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.

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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.

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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.

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Great! I was thinking that there is a more popular word for the 1st order Taylor's expansion (like derivative or a differential), but the lack of coffee didn't allow me to remember that :) I wonder, by the way - if there is a natural way to prove this limit just based on sequences, and not on analysis involving limits of functions. –  Ilya May 30 '12 at 12:58
There are things one could do using the binomial expansion, the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ pops out. More fundamental in a sense, but also a lot more work. –  André Nicolas May 30 '12 at 13:06

Depends on your knowledge, e.g. with Taylor expansion you have $2^x-1\sim x\log2$ for $x\to 0$, so $$\lim\limits_{n\to\infty}n(\sqrt[n]{2}-1) = \lim\limits_{x\to 0}\frac1x(2^x-1) = \lim\limits_{x\to 0}\frac1x(x\log2) =\log 2.$$

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