Check the continuity of the function $f$ where $$ f(x)=\begin{cases} 3\quad,\text{ if }0\leq x\leq1\\ 4\quad,\text{ if }1< x<3\\ 5\quad,\text{ if }3\leq x\leq10\\ \end{cases} $$
Case 1: At $x$=$1$ $$ \lim_\limits{x\to{1^+}}f(x)=\lim_\limits{x\to{1^+}}4=4\\ \lim_\limits{x\to{1^-}}f(x)=\lim_\limits{x\to{1^-}}3=3\neq4=\lim_\limits{x\to{1^+}}f(x) $$ $\therefore$ $f$ is not continuous at $x=1$
Case 2: At $x$=$3$ $$ \lim_\limits{x\to{3^+}}f(x)=\lim_\limits{x\to{3^+}}5=5\\ \lim_\limits{x\to{3^-}}f(x)=\lim_\limits{x\to{3^-}}4=4\neq5=\lim_\limits{x\to{3^+}}f(x) $$ $\therefore$ $f$ is not continuous at $x=3$
Case 3: Let $x$=$c$ where $1<c<3$ $$ f(c)=4\\ \lim_\limits{x\to{c}}f(x)=\lim_\limits{x\to{c}}4=4=f(c)\\ $$ $\therefore$ $f$ is continuous at $1<x<3$
What about Case 4 and Case 5 ?
where,
Case 4: $0\leq x<1$
Case 5: $3<x\leq 10$
do I need to verify continuity in the open interals $(0,1)$ and $(3,10)$ and then $\lim_\limits{x\to{0^+}}f(x)=f(0)$ and $\lim_\limits{x\to{10^-}}f(x)=f(10)$ ?
My Attempt
Case 4: $0\leq x<1$
Let $x=c$ where $c\in(0,1)$ $$ \lim_\limits{x\to c}f(x)=\lim_\limits{x\to c}3=3=f(c) $$ $\therefore$ $f$ is continuous for $x\in(0,1)$ And $$ \lim_\limits{x\to 0^+}f(x)=3=f(0) $$ $\therefore$ $f$ is continuous at $0$
Case 5: $3<x\leq10$
Let $x=c$ where $c\in(3,10)$ $$ \lim_\limits{x\to c}f(x)=\lim_\limits{x\to c}5=5=f(c) $$ $\therefore$ $f$ is continuous for $x\in(3,10)$ And $$ \lim_\limits{x\to 10^-}f(x)=5=f(10) $$ $\therefore$ $f$ is continuous at $10$
Is it the right way to approach the given problem ?
In those cases, for each $\varepsilon>0$ you take $\delta>0$ such that $(x-\delta,x+\delta)$ is a subset of the interval that you're dealing with ($[0,1)$ or $(3,10]$). Then$$|y-x|<\delta\iff y\in(x-\delta,x+\delta)\implies\bigl|f(y)-f(x)\bigr|=0<\varepsilon.$$So, for instance, in the Case 4 this proves that$$|y-x|<\delta\implies\bigl|f(y)-3\bigr|=0\text{ (since $f(y)=3$)}<\varepsilon.$$