Assuming $$\lim_{x \to x_0} f(x) = f(x_0)$$
As $x$ approaches $x_0\ $i.e. $(x\to x_0)$, can we always say $f(x)$ approaches $f(x_0)\ $ i.e. $(f(x)\to f(x_0))$? From what I understand, this is the motivation for the $\epsilon$-$\delta$ definition of limit.
The $\epsilon-\delta$ definition states that- $\forall \epsilon$, $\exists \delta$, such that:
$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.
Approach is apparently an informal idea, and not what the limit defines. So the idea of $f(x)$ getting closer to $f(x_0)$, as $x$ gets closer to $x_0$ is inaccurate and not always true.
Update: The statement apparently is true, when $f(x)$ is continuous at $x_0$. Though I have been unable to find a general proof for it yet.
Yes. If $\lim_{h\to 0}f(x_0+h)=L$ then for every $\varepsilon>0$ there exists $\delta>0$ such that whenever $|h|<\delta$ then $|f(x_0+h)-L|<\varepsilon$.
This means, for the same value of $\varepsilon$ and $\delta$, if $|x-x_0|<\delta$ then, setting $h=x-x_0$ we have $|f(x)-L|=|f(x_0+h)-L|<\varepsilon$, so the definition of $f(x)\to L$ as $x\to x_0$ is satisfied.
Note that this is actually slightly stronger than saying $f(x)\to L$ as $x\to x_0$, as $\lim_{h\to 0}f(x_0+h)=L$ also implies that $f(x_0)=L$, whereas in the definition of "$f(x)\to L$ as $x\to x_0$" it doesn't matter whether $f(x_0)$ even exists (or, if it does, what value it takes).