**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 \(| f(x_1) – f(x_2) | > \epsilon \) .

## 1 comment