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Uniform Continuity

Problem: Let f: R --> R be defined by f(x) = sin (x^3) . Then f is continuous but not uniformly continuous.

Discussion:

True

It is sufficient to show that there exists an epsilon > 0 such that for all \delta > 0 there exist x_1 , x_2 \in R such that | x_1 - x_2 |< \delta implies  | f(x_1) - f(x_2) | > \epsilon .

Assume epsilon = 0.99 . Let x_1 = (k \pi )^{\frac{1}{3}} and x_2 = (k \pi + \frac{\pi}{2} )^{\frac{1}{3}} .

Hence |x_1 - x_2 | = (k \pi + \frac{\pi}{2})^{\frac{1}{3}} - (k \pi)^{\frac{1}{3}} < \frac {1}{k^{2/3}} . (This is achieved by some simple algebra like rationalization )

Now if we take k > \frac {1}{\delta ^{3/2}} then |x_1 - x_2| < \frac {1}{k^{2/3}} < \delta but |f(x_1) - f(x_2) | = 1 > 0.99 = \epsilon

Hence there exists an \epsilon (= 0.99) such that for any value of \delta >0 we will get x_1 , x_2 such that | x_1 - x_2 |< \delta implies  | f(x_1) - f(x_2) | > \epsilon .

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