Proving $\lim_{n\rightarrow\infty} n(2^{1/n} - 1) = \log 2$ How do you prove $\lim_{n\rightarrow\infty} n(2^{1/n} - 1) = \log 2$ ?
Background: in computer science, if you allocate CPU time to $n$ processes by rate-monotonic scheduling, all the processes get sufficient amount of time when the quantity $U$ called CPU utilization is less than or equal to $n(2^{1/n} - 1)$.  It is monotonously decreasing and tends to $\log 2$ when $n\rightarrow\infty$, so if $U \le \log 2$, you can be sure all the processes will be given sufficient CPU time.
 A: Let $h=1/n$. It is enough to show that
$$\lim_{h\to 0}\frac{2^h-1}{h}=\log 2.$$
This just says that the derivative of $2^x$ at $x=0$ is $\log 2$, which is easy to check.
Remark: Equivalently, let $x=1/n$, rewrite our expression as $\frac{e^{(\ln 2) x}}{x}$, and find the limit of this as $x\to 0$ using L'Hospital's Rule. But reducing to the definition of the derivative has a more "elementary" feel. 
The mathematically natural approach is the one by Ilya. The Taylor expansion of $2^x$, that is, of $e^{(\ln 2)x}$, tells us about the "local" behaviour of $2^x$ near $x=0$, so it is the right tool to use. The derivative also tells us about local behaviour, but the Taylor expansion is finer-grained.
A: Depends on your knowledge, e.g. with Taylor expansion you have $2^x-1\sim x\log2$ for $x\to 0$, so
$$
\lim\limits_{n\to\infty}n(\sqrt[n]{2}-1) = \lim\limits_{x\to 0}\frac1x(2^x-1) = \lim\limits_{x\to 0}\frac1x(x\log2) =\log 2.
$$
A: We have 
$$\lim_{n\to\infty} n(2^{1/n} - 1) = \lim_{t\downarrow 0}{2^t - 1\over t}.$$
Put $f(t) = 2^t$; then the limit is just $f'(0)$.  Since $f'(x) = 2^x\log(2)$, we are done.
