt[n]{y}-1\right)=\log\left(y\right)\right)$ $\endgroup$ Nov 4 '14 at 18:10
  • $\begingroup$ 1) If you replace $\;y=\exp(x)\;$ , then $\;x=\log y\;$ , and thus the first expression equals in fact $\;y=\left(1+\frac{\log y}n\right)^n\;$ . You can't substitute for $\;x\;$ in one part of the expression but not in another part of the same expression. $\endgroup$
    – Timbuc
    Nov 4 '14 at 20:17

Defining $\displaystyle a_n := \sqrt[n]{n} - 1$, we have $a_n > 1$ and the well-known result$\displaystyle \lim_{n \to \infty}a_n = \lim_{n \to \infty}\sqrt[n]{n} - 1 =0.$


$$n = (1+a_n)^n, \\ \log n = n \log (1 + a_n).$$

Using the inequality $\displaystyle \frac{a_n}{1+a_n} \leqslant \log (1+a_n) \leqslant a_n$, we find

$$\frac{na_n}{1+a_n} \leqslant \log n \leqslant na_n,\\ 1 \leqslant \frac{n a_n}{\log n}\leqslant 1 + a_n, \\ 1 \leqslant \frac{n (\sqrt[n]{n} -1)}{\log n}\leqslant \sqrt[n]{n} .$$

Tt follows from the squeeze principle that

$$\lim_{n \to \infty}\frac{n (\sqrt[n]{n} -1)}{\log n}=1.$$

Alternatively, using L'Hospital's rule for $x \in \mathbb{R}$,

$$\lim_{x\to\infty}\frac{x\left(\sqrt[x]{x}-1\right)}{\log x}=\lim_{x\to\infty}\frac{\exp{(x^{-1}\log x})-1}{x^{-1}\log x}\\=\lim_{x\to\infty}\frac{\exp{(x^{-1}\log x)})\frac{d}{dx}(x^{-1}\log x)}{\frac{d}{dx}(x^{-1} \log x )}\\=\lim_{x\to\infty}\exp(x^{-1} \log x)=1$$

Then it is not difficult to show, for $n \in \mathbb{N}$,



$\begin{array}\\ \dfrac{n\left(\sqrt[n]{n}-1\right)}{\log n} &=\dfrac{n\left(e^{\ln(n)/n}-1\right)}{\log n}\\ &=\dfrac{n\left(1+\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})-1\right)}{\log n}\\ &=\dfrac{n\left(\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})\right)}{\log n}\\ &=\dfrac{\left(\ln(n)+O(\frac{\ln^2(n)}{n})\right)}{\log n}\\ &=1+O(\frac{\ln(n)}{n})\\ &\to 1\\ \end{array} $


rewrite your term in the form $\frac{n^{1/n}-1}{\log(n)}{n}$ and use L'Hospital the result is $1$


We can use approximations : $$\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n} = \lim_{n\to\infty}\frac{\left(\sqrt[n]{n}-1\right)}{\frac{\log n}{n}}=\lim_{n\to\infty}\frac{\left(n^{\frac{1}{n}}-1\right)}{\log n^{\frac{1}{n}}}$$

Now, observe that $\lim_{n\to\infty} \frac{\log n}{n} = 0$ and hence $\lim_{n\to\infty}n^{\frac{1}{n}}=1$. And we know $\log (x) \approx x-1$ when $x$ is close to $1$. Therefore $\log n^{\frac{1}{n}} \approx \left(n^{\frac{1}{n}}-1\right)$ and hence the required limit will be $1$


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