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.
 A: Let $x = 1/n$. Then you want the limit of 
$$
\frac{a^x-1}{x} = \frac{e^{x \log(a)} - 1}{x} = \frac{1 + x \log(a) + R(x \log a) -1}{x}
= \log (a) + \log a \left[\frac{R(x \log a)}{x \log a} \right] $$
Now 
$$ \frac{R(x \log a)}{x \log a} \rightarrow 0 \text { as } x\rightarrow 0$$
Where I have used the fact that $$e^{\theta} = 1 + \theta + R(\theta),
~~~ \text { and } \lim_{\theta->0} \frac{R(\theta)}{\theta} = 0$$
Here $R(\theta)$ is
$$
R(\theta)  = \sum_{k=2}^\infty \frac{\theta^k}{k!}$$
A: 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
$$
\lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)
$$
A: Another way is to notice that this is a definition of derivative (if you are allowed to use it):
$$
\lim_{v \to 0}\frac{a^v-1}{v} = a^v \log a\Bigg|_{v=0}=\log a
$$
