$\lim_{h\to 0} f(x_0+h)-f(x_0) = 0$ implies $f$ is continuous at $x_0$? $\lim_{h\to 0} f(x_0+h)-f(x_0) = 0$ implies $f$ is continuous at $x_0$?
Is it always right to claim this?
I am familiar with the formal definitions (delta-epsilon and sequences) but never been taught this approach.
 A: Yes this is in fact the definition of the continuity of $f$ on $x_0$.
A: $f$ is continuous at $x_0$ if:
$$\forall \epsilon \exists \delta:\forall x: |x-x_0| < \delta \implies |f(x)-f(x_0)| < \epsilon$$
The limit $\lim_{h\to0} F(h)$ is equal to $0$ if:
$$\forall \epsilon\exists\delta: \forall h: |h|<\delta \implies |F(h) - F(0)|<\epsilon$$
You can, using these definitions, rewrite your limit so it will be the same as the definition of continuity.
A: Yes. 
It is the definition of continuous function at some point.
A: Writing
$$\lim_{x\to x_0} f(x)=l$$
or
$$\lim_{h\to0} f(x_0+h)=l$$
is exactly the same thing. Suppose the first limit is given. We want to see that the second limit is correct, that is, for all $\varepsilon>0$, there is $\delta>0$ such that $0<|h|<\delta$ implies $|f(x_0+h)-l|<\varepsilon$.
If we take $\varepsilon>0$, by hypothesis we can find $\delta>0$ such that $0<|x-x_0|<\delta$ implies $|f(x)-l|<\varepsilon$. If $0<|h|<\delta$, then $0<|(x_0+h)-x_0|<\delta$ and so, by hypothesis, $|f(x_0+h)-l|<\varepsilon$.
The converse is very similar and you should work out it in detail.
Now, suppose $\lim_{h\to0}(f(x_0+h)-f(x_0))=0$; then, by the theorems on limits,
$$
\lim_{h\to0} f(x_0+h)=
\lim_{h\to0}[(f(x_0+h)-f(x_0))+f(x_0)]=
\lim_{h\to0}(f(x_0+h)-f(x_0))+\lim_{h\to0}f(x_0)=0+f(x_0)=f(x_0)
$$
so, by what we saw above, $\lim_{x\to x_0}f(x)=f(x_0)$.
Now assume $\lim_{x\to x_0}f(x)=f(x_0)$ and try proving that $\lim_{h\to0}(f(x_0+h)-f(x_0))=0$ (it's very similar).

Actually, this is a special case of a general technique for limits: changing the variable, which can be done when the change of variables is continuous (with continuous inverse).
