I was asked to pick a function $f$ for which $\lim_{x\to c^-} f(x) \neq \lim_{x\to c^+} f(x)$ for some $c$. I used $f(x)=\sqrt{x-2}$ with $c=2$ as an example of such a function.

My question is the following. Since $f$ is undefined for $x<2$ and the left-hand limit as $x$ approaches $2$ is not defined, does that make it unequal to $0$, the right-hand limit at the same point?

I got the answer wrong when I said that, so an explanation would help. I also had to explain why it was not continuous and I said because it was undefined for $x<2$.

  • $\begingroup$ The problem with you example is that your function doesn't have a left limit. Your function is undefined for $x<2$, at least in the real numbers. $\endgroup$ – Enigma Dec 7 '14 at 20:12
  • $\begingroup$ This is basically a duplicate. The answer is no. $\endgroup$ – Git Gud Dec 7 '14 at 20:16
  • $\begingroup$ so I guess my mistake was thinking if it does not exist, then it is not equal? $\endgroup$ – Sara Dec 7 '14 at 20:23
  • 1
    $\begingroup$ @Sarah Yes. But even if non-existence implied non-equality, part of solving a problem is understanding what is being asked. An argument could be made that you were supposed to provide a counter example in which both limits exist, even if the problem didn't strictly ask for this. But this is another matter... $\endgroup$ – Git Gud Dec 7 '14 at 20:25

Your reasoning seems to be:

  1. For a limit to exist at a point $c$ it must have equal left and right side limit values at $c$.

  2. "not defined" vs $0$ means not equal.

  3. If $f$ has no limit at $c$, then $f$ is not continous at $c$.

That is for

  1. not usable, the one sided limits must exist both as well for this argument

  2. fishy, because the problem context requires binary equality / inequality relations for elements from $\mathbb{R} \cup \{ -\infty, +\infty \}$, with both arguments of the relations being defined, there is no custom to have an undefined or $\bot$ special value here

  3. fine

So it breaks down with 1.

In fact $ \lim\limits_{x \to 2} \sqrt{x - 2}. = 0 $ and $\sqrt{x-2}$ is continous at $x=2$.

A simple answer would have been a function with a jump at $c$, like $$ f(x) = \left\{ \begin{matrix} 0 \quad \mbox{ for } x \le c \\ 1 \quad \mbox{ for } x > c \end{matrix} \right. $$


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