For each $\epsilon>0$ there is such $\delta>0$, s.t. $|x-x_0|<\epsilon$ $\Rightarrow$ $|f(x)-f(x_0)|<\delta$. What does it mean? We know  that $\lim_\limits{x\to x_0}f(x)=f(x_0)$ means that: for each $\epsilon>0$ there is such $\delta>0$, s.t. $0<|x-x_0|<\delta$ $\Rightarrow$ $|f(x)-f(x_0)|<\epsilon$. 
what does the following expression mean?
for each $\epsilon>0$ there is such $\delta>0$, s.t. $|x-x_0|<\epsilon$ $\Rightarrow$ $|f(x)-f(x_0)|<\delta$.


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*Is $f$ continuous at $x_0$?

*What quality of $f$ does the above-mentioned definition describe?  


The lecturer gave is this new definition and told us to think about it, is that some kind of introduction to Uniform Continuity? 
 A: What it states is that for all $\epsilon > 0$ there exists $\delta > 0$ such that the value of $f$ on the interval $]x_0 - \epsilon, x_0 + \epsilon[$ is in the interval $]f(x_0) - \delta, f(x_0) + \delta[$.
The answer is that $f$ does not have to be continuous at $x_0$. Can you see why?
A: There are many exercises based on randomly modifying the $\epsilon$-$\delta$ definition of limit to see what properties of a function they define.  This is one of them.
The modification here is that the roles of $\epsilon$ and $\delta$ are reversed.  The condition is stronger for larger $\epsilon$.   It is saying that $f$ is bounded on larger and larger intervals surrounding $x_0$, and in fact is bounded on every finite interval.  Two things to notice are that this property is weaker than being bounded on $\mathbb{R}$ and that it does not depend on $x_0$ although the definition was stated with $x_0$.
Nobody would express boundedness on finite intervals in this way deliberately, it is only for purposes of exercise on the definition of limit.
A: $f$ might not be continuous.
Take the example of the function defined as $$f(x)=\begin{cases}
0 & \text{ if } x= 0\\
\sin \frac{1}{x} & \text{ else}
\end{cases}$$
$f$ satisfies your criteria at $0$ (with $\delta=2$ for all $\epsilon$).
$f$ is always locally bounded at $x_0$.
A: For the sake of defining limits, suppose $|f(x_0)|< \infty.$
Now consider all $x$ within some vicinity ($\epsilon$) of $x_0.$ The condition states that for these $x,$ the values of $f(x)$ differ from $f(x_0)$ by at most $\delta.$ Now $\epsilon$ can be any positive number, so this means $f$ cannot be "infinite distance" from $f(x_0),$ if that makes sense, except when $|x|=\infty$. In other words, $f$ is bounded near $x_0$ (near meaning within finite distance from it). So basically, the statement implies $|f(x)|<\infty$ when $|x|<\infty.$
Also, this does not imply $f$ is continuous. Consider, for example,
$$f(x) = \begin{cases} 0 &\mbox{if } x \leq x_0 \\1 &\mbox{otherwise} \end{cases}.$$ Clearly, $f(x)$ is locally bounded (in fact, globally bounded), but isn't continuous.
