Confusion in understanding continuous function Okay so I've started to study derivatives and there is this idea of continuity. The book says "a real valued function is considered continuous at a point iff the graph of a function has no break at the point of consideration, which is so iff the values of the function at the neighbouring points are close enough to the value of the the function at the given point"

So what I dont understand is that why is it that values of the function at the neighbouring points should be close enough to the value of the function at the given point, isn't it enough if they are defined why do they have to be close enough the value of the function at the given point?
Intuitively, the idea is that the closer and closer you get to $x$, the closer and closer $y$ gets to $f(x)$. A small neighborhood of $x$ must correspond to a small neighborhood of $f(x)$.

When there is a discontinuity, the image of a neighborhood of $x$ always includes that discontinuity and that blocks it from being as small as you want.
If the values of the function in points $x$ close enough to a specific point $x_0$ are not arbitrarily close to $f(x_0)$ the graph will be (usually) broken.

This definition is not completely rigorous, as it assumes we have intuitive understanding what a "graph with no break is", and it can create some misunderstandings. For example, it can be proven that the graph of function $$ f(x) = \left\{\begin{array}{l} \sin\frac{1}{x} &amp; \text{for }x\neq 0 \\ 0 &amp;\text{for }x=0 \end{array}\right. $$
is connected, but the function is still non-continuous.

It's better to consider "iff the values of the function at the neighbouring points are close enough to the value of the the function at the given point" as the definition, or, rigorously speaking 


  Function $F : X \rightarrow Y$ is called continuous at point $x_0\in X$, iff $$ \forall \epsilon&gt;0 \exists \delta&gt;0 \forall x\in X: \Big((|x-x_0|&lt;\delta) \Rightarrow \big(|f(x)-f(x_0)|&lt; \epsilon\big)\Big)$$


which means: "For arbitrary distance $\epsilon$ I can choose a neighbourhood of $x_0$ small enough for the values of function to be within distance of $\epsilon$ from the value of function at point $x_0$".