Proving discontinuity - almost there again This time the function is different (of course).
$g:\mathbb{R}\rightarrow\mathbb{R}$
$h:\mathbb{R}\rightarrow\mathbb{R}$
and both are continuous.
I wish to show that $f:\mathbb{R}\rightarrow\mathbb{R}$ defined as $f(x)=h(x)$ if $x<c$ and $f(x)=g(x)$ if $x\ge c$ is continuous if and only if $h(c)=g(c)$.
I have shown that $h(c)=g(c)\implies f(x)$ is continuous. I now want to go the other way, show it is discontinuous if $g(c)\ne h(c)$
I will have a $|f(x)-f(c)|$ in it and this is $=|f(x)-h(c)$| so I want to take $f(x)$ on the side $<c$ but really close to $c$ so $f(x)-h(c)$ becomes really close to $g(c)-h(c)$ which are different, thus I can find and epsilon. 
From this I would pick $\epsilon=\frac{1}{2}|g(c)-h(c)|$ because that is in between both and as they are not equal there is a point between them. This gives me a nice symmetry I may exploit (it's half way between either one) 
I'm not sure how to complete this proof.
As before I sense I am close! Not quite there though.
Addendum
I might be able to use the "left side of c" (which there is because $\delta>0$) to get some $\epsilon_1$ (subscript because it may not be the same one) close to $g(c)$ I'll mess with this.
 A: You are indeed very close. Choose $\epsilon= 1/2|g(c)-h(c)|$ as you wanted. Because $h$ is continuous, there exists $\delta>0$ such that $h(]c-\delta,c+\delta[)\subset ]h(c)-\epsilon,h(c)+\epsilon[$. Hence $f(]c-\delta,c[)=h(]c-\delta,c[)$ is at distance $\epsilon$ from $f(c)$. This shows that $f$ is not continuous.
A: You need to show that $\exists\epsilon$ such that $\forall\delta\exists x$ such that$ |x-c|<\delta$ but $|f(x)-f(c)|\geq\epsilon$.
Since $h$ is continuous, $\forall\epsilon_1>0\exists\delta_1$ such that $|x-c|<\delta_1\Rightarrow |h(x)-h(c)|<\epsilon_1$.
Pick $\epsilon=\epsilon_1=\frac{|g(c)-h(c)|}{2}$. Let $\delta_1$ be the one that the definition of continuity of $h$ says exists. Then for every $x>c$ such that $|x-c|<\delta_1$, the definition of continuity of $h$ asserts that $|h(x)-h(c)|<\frac{|g(c)-h(c)|}{2}$. But then $$f(x)=h(x)>h(c)-\frac{|g(c)-h(c)|}{2}>g(c)+\frac{|g(c)-h(c)|}{2}$$, but the definition of continuity of $f$ tells us we need that inequality to go the other way.
Now we need to show that $\forall\delta$ some such $x$ exists, but the above is about $\forall x>c$ such that $|x-c|<\delta_1$, so $\forall \delta>0$ one can find such an x.
A: Define $(x_n)$ and $(y_n)$ by $x_n:=c-\frac{1}{n}$ and $y_n:=c+\frac{1}{n}$. Continuity of $g$ and $h$ gives us: $$\lim \limits_{n\to \infty}f(x_n)=\lim \limits_{n\to \infty}h(x_n)=h(c)\neq g(c)=\lim \limits_{n\to \infty}g(y_n)=\lim \limits_{n\to \infty}f(y_n)$$
Therefore $\lim \limits_{x\to c}f(x)$ doesn't exist and $f$ is not continuous.
