$\epsilon -\delta $ criteria for piecewise defined function Let's assume we have the following function
$$f(x):=\begin{cases} 
      h(x) & x<a \\
      i(x) & x=a \\
      j(x) & x>a 
   \end{cases}
$$
and we want to show it is continuous. 
How would we use the criteria?
Thanks in advance. I am only interested in the approach, as the definition says ... $|f(x)-f(a)|< \varepsilon $
 A: For any $\varepsilon>0$ there exists a $\delta>0$, that
$$|h(x)-i(a)|<\varepsilon,\ \text{ for }a-\delta<x<a $$
$$|j(x)-i(a)|<\varepsilon, \ \text{ for }a<x<a+\delta $$
Hard to say anything else.
Further explanation.
$$f\ \text{ is continuous at }a\iff \forall_{\varepsilon>0}\exists_{\delta>0}\forall_{x\in (a-\delta, \ a+\delta)} \left| f(x)-f(a)\right| <\varepsilon \iff \\ \forall_{\varepsilon>0}\exists_{\delta>0}\left(\forall_{x\in (a-\delta, \ a)} \left| f(x)-f(a)\right| <\varepsilon \ \land \forall_{x\in (a, \ a+\delta)} \left| f(x)-f(a)\right| <\varepsilon \ \land \left| f(a)-f(a)\right| <\varepsilon \right ) \\ \iff \forall_{\varepsilon>0}\exists_{\delta>0}\left(\forall_{x\in (a-\delta, \ a)} \left| h(x)-i(a)\right| <\varepsilon \ \land \forall_{x\in (a, \ a+\delta)} \left| j(x)-i(a)\right| <\varepsilon  \right )  $$
It's all just matter of dividing the last quantifier in parts.
A: To show it is continuous, we do the following.
For  $x<a$, it is continuous whenever the function $h$ is
For $x=a$, we have to be able to bound $|h(x)-i(a)| < \epsilon$ whenever $x <a$ and $|x-a| < \delta$. This is the same as asking that $\lim_{x\to a} h(x) = i(a)$
We also have to ask the same when we approach from the right.
whenever $x>a$ and $|x-a|<\delta$ we must have $|j(x)-i(a)|<\epsilon$, which is the same as $\lim_{x \to a} j(x) = i(a)$
For $x>a$, $f$ is continuous whenever $j$ is
A: Show that
$$\lim_{x\to a^-}h(x)=i(a)=\lim_{x\to a^+}j(x)$$
using the $\delta-\epsilon$ formalism.
