Let $f(x)=\begin{cases} 2x + 3&\text{ for }x\geq 1,\\ -x+5 &\text{ for }x<1. \end{cases}$

$f$ is continuous from the right at $x\geq1$. The proof would be:

Let $\epsilon>0$ be arbitrary.

Let $x_0\geq1$.

Let $\delta=\epsilon/2$.

Let $x\in R$ and $x_0\leq x<x_0+\delta$. Thus $x\geq1$.

Thus $|f(x)-f(x_0)|=|2x+3-2x_0-3|=|2x-2x_0|=2|x-x_0|<2\delta=2\epsilon/2=\epsilon$.

This comes from the definition for continuity from the right:$\forall\epsilon>0\; \exists\delta>0$ such that if $x\in I$ and $x_0\leq x<x_0+\delta$ then $|f(x)−f(x_0)|<\epsilon$.

Prove $f$ is discontinuous from the left at $x=1$ using the definition: $\exists\epsilon>0\; \forall\delta>0$ such that if $x\in I$ and $x_0-\delta<x\leq\ x_0$ then $|f(x)−f(x_0)|\geq\epsilon$.

I can't seem to find $\epsilon$. But I think the proof would go like:

Let $\epsilon$ = ?

Let $\delta >0$ be arbitrary.

Let $x_0<1$.

Let $x\in I$ that is to say $x<1$ and $x_0-\delta<x\leq\ x_0$

Then from there is figuring out $|f(x)−f(x_0)|\geq\epsilon$ which I don't get because $|f(x)−f(x_0)|=|-x+5 +x_0-5|=|-x+x_0|=|x-x_0|<\delta$.

So would you set $\epsilon\leq\delta$?

I know my definitions are correct. My teacher has drilled them into our brains. $I$ stands for the domain of $f$ which is the reals or $R$ except the domain is split in two. And by "from the right" and "from the left" I mean that the space between $x$ and $x_0$ denoted as $\delta$, or $|x-x_0|<\delta$, is only calculated on one side, either adding or subtracting $\delta$, not by doing both which would be $x_0-\delta<x<x_0+\delta$.

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    $\begingroup$ Is this really how the question reads, word for word, in its entirety? Also, what is $I$? Also also, your definition of discontinuous is wrong. $\endgroup$
    – dfeuer
    Nov 3, 2013 at 0:04
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    $\begingroup$ I think your problem is not complete. The function $f$ should be also defined on $x\geq 1$. $\endgroup$ Nov 3, 2013 at 0:05
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    $\begingroup$ the function is continuous everywhere fella $\endgroup$
    – ILoveMath
    Nov 3, 2013 at 0:06
  • $\begingroup$ @WorawitTepsan It looks like a $\tt new$ definition of discontinuity: "It is not defined 'somewhere'...". $\endgroup$ Nov 3, 2013 at 0:08
  • $\begingroup$ @Maddy : I am not going to downvote your question, because it looks you put a lot of work into it. But you have to define what $f(x)$ is for $x \geq 1$. And put that into your question, not just a comment. $\endgroup$ Nov 3, 2013 at 0:14

1 Answer 1


Hint 1: Show that $\lim_{x\rightarrow 1^{-}}\left(f(x)\right)$=4. However $f(1)=5$ which is not equal to the left limit. Showing this is sufficient to say that $f$ is not left continuous at 1. Hope this helps.

Hint 2 : Take $ϵ=1/2$. Let $δ>0$. What I want you to do is to prove the converse of the definition of left continuity. Suppose $x<1$. Observe that $|f(x)-f(1)|=|-x+5-2(1)-3|=|-x|=|x|$. Clearly which ever interval $(1-δ,1)$ that is formed depending on $δ$ there always exists a $x$ that is in $(1-δ,1)$ such that $|x|>1/2$ ($|f(x)-f(1)|>1/2$). This is not written in the most formal way I hope you can figure out the whole answer.

  • $\begingroup$ Thanks for the help, but we can't use limits. Could you show me using the definitions? $\endgroup$
    – Maddy
    Nov 3, 2013 at 9:15
  • $\begingroup$ I get where you are going with this, but the key my teacher wants is to be able to show $|f(x)-f(x_0)|\geq\epsilon$. Then I'm home free. $\endgroup$
    – Maddy
    Nov 3, 2013 at 9:24
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    $\begingroup$ $\epsilon = 1$ will do, try it, the reason is that no matter how close you choose the values of $x$ around $x=1$, you can't get the difference of the function on both sides to be less than 1. That is the essence of the proof by $\epsilon$-$\delta$ method. $\endgroup$ Nov 3, 2013 at 9:56
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    $\begingroup$ @Maddy I added another hint. Hope it helps. $\endgroup$
    – Heisenberg
    Nov 3, 2013 at 12:05
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    $\begingroup$ @Maddy If you think the answer is right you can accept it by clicking on the tick mark. It would benefit others as well. Glad I was of some help. $\endgroup$
    – Heisenberg
    Nov 4, 2013 at 2:28

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