I would like to show where the following function is continuous.

$$f(x)=\begin{cases}13&\text{if }\;0\le x\leq 1\\x&\text{if }\;1<x\le 2\end{cases}$$

I am pretty sure its continuous on $[0,1)\cup(1,2]$

I know from elementary calculus that this thing is not continuous at $x=1$ because the right side limit is $1$ and the left side limit is $13$, but I would like to prove it.

Using the sequence version of continuity of a function,

Consider $x_n = 1+\frac{1}{n}$, then $x_n \rightarrow 1$, and $f(1)=13$ but $\forall n\in \mathbb{N}$ $(f(1+\frac{1}{n}) = 1+\frac{1}{n} \ne 13)$ thus $f$ is not continuous at $x=1$.

Is this correct?

  • 1
    $\begingroup$ Not quite. You have to show that $f(x_n)$ does not equal $13$ in the limit. What you've stated is that it doesn't equal $13$ for any value of $n$. $\endgroup$ – Callus - Reinstate Monica Oct 31 '14 at 8:36
  • $\begingroup$ how do I go about doing that @callus ? $\endgroup$ – tmpys Oct 31 '14 at 8:56
  • $\begingroup$ I think the easiest way is to show that $|f(x_n)-13|$ is always greater than some non-zero number. In this case, you can choose a pretty large number, like $5$ for example. $\endgroup$ – Callus - Reinstate Monica Oct 31 '14 at 11:47

Since $f(1)=13$, just remark that for any $\delta>0$, all the points of the interval $(1-\delta,1)$ are mapped to values $<1$. Therefore, pick $\epsilon=1$ and remark that any neighborhood $(1-\delta,1+\delta)$ contains points that are mapped to values that do not fall into the neighborhood $(13-1,13+1)$. You have proved that some neighborhood of $f(1)$ cannot be associated to any neighborhood of $1$ as in the definition of continuity.

If you prefer sequences, then write $x_n=1-1/n$ and remark that $\lim_n f(x_n)=1 \neq f(1)$.

  • $\begingroup$ but f(1) = 13 by definition $\endgroup$ – tmpys Oct 31 '14 at 9:14
  • $\begingroup$ and if x_n = 1+1/n then x_n is in (1,2] $\endgroup$ – tmpys Oct 31 '14 at 9:17
  • $\begingroup$ But I could say that $f(x_n) = 1+1/n$ for all $n$. Suppose $f(x_n)$ converges to $f(1) = 13$. Let $\epsilon = 1$ then there exists $N$ s.t. $n\ge N$ implies $|1+1/n-13|=|1/n-12|<1$ a contradiction because $1\le n\implies 1/n \le 1$ for all $n$. Is this right? $\endgroup$ – tmpys Oct 31 '14 at 9:29
  • $\begingroup$ Sorry, I read the $=$ on the wrong half of the definition of $f$. $\endgroup$ – Siminore Oct 31 '14 at 9:58
  • $\begingroup$ Is what I did ok? $\endgroup$ – tmpys Oct 31 '14 at 10:02

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