Proof $\lim\limits_{n \rightarrow \infty}n(a^{\frac{1}{n}}-1)=\log a$ I want to show that for all $a \in \mathbb{R }$
$$\lim_{n \rightarrow \infty}n(a^{\frac{1}{n}}-1)=\log a$$
So far I've got $\lim\limits_{n \rightarrow \infty}ne^{(\frac{1}{n}\log a)}-n$, but when i go on to rearrange this, i come  after a few steps back to the beginning...
We have no L'Hôpital and no differential and integral calculus so far.

This is essentially the inverse of
$$
e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n
$$
The sequence of functions $f_n(x)=\left(1+\frac xn\right)^n$ converge equicontinuously; simply note that $f_n'(x)=\left(1+\frac xn\right)^{n-1}\sim e^x$. Thus, we get
$$
\lim_{n\to\infty}n\left(e^{x/n}-1\right)=x
$$
which is the same as
$$
\lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)
$$
