4
$\begingroup$

Given $$f(x)=\begin{cases} 2x+3, x\ge 1 \\ 2x-2, x\lt 1 \end{cases}$$

Show it is discontinuous at x=1, using epsilon delta. Any pushes in the right direction would be greatly appreciated.

$\endgroup$
1
$\begingroup$

To show explicitly, you must show that there exists an $\epsilon>0$ such that for all $\delta>0$, you can find some $x \in B(1,\delta)$ (that is, $|x-1| < \delta$) such that $f(x) \notin B(f(1),\epsilon)$ (that is, $|f(x)-5| \ge \epsilon$).

If you plot $f$ you will see that $f(x) \le 0$ for all $x < 1$, so we can choose $\epsilon = 2$. Let $\delta>0$ be arbitrary, then $f(1-\frac{1}{2}\delta) \le 0$, and so $|f(1-\frac{1}{2}\delta)-5| \ge 3 > \epsilon$, as required.

Hence $f$ is discontinuous at $x=1$.

$\endgroup$
6
$\begingroup$

We have $f(1)=5$. So to show that $f$ is not continuous at $x=1$, it is enough to show that it is not true that $\lim_{x\to 1} f(x)= 5$.

Suppose to the contrary that the limit exists and is equal to $5$. Then for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-1|\lt\delta$, then $|f(x)-5|\lt\epsilon$.

Pick $\epsilon=\frac{1}{2}$. We show there is no $\delta$ with the required property.

If $x\lt 1$, then $f(x)\lt 1$. In particular, if $x\lt 1$, then $|f(x)-5|\gt 4$. It follows that there is no $\delta$ such that $|x-1|\lt \delta$ guarantees that $|f(x)-5|\lt \epsilon$.

Remark: The idea is basically geometric. We are showing that there are points arbitrarily near $x=1$ at which the function value is not close to $5$. Once one has the geometry under control, writing out the details in $\epsilon$-$\delta$ language is a translation job.

$\endgroup$
0
$\begingroup$

We need to show that there exists a fixed $\varepsilon>0$ such that for any chosen $\delta>0$ there is an $x$ satisfying $|x-1|<\delta$, but $|f(x)-f(1)|\ge\varepsilon$. Then $f$ cannot be continuous at $1$.

Try fixing $\varepsilon=1$. Now let us look for a suitable $x$ candidate. If we choose $x$ such that $x<1$ and $|x-1|<\delta$ then can you see that $f(x)<0$? This inequality follows: $$|f(x)-f(1)|\ge |0-f(1)|=|f(1)|=|5|=5>1=\varepsilon$$ and we are done.

$\endgroup$

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.