A function f is continuous on $\mathbb{R}$ and $f(n)=0$ for all integers $n$, then $f$ may not always uniformly continuous. I tried to find such example that f is not uniformly continuous. But if I think it by graphically, then this function return to x axis every time at integers with a continuous arc. I don't understand how do it possible to be not uniformly continuous. Please help me here.
 A: You just need it to get steeper and steeper as you go along, so something like 
$$f(x)=x\sin(\pi x)$$

A: Hint: For each $n$ draw a triangle with vertices $(n-\frac{2}{n},0) , (n-\frac{1}{n},1), (n,0)$. 
A: Uniformly continuous means that $|f(x)-f(y)|$ is controlled uniformly by $|x-y|$. So to "make" a function non-uniformly continuous, one attempt would be to ensure the following : for every $\epsilon$ small enough and $K$ large enough, ensure that there are a pair of points $x,y$ such that $|x-y| < \epsilon$ but $|f(x) - f(y)| > K$. Then such a function won't be uniformly continuous.
By the mean value theorem (assuming the function is differentiable wherever required) this would happen if the derivative of $f$ can be found to be arbitrarily large in places. 
For example, since $f$ is pinned to zero at the integers, it can do what it wants between them as long as it is continuous. The idea is simple : the graph of $f$ should be very "slopy" so that $f$ changes very fast in these parts. Such a  slope is clearly introduced by a graph looking like a steep triangle!
So the idea is to create an $f$ whose graph looks like triangles getting progressively steep, with base resting on the $x$-axis. Try to draw this for yourself!
(Others have provided the details : I provide the reason why this construction works).
A: Another example, which is continuous but not uniformly continuous with the above condition is $x^{2}\sin(\pi x)$. One fact that may help you is that if $f$ is an u.c function from $\mathbb{R}$ to $\mathbb{R}$, then the absolute of $f$ is somehow linear that is there exists $a,b$ in $\mathbb{R}^{+}$ such that $|f| \leq $a|x|+b$. This may give you some intuition on looking for functions that are continuous but not u.c.
