How to prove this continuous function? let $a<b<c$. Suppose that $f$ is continuous on $[a,b]$, that $g$ is continuous on $[b,c]$, and that $f(b)=g(b)$. 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 $[a,c]$.
This is my idea: $f$ is continuous on $[a,b]$, then $\lim_{a\to b} f(x) $. $g$ is continuous, then $\lim_{b\to c} g(x)$. since $f(x)=g(x)$ where $x \in [a,b]$, then $\lim_{a\to b} f(x)+\lim_{b\to c} g(x) = \lim_{a\to b} h(x)+\lim_{b\to c} h(x)$, that is, $\lim_{a\to c} (f+g)(x)=\lim_{a\to c} h(x)$. Since continuous + continuous = continuous, therefore $h$ is continuous on $[a,c]$.
 A: In general, given a function $h$ on a closed interval $[a,c]$ with $a<c,$ $h$ is continuous if and only if the following three things are true:


*

*$h(a)=\lim\limits_{x\to a^+}h(x),$

*$h(c)=\lim\limits_{x\to c^-}h(x),$ and

*for every $x_0$ with $a<x_0<c,$ we have $h(x_0)=\lim\limits_{x\to x_0^+}h(x)=\lim\limits_{x\to x_0^-}h(x).$


Now, from the fact that $f$ is continuous on $[a,b],$ we can conclude that


*

*$f(a)=\lim\limits_{x\to a^+}f(x),$

*$f(b)=\lim\limits_{x\to b^-}f(x),$ and

*for every $x_0$ with $a<x_0<b,$ we have $f(x_0)=\lim\limits_{x\to x_0^+}f(x)=\lim\limits_{x\to x_0^-}f(x).$


Similarly, we can conclude that


*

*$g(b)=\lim\limits_{x\to b^+}g(x),$

*$g(c)=\lim\limits_{x\to c^-}g(x),$ and

*for every $x_0$ with $b<x_0<c,$ we have $g(x_0)=\lim\limits_{x\to x_0^+}g(x)=\lim\limits_{x\to x_0^-}g(x).$


Now, try to apply the definition of $h$ and use the six concluded facts above to prove statements 1 through 3. Statements 1 and 2 should be easy. For statement 3, you'll need to split it into three cases, depending on whether $x_0<b,$ $x_0>b,$ or $x_0=b.$
A: By its definition $h$ is continuous everywhere except maybe at $b$ (on either side of $b$, take $\epsilon$ small enough that it doesn't encroach on the other interval.)
As $f(b)=g(b)$, at $b$ you will find an $\epsilon_f$ on the left and an $\epsilon_g$ on the right that prove continuity of their respective functions, and take $\epsilon$ being the smallest of the two to witness continuity of $h$.
A: Hints:
The continuity on [a,b) and (b,c] of $h(x)$ follows easily from the piecewise definition, and the point $b$ can be proved continuous by showing that
$$\lim_{x\to b^-}h(x) = \lim_{x\to b^+}h(x)$$
A: You're a bit confused, it seems.
The sentence

$f$ is continuous on $[a,b]$, then $\lim\limits_{a\to b}f(x)$

has no meaning.
You want to prove continuity of $h$ at every point of $[a,c]$. If $x\in[a,c]$ and $a\le x<b$, then $h$ coincides with $f$ in a neighborhood of $x$, hence it is continuous at $x$. Similarly, if $b<x\le c$, then $h$ coincides with $g$ in a neighborhood of $x$, so it is continuous at $x$.
Thus the only problem is at $b$. However
$$
\lim_{x\to b^-}h(x)=\lim_{x\to b^-}f(x)=f(b)
$$
by continuity of $f$; on the other hand,
$$
\lim_{x\to b^+}h(x)=\lim_{x\to b^+}g(x)=g(b).
$$
Since $f(b)=g(b)$ by hypothesis, the function $h$ is also continuous at $b$.
