I want to prove that the following function: $f(x)=x\sin(x)$ isn't uniformly continuous
So, I need to find 2 sequences $x_n$ and $y_n$ such that $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\geq\epsilon_0$ where $\epsilon_0$ is a positive number (const.).
Any help with this problem? Plus I would be glad if someone could write the solution more formally and clean (I wrote most of it but couldn't find the sequences)
Best Answer
hint
Take $$x_n=2n^2\pi$$ $$y_n=x_n+\frac 1n$$
it is clear that $$\lim_{n\to +\infty}(y_n-x_n)=0$$ and, by MVT,
$$f(y_n)-f(x_n)=(y_n-x_n)f'(c_n)$$ $$=\frac 1n(\sin(c_n)+c_n\cos(c_n))$$
with $$2n^2\pi<c_n<2n^2\pi+\frac 1n$$
and $$2n\pi<\frac{c_n}{n}$$ $$0<\sin(c_n)<\sin(\frac 1n)$$ $$\cos(\frac 1n)<\cos(c_n)<1$$ thus $$\lim_{n\to+\infty}(f(y_n)-f(x_n)=+\infty$$