Is this composition of functions $h(x)=\begin{cases}f(x)& -1\leq x\leq 0,\\g(x), &\;\;\;\;\;\;\;\;\;x>0\end{cases}$ continuous?

Question

Given that $f:[-1,0]\to\Bbb{R},\;\;g:[0,1]\to\Bbb{R},$ $f,g$ continuous with $f(0)=g(0).$

Let $h:[-1,1]\to\Bbb{R}$ be defined as

$$h(x)=\begin{cases}f(x)& -1\leq x\leq 0,\\g(x), &\;\;\;\;\;\;\;\;\;x>0.\end{cases}$$

How do I show that this $h(x)$ is continuous? That is:

$$\lim\limits_{x\to 0^+}h(x)=g(0)=f(0)=h(0)=\lim\limits_{x\to 0^-}h(x).$$

$``\epsilon-\delta"$ or "limit" approaches will be highly appreciated.

Answer 1

1) $f$ is left continuous at $x=0$.

Let $\epsilon >0$ be given.

There is a $\delta_1 >0$ such that for $x \in [-1,0]$

$|x| \lt \delta_1$ implies $|f(x)|\lt \epsilon.$

2) $g$ is right continuos at $x=0$.

There is a $\delta_2 >0$ such that for $x \in [0,1]$

$|x| \lt \delta_2$ implies $|g(x)|< \epsilon.$

Choose $\delta =\min(\delta_1,\delta_2)$.

Then for $x \in [-1,1] $:

$|x| \lt \delta$ implies $|h(x)| \lt \epsilon.$

Answer 2

Using $\epsilon-\delta$ definition:

For any $x\in[-1,0]$ $h(x)=f(x)$ and hence is continuous. If $x\in[0,1]$ then $h(x)=g(x)$ is continuous.[Clear by definition].

Now we check the continuity of $h(x)$ at $x=0$.

Let, $\epsilon>0$ be given, now $|h(x)-h(0)|<\epsilon\implies|f(x)-f(0)|<\epsilon$ $\forall x\in[-1,0]$ and so $\exists \delta_1>0$ s.t. $|h(x)-h(0)|<\epsilon$ whenever $0-x<\delta_1$ since, $f$ is continuous and if $x \in [0,1]$ then $|h(x)-h(0)|<\epsilon\implies|g(x)-g(0)|<\epsilon$ $\forall x\in[0,1]$ and so $\exists \delta_2>0$ s.t. $|h(x)-h(0)|<\epsilon$ whenever $x-0<\delta_2$ since, $g$ is continuous with $h(0)=f(0)=g(0)$ Hence $h$ is continuous.