Proving a piecewise function is discontinuous at a point

Question

Give a $\delta,\epsilon$ proof that the following function, $f(x) = \begin{cases} x, & \text{if $x \geq 0$} \\ x+3, & \text{if $x < 0$} \end{cases}\quad \quad $
is discontinuous at $x=0$.

Attempt:

If $f(x)$ is not continuous at $x=0$, then $\exists \epsilon>0$ such that $\forall \delta > 0$, $\exists x$ such that $0<|x-0|<\delta$ and $|f(x)-f(0)|=|x+3| \geq \epsilon.$

$\quad$Choose $\epsilon=1$. We must show that for a given $\delta$, then $\exists x$ such that $|x+3|\geq1$.

$\quad \quad$Let $a=x+3,b=-x,a+b=3$. By the Triangle Inequality, we have:

$\begin{align} \quad \quad \quad \quad|a+b|\leq|a|+|b|&\Rightarrow |3|\leq|x+3|+|-x| \\ &\Rightarrow 3-|-x|\leq|x+3| \end{align}\\$

$\begin{align} \quad \quad \text{Let } \delta\leq1 &\Rightarrow 0 < x <1\\ &\Rightarrow 3 < x+3 < 4 \end{align}\\$

$\rule{18cm}{.03cm}$

I am unsure as to how to continue with the proof. I believe that after manipulating our $3<x+3<4$, we can use it to create an inequality such that $\epsilon=1<3...\leq|x+3|.$ Which would lead us to the following result:

If $\delta\leq1$, every $x$ where $|x|<\delta$ has $|x+3|\geq \epsilon=1$.

If $\delta >1$, then $(x \bigr ||x|<1)$ is contained in the set of $x$, where $|x|<\delta$. Thus the set of points where $\delta>1$ contains points where $|x+3|\geq \epsilon=1$.

Therefore $f(x)$ is discontinuous at $x=0$.

Answer 1

You cannot start the proof with 'If $f$ is not continuous at $0$' because that is what you want to prove.

Suppose $f$ is continuous at $0$. Then there exists $\delta >0$ such that $|f(x)-f(0)| <1$ if $|x-0| <\delta$. Now take $x=-\frac 1n$ with $n >\frac 1{\delta}$. Then $|x| <\delta$ but $|f(x)-f(0)|=3-\frac 1 n >1$. This contradcition proves the result.

Answer 2

I only read the first paragraph of the attempt. It doesn’t make sense: it appears to be a misunderstanding of the definition of limit.

The statement “$f(x)$ is discontinuous at $x = 0$” means that $\lim \limits_{x \to 0} f(x)$ does not exist or does exist but is not equal to $f(0)$. In fact, this limit does not exist, as is shown below.

The proof in this answer is not the simplest, but I tried to write it in a way that explains the nature of limits generally.

Intuitive explanation

In order for $\lim \limits_{x \to 0} f(x)$ to exist, $f(x)$ would have to get close to some value (call it $L$) as $x$ got close to $0$. More specifically, we would need to be able to make $f(x)$ get as close as we like to $L$ by making $x$ close enough to $0$.

For this choice of $f$, it is actually the opposite. No matter how close to $0$ we require $x$ to be, we can always find two values of $x$ where the corresponding values of $f(x)$ are far from each other, and therefore can’t both be close to $L$.

Formal proof

By definition, $\lim \limits_{x \to 0} f(x) = L$ is equivalent to $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x - 0| < \delta \implies |f(x) - L| < \epsilon$. We just need to show that for some $\epsilon$, there is no suitable $\delta$.

Now consider $\epsilon < \frac{3}{2}$ and any value of $\delta$. Let $\delta' = \min\{\frac{\delta}{2}, \frac{3 - 2\epsilon}{2}\}$, $x_1 = -\delta'$ and $x_2 = \delta'$. Then $|x_1 - 0| < \delta$ and $|x_2 - 0| < \delta$, but $|f(x_2) - f(x_1)| \geq 2\epsilon$, so it is impossible for $|f(x_1) - L| < \epsilon$ and $|f(x_2) - L| < \epsilon$ to both be true.