1. ## Not Uniform Continuous?

Consider the function $f(x)=xsinx$

Now, $f'(x)=xcosx+sinx$, which is not bounded, so it is not uniformly continuous.

So pick $\epsilon = 1$, let $\delta > 0$, for every $x,y \in \mathbb {R} , |x-y| < \delta$, we need $|xsinx-ysiny| \geq \epsilon$, but how should I go about that? Thanks.

2. " $f$ is not uniformly continuous" means:
$\exists\varepsilon_0>0, \forall \delta>0, \exists x,y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\varepsilon_0$.

I let you get convinced yourself that a proof would consist in finding a sequence a couples $(x_n,y_n)_{n\geq 1}$ such that $|x_n-y_n|\to_n 0$ and $|f(x_n)-f(y_n)|>\varepsilon_0$ for some $\varepsilon_0$ (or even $|f(x_n)-f(y_n)|\to_n+\infty$, if possible).

In order to choose those couples, look at (or just imagine) what the graph looks like: the $x_n,y_n$ must be close but the $f(x_n),f(y_n)$ must be way apart from each other.

$x_n=n\pi$ is a possible start.

3. Thanks for the hint. So for the choice that $x_n = n \pi$, we would have $f(x_n) = 0 \ \ \ \forall n$, but any point right afterward would get larger as n increases.

Pick $y_n = x_n + \frac { \delta}{2}$, then $|x_n - y_n | < \delta$, but $|f(x_n)-f(y_n)| \rightarrow \infty$

Is this right?

Pick $y_n = x_n + \frac { \delta}{2}$, then $|x_n - y_n | < \delta$, but $|f(x_n)-f(y_n)| \rightarrow \infty$

Is this right?
No, because for every $\delta>0$ you must have $|x_n-y_n|<\delta$ for some $n\geq 0$. This is why I said you must have $|x_n-y_n|\to_n 0$.

You can write $y_n=x_n+u_n=n\pi+u_n$ where $u_n\to_n 0^+$ is to be determined.

You have $|f(y_n)-f(x_n)|=|f(y_n)|=|(n\pi+u_n)\sin(n\pi+u_n)|=|(n\pi +u_n)(-1)^n\sin(u_n)|$ $\sim n\pi\sin(u_n)\sim n\pi u_n$ as $n\to\infty$. So that you must choose $u_n\to_n 0$ such that $\liminf_n n u_n>0$. For instance, $u_n=\frac{1}{n}$ (hence $\varepsilon_0$ must be small enough), or $u_n=\frac{1}{\sqrt{n}}$ (and any $\varepsilon_0$ is adequate).

You can do a more explicit proof using the convexity inequality $\sin(u_n)\geq\frac{2u_n}{\pi}$ which holds as soon as $0\leq u_n\leq \frac{\pi}{2}$. This yields to the same possibilities for the choice of $(u_n)_n$.

5. Thanks for the help. But one thing I'm not clear is, why is it enough to show that $
|x_n-y_n|\to_n 0
$
implies $
|f(x_n)-f(y_n)|>\varepsilon_0
$
?

6. Explicitating my first post as much as I can:

You know that " $f$ is not uniformly continuous" means:
$\exists\varepsilon_0>0, \forall \delta>0, \exists x,y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\varepsilon_0$.
You obtain this by taking the logical contrary of the definition of "uniformly continuous".

Suppose we found a sequence of couples $(x_n,y_n)_{n\geq 1}$ and an $\varepsilon_0>0$ such that $|x_n-y_n|\to_n 0$ and, for all $n\geq 1$, $|f(x_n)-f(y_n)|>\varepsilon_0$.

Then, for every $\delta>0$, there is an $n$ such that $|x_n-y_n|<\delta$ (because of the convergence to 0). And for this same $n$, we have $|f(x_n)-f(y_n)|>\varepsilon_0$. As a consequence, we have realized the above definition of "not uniformly continuous" with $x_n,y_n$ in place of $x,y$.