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I am trying to prove why this function is discontinuous based on the three conditions

function exists at $x=a$ (In other words, $f(a)$ is a real number)

the limit of the function exists at $x=a$. (That is, $\lim_{x\to a}f(x)$ is a real number)

The two values are equal (That is, $\lim_{x\to a}f(x)=f(a)$.)

$$f(x)\begin{cases} x^2 & \text{if $x<2$} \\ 3x-2 & \text{if $x>2$} \\ \end{cases}$$

at $x=2$ (I don't know how to put this to the right of the branches, which is how my book shows it)

From what I gather, if I am looking at this correctly: Conditions 1 and 2 are not met, but condition 3 is

$f(2)$ both parts of the function $x^2=4$ and $3x-2=4$ equal 4 but only $3x-2$ meets the if $x>2$. So the $f(2)=4 \ne2$

The third condition is true because they are equal?

$lim_{x\to 2}(x^2)=4$

$lim_{x\to 2}(3x-2)=4$

So, I think I am on the right track, but not positive.

David K
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xerxes2985
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  • What are the relations between those three conditions? What happens when $x<a$? – Marra Nov 13 '17 at 22:17
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    Both branches (left and right) approach $4=(2)^2=3*2-2$. So what is $f(2)$ defined to be? If it equals one of the branch values, then the function is continuous. – AnyAD Nov 13 '17 at 22:20
  • It's very hard to understand what you're writing because you have little disconnected fragments of statements and other completely ungrammatical phrases all over the place. (For example one line that just says, "at $x = 2$". What happens at $x=2$?) – David K Nov 13 '17 at 22:22
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    BTW, write \lim_{x \to 2} in MathJax to display $\lim_{x \to 2}.$ (I suppose that's what your $\lim_{x>2}$ and $\lim_{x->2}$ were supposed to be?) – David K Nov 13 '17 at 22:24
  • As interpolated from what you have written your definition of continuity is only continuity from the right. It is not clear whether $f(2)$ has been defined. The correct definition of continuity at $a$ in these kinds of split definition cases is 1) $f$ is defined at $a$ 2) continuity from the right at $a$ and 3) continuity from the left at $a$. In both 2) and 3) continuity is determined based on the independent definition of $f(a)$ not the values of the functions on the left and right. – Stephen Meskin Nov 13 '17 at 22:33
  • @Marra : I think OP is using ">" to mean "$\to$" – MPW Nov 13 '17 at 22:33
  • @DavidK Sorry, I cannot write it in the equation, I am not sure how.. I'll try to edit the post so it looks more like the actual question. – xerxes2985 Nov 13 '17 at 22:39
  • @Marra yes. I am not too familiar with the proper syntax.. learning as I go. – xerxes2985 Nov 13 '17 at 22:43
  • I apologize, yes $f(2)$ is not defined at all (I was attempting to guess what the $f(2)$ is. The given value of $x=2$ is provided. The book says to determine which rules are being violated to prove why the function is discontinuous. – xerxes2985 Nov 13 '17 at 22:46
  • Details matter. If the book says, "Show that $f$ is not continuous at $x=2,$" but you tell us, "I need to prove that $f$ is discontinuous," you're likely to get some confusing answers and comments. Those two concepts are not the same! Please review exactly what the question in the book is asking. – David K Nov 14 '17 at 03:54
  • The book tells me it’s discontinuous, and says to explain what conditions it violates. – xerxes2985 Nov 17 '17 at 20:44

2 Answers2

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Some books do not include the first criteria.

A function is continuous if $\lim_\limits{x\to a} f(x) = f(a)$ for all $a$ in the domain of $f$

This also concludes that the limit exists at all $a$ in the domain.

If the limit exists then $\lim_\limits{x\to a^-} f(x) = \lim_\limits{x\to a^+} f(x)$

By this definition $f(x) = \frac 1x$ is continuous everywhere the function is defined.

But, using the definition you have provided.

$2$ is not in the domain if $f(x)$ and the function is not continuous (and you can stop here).

As for the limit.

$\lim_\limits{x\to a^-} = \lim_\limits{x\to a^+} = 4$

The limit is defined. Condition 2 is met. Condition's 1 and 3 have not been met.

Doug M
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  • Good answer. You may want to change reference to the point $f(a)$ at the end of the 2nd line to just the function $f$ or $f(x)$ – Stephen Meskin Nov 13 '17 at 22:42
  • Actually, according to the definition given in this answer, $f$ is continuous. Since $f(2)$ is not defined, $2$ is not in the domain of $f,$ so the "$a$" in the definition must be another number. And it turns out $f$ is continuous at $a$ if $a$ is any real number other than $2,$ that is, if $a$ is any number in the domain of $f.$ – David K Nov 14 '17 at 03:52
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You are right that condition 1 is not met. Since continuity requires all three conditions to be met, you already have enough information to conclude that the function is discontinuous at $2$.

On the other hand, condition 2 is met. As for condition 3, it makes no sense unless conditions 1 and 2 are both met, since it talks about "the two values" obtained from those conditions. But nothing in the present paragraph alters the fact that the function is discontinuous at $2$ simply because it isn't defined at $2$.

Andreas Blass
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