# Proving a piecewise function is discontinuous at a point

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 $$3, we can use it to create an inequality such that $$\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 $$x=0$$.

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.

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.