$x\sin(1/x)$ at $ x$ very close to zero the function $x\sin(1/x)$ goes up and down increasingly fast as $x$ approaches $0$. Is there any way to tell whether the function goes up or down at the closest point to zero? 
 A: Strictly speaking, $x\sin(1/x)$ is not defined when $x=0$, but it makes sense to define the function value as $0$ there, thus considering the function
$$ f(x) = \begin{cases} 0 & \text{when }x=0 \\ x\sin(1/x) & \text{otherwise} \end{cases} $$
This produces a continuous function, but it is not differentiable at $0$ and so doesn't have a well-defined slope at that point. (At least by the usual definition, but it is hard to imagine a principled definition of "slope" that would give a definite result for this situation).
In particular, the function is neither increasing nor decreasing in any neighborhood of $0$.
A: There is no "closest point to 0."
A: As was rightly noted, there is no point "closest to zero".
That said, perhaps this will be of some help if you are willing to consider the more "formal" approach of reaching a number in the line as close as we can, that is, using limits.
Let $$u=\frac1x$$
and consider the limit $$\lim_{u\rightarrow\infty}\frac{\sin u}{u}$$
This limit is equal to zero as you can see by the fact that $$-1\le\sin u\le1\Rightarrow\\-\frac{1}u\le \frac{\sin u}{u}\le\frac{1}u$$
Since $$\lim_{u\rightarrow \infty} \frac{1}{u}=0$$ you get $$\lim_{u\rightarrow\infty}\frac{\sin u}{u}=0$$
