Problem: Every differentiable function f: (0, 1) –> [0, 1] is uniformly continuous.

**Discussion;**

**False**

Note that every differentiable function f: [0,1] –> (0, 1) is uniformly continuous by virtue of uniform continuity theorem which says every continuous map from closed bounded interval to R is uniformly continuous. However in this case the domain is an open interval.

We can easily find counter example such as \(f(x) = \sin ( \frac {1}{x} ) \). Intuitively speaking the function oscillates (between -1 and 1) faster and faster as we get close to x = 0. Hence we can get two arbitrarily close values of x such that their functional value’s difference equals a particular number (say 1) therefore exceeding any \(\epsilon < 1 \)

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