How to find $\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)$?

Question

How to compute the following limit?

$\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)$

We know that it's got something to do with $\ln$ or $\exp$.

We know that $\lim\limits_{n\to\infty} \sqrt[n]{a} = 1$ but it seems to not to be true that therefore $\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1) = \lim\limits_{n\to\infty} n·(1-1) = 0$.

What we know is

$\lim\limits_{n\to\infty} (1-\frac{1}{n})^n = \lim\limits_{n\to\infty} (1 + \frac{1}{n})^n \lim\limits_{n\to\infty} (1-\frac{1}{n})^{n+1} = e$

and

$\lim\limits_{n\to\infty} (1+\frac{x}{n})^n = \exp(x)$

Answer 1

Substitute $t=\frac{1}{n}$ , hence :

$L=\displaystyle \lim_{t \to 0} \frac{a^t-1}{t}$

Now , apply L'Hopital rule ,hence :

$L=\displaystyle \lim_{t \to 0} a^t \cdot \ln a=\ln a$

Answer 2

Using Taylor's expansion, $$ n(\sqrt[n]{a}-1)=n(e^{\frac1n\log a}-1)=n(1+\frac1n\,\log a+o(\frac1{n^2})-1)=\log a + o(\frac1n)\xrightarrow{n\to\infty}\log a. $$

Answer 3

Set $x_n=n(\sqrt[n]{a}-1)$ for all $n$. Rearranging this you get $(\frac{x_n}{n}+1)^n=a$ for all $n \in \mathbb{N}$. But this also means that $$(\frac{x_n}{n}+1)^n \rightarrow a=e^{\ln(a)}$$ as $n\rightarrow \infty$. Hence we must have $x_n \rightarrow \ln(a)$ as $n \rightarrow \infty$.