# If a limit does not exist does that make it unequal to some given value?

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 $x<2$.

• 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. – Enigma Dec 7 '14 at 20:12
• This is basically a duplicate. The answer is no. – Git Gud Dec 7 '14 at 20:16
• so I guess my mistake was thinking if it does not exist, then it is not equal? – Sara Dec 7 '14 at 20:23
• @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... – Git Gud Dec 7 '14 at 20:25

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 $$f(x) = \left\{ \begin{matrix} 0 \quad \mbox{ for } x \le c \\ 1 \quad \mbox{ for } x > c \end{matrix} \right.$$