Question about continuity This is a question from my math book: 

Let $a< b < c$. Suppose that $f$ is continuous on $[a,b]$, $g$ is continuous on $[b,c]$, and $f(b) = g(b)$. Define $h$ on $[a,c]$ by $h(x) = f(x)$ for $x\in [a,b]$ and $h(x) = g(x)$ for $x \in [b,c]$. Prove that $h$ is continuous on $[a,c]$. 

What I want to do is prove that $h$ is continuous on $[a,c]$ but not at $b$. 
I'm thinking that I have to pick an $α$ such that $a<α< b$, and then show that $h(x)$ is continuous at $α$. So basically, I want to show that $∀\epsilon>0, ∃ δ>0$ such that if $x\in [a,c]$ and $|x-α|<δ$ then $|h(x)-h(α)|<\epsilon$. And from the given information, I know that $f:[a,b]\to \mathbb{R}$, and $∀ \epsilon>0, ∃ δ>0$ such that if $x\in [a,b]$ and $|x-α|<δ_f$ then $|f(x)-f(α)|<\epsilon$.
How do I connect these two definitions to find $δ$? And is there anything else I need to prove? Like are there any cases I should be making? 
Thanks in advance. 
 A: What you have to prove is precisely that $h$ is continuous at $b$. Because at any point $d$ other than $b$, if $d<b$ then $h(x)=f(x)$ for all $x$ in a small interval around $d$, and so the continuity of $f$ implies the continuity of $h$ for every $d\in[a,b)$. Similarly one deduces that $h$ is continuous on $(b,c]$ by using the continuity of $g$. 
For the continuity at $b$. Fix $\varepsilon>0$. By the continuity of $f$ at $b$, there exists $\delta_1>0$ such that if $x<b$ and $b-x<\delta_1$, then $|f(x)-f(b)|<\varepsilon$. Similarly, there exists $\delta_2>0$ such that if $x>b$ and $x-b<\delta_2$, then $|g(x)-g(b)|<\varepsilon$. 
Let $\delta=\min\{\delta_1,\delta_2\}$. Now, if $|x-b|<\delta$, we consider two cases: first, if $x<b$, then $b-x=|x-b|<\delta\leq\delta_1$, and so 
$$
|h(x)-h(b)|=|f(x)-f(b)|<\varepsilon;
$$
if $x>b$, then $x-b=|x-b|<\delta\leq\delta_2$, and so 
$$
|h(x)-h(b)|=|g(x)-g(b)|<\varepsilon.
$$
So $h$ is continuous at $b$. 
A: We can prove that $h$ is continuous on $[a,c]$ in three steps:
First, we prove that $h$ is continuous on $[a,b)$. Since $f$ is continuous on $[a,b)$, given any $x\in [a,b)$ and $\epsilon>0$ we have some $\delta>0$ such that for $y\in [a,b)$ we have $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$. Let $\delta'=\min\{\delta,b-x\}$. Since $h=f$ on $[a,b)$, we have that for $y\in [a,c]$, $|x-y|<\delta'$ implies that $y\in [a,b)$ and $|x-y|<\delta$ so $|h(x)-h(y)|<\epsilon$, thus $h$ is continuous at $x$. Hence $h$ is continuous on $[a,b)$.
Second, we prove that $h$ is continuous on $(b,c]$. This proof is nearly identical to the previous one.
Finally, we prove that $h$ is continuous at $b$. Since $f$ and $g$ are continuous at $b$, given any $\epsilon>0$ we have some $\delta,\delta'>0$ such that for $y\in [a,b]$ if $|b-y|<\delta$ then $|f(b)-f(y)|<\epsilon$ and for $y\in [b,c]$ if $|b-y|<\delta'$ then $|g(b)-g(y)|<\epsilon$. Thus we can let $\delta''=\min\{\delta,\delta'\}$ and we get that if $y\in [a,c]$ and $|b-y|<\delta''$ then either $y\in [a,b]$ and $|b-y|<\delta$ in which case $|h(b)-h(y)|=|f(b)-f(y)|<\epsilon$, or $y\in [b,c]$ and $|b-y|<\delta'$ in which case $|h(b)-h(y)|=|g(b)-g(y)|<\epsilon$. Thus $h$ is continuous at $b$.
