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Suppose $f$ is a real-valued function and $f(x) \geq 1$ for every real $x \in [0,1]$.

Why we can always find a unique positive integer $n$ such that $2^{n} \leq f(x) < 2^{n+1}$ ?

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    @User: Take $f(x)=e^{x}$. Clearly $f(x) \geq 1$ for every $x \in [0,1]$. But as $x \to \infty$, $e^{x} \to \infty$.2010-11-17
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    Do you mean for any given $x$? Or for all $x$ at once? The second statement is false; the first is true because $f(x)\ge 1=2^0$ and $\{[2^n,2^{n+1})\}$ partitions $[1,\infty)$.2010-11-17

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