Finding $\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$ I am trying to find the limit of:
$$\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$$
I know it should be very simple but I don't seem to get it.
Thanks in advance for any help!

It should be noted that we can define the (natural) logarithm of a number $x>0$ as
$$L(x)=\lim_{h\to 0}\frac{x^h-1}{h}$$
Although it may take a time, it is not awfully complicated to prove this function indeed has the defining properties of the logarithm.
$$\tag 1 L(xy)=L(x)+L(y)$$
$$\tag 2 L(x^a)=a L(x)$$
$$\tag 3 1-\frac 1 x \leq L(x)\leq x-1$$
$$\tag 4\lim_{x\to 0}\frac{L(x+1)}{x}=1$$
$$\tag 5 L'(x)=\frac 1 x$$
Note that $1\Rightarrow 2 \;(a\in \Bbb Z)\Rightarrow 3\Rightarrow 4\Rightarrow 5$.
On the other hand, a naïve use of L'Höpital's rule, gives
$$\lim_{h\to 0}\frac{x^h-1}{h}\mathop=\limits^{\frac 0 0}\lim_{h\to 0}\frac{x^h\log x}{1}=\log x$$
If you're interested in the proofs, just let me know.
