# Thread: sin(x^2) not uniformly continuous

1. ## sin(x^2) not uniformly continuous

Can someone please explain to me why sin(x^2) is not uniformly continuous.

2. Originally Posted by summerset353
Can someone please explain to me why sin(x^2) is not uniformly continuous.
On what set?

3. the set being R

4. The reason is that the function is contionuos and bounded but its derivative is unbounded...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. that's false.

consider $\displaystyle f(t)=\sqrt t$ being uniformly continuous on $\displaystyle [0,\infty)$ but its derivative is not bounded.

6. Krizalid is right. Having bounded derivative is a sufficient condition for being uniformly continuous, but it is not necessary.

Consider the sequences $\displaystyle x_n=\sqrt{2\pi n}$, $\displaystyle y_n=\sqrt{2\pi n+\pi/2}$. We have that $\displaystyle |y_n-x_n|$
tends to 0 and $\displaystyle f(y_n)-f(x_n)=1$. From this fact it follows that $\displaystyle f$ is not uniformly continuous.

7. Originally Posted by summerset353
Can someone please explain to me why sin(x^2) is not uniformly continuous.
Let $\displaystyle f(x) = \sin(x^2)$. Let $\displaystyle x_n = \sqrt{(2n+\tfrac12)\pi}$ and $\displaystyle y_n = \sqrt{(2n+\tfrac32)\pi}$. Notice that $\displaystyle x_n$ and $\displaystyle y_n$ can be arbitrarily close (for n large enough) and yet $\displaystyle |f(x_n) - f(y_n)|=2$.

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