0
$\begingroup$

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 $x=0$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.