Check continuity with $\epsilon$-$\delta$ criterion We have the function $$f(x)=\begin{cases}x^2 , & x\geq 1 \\ x-1 , & x<1\end{cases}$$ Is the function continuous at $x_0=1$ ?
To check that using the $\epsilon$-$\delta$ criterion we have the following :
Let $\epsilon>0$ arbitrarily.
$$|f(x)-f(x_0)|=|f(x)-f(1)|=|f(x)-1|$$ Do we take cases if $x<1$ and if $x>1$ ?
I mean do we do the following ?
For $x<1$ : $$|f(x)-f(x_0)|=|f(x)-f(1)|=|f(x)-1|=|x-1-1|=|x-2|$$
For $x>1$ : $$|f(x)-f(x_0)|=|f(x)-f(1)|=|f(x)-1|=|x^2-1|=|x+1|\cdot |x-1|$$ For this one we cannot find an upper bound since $|x+1|$ is not bounded, right?
 A: We want to determine if $f$ is continuous at $x_0=1$.
We note that for $x\ge 1$, $f(x)=x^2$.  So, as $x$ approaches $1$ from the right, the limit is $1$.  This is also the value $f(1)=1$.
And for $x<1$, $f(x)=x-1$.  So, as $x$ approaches $1$ from the left, the limit is $0$.
So, $f$ is not continuous at $1$.
To show this using a $\varepsilon-\delta$ proof, we need to find a value of $\varepsilon>0$ such that for any $\delta>0$, there exists a number $x_\delta \in [x_0-\delta,x_0+\delta]$, such that $|f(x_\delta)-f(x_0)|\ge \varepsilon$.  This is the negation of continuity of $f(x)$ at $x_0$.
Since the limits of $f$ from the right of $1$ and left of $1$ differ by $1$, lets test with $\varepsilon=1/2$ (any number between $0$ and $1$ will suffice).
Then, for any given $\delta>0$, we can find a number $x_\delta\in [1-\delta,1+\delta]$ such that  $|f(x_\delta)-f(1)|=|f(x_\delta)-1|\ge \varepsilon = 1/2$.  In fact, any number $1-\delta<x_\delta<1$ suffices.  Then, we have $|f(x_\delta)-1|=|x_\delta-2|\ge 1/2$ for $1-\delta<x_\delta<1$ and conclude that $f$ is discontinuous at $1$.
A: $f$ is discontinuous at $1$ if $\exists\epsilon>0$ such that $\forall \delta>0, \exists x_0\in \Bbb{R} $ with $|x_0-1|<\delta$ but $|f(x_0)-1|\ge \epsilon$

Let $\delta>0$ be given. Now by Archimedean property of Real numbers, we can choose $n\in\Bbb{N}$ such that $\frac{1}{n}<\delta$
Let $x_0=1-\frac{1}{n}$ , then $|x_0-1|=|1-\frac{1}{n}-1|=\frac{1}{n}<\delta$
But
$\begin{align}|f(x_0) -1|&=|1-\frac{1}{n}-1-1|\\&=1+\frac{1}{n}>1\end{align}$
Choose $\epsilon=1$ , then $\forall\delta>0, \exists x_0=1-\frac{1}{n}\in (1-\delta, 1+\delta) $ but $|f(x_0)-1|>\epsilon$
