Question is to prove that :
$f(x)=\sin x^3$ for $x\in \mathbb{R}$ is not uniformly continuous.
What would my first observation in checking uniform continuity is to check if its derivative is bounded.
In this case its derivative $f'(x)=3x^2\sin x^3$ which is unbounded. So, I can not rely on this.
I tried with definition :
$|f(x)-f(y)|=|\sin x^3-\sin y^3|=|2\cos (\frac{x^3+y^3}{2})\sin (\frac{x^3-y^3}{2})|\leq 2 |\sin (\frac{x^3-y^3}{2})|$
I see that $\sin x \leq x$.. This may not imply $|\sin x|\leq |x|$ but i am assuming it...
So, $|f(x)-f(y)|\leq 2 |\sin (\frac{x^3-y^3}{2})|\leq 2\frac{|x^3-y^3|}{2}=|x^3-y^3|=|x-y||x^2+xy+y^2|$.
Though $|x-y|$ is small i should make $|f(x)-f(y)|$ considerably large.
I now have $|f(x)-f(y)|\leq 2 |x-y||x^2+xy+y^2|$
I some how can sense that I can make $|x^2+xy+y^2|$ large enough though $|x-y|$ is small but not so sure how to make this in $\epsilon-\delta$ case.
Please help me to solve this.
Thank you.
EDIT : I know that uniform continuous functions takes cauchy sequence to cauchy sequence and tried to use it in this case. But, I could not find correct cauchy sequence that would help me to go through this.
I would be thankful if some one can help me in this way too.
Let $x_n=(n\pi+\pi/2)^{1/3}$, then $x_{n+1}-x_n\to0$ but $f(x_{n+1})-f(x_n)=\pm2$ hence $f(x_{n+1})-f(x_n)$ does not converge to $0$. Therefore, no uniform continuity.