# Can a limit exist at $x\to a$ but not equal to $f(a)$?

In the calculus course I'm taking, the conditions for a continuous function are said to be when

a.) $$\lim_{x\to a}f(x)$$ exists
and
b.) $$\lim_{x\to a}f(x)$$ = $$f(a)$$

And it seems a little redundant, since if a limit exists at $$x\to a$$, wouldn't it always equal to $$f(a)$$? Are there situations where that is not the case?

I apologize if this was a bad question/ formatted bad, I'm pretty new to math and this site in general.

• We use limits to study the behaviour of he function "near" the point of interest but excluding it (i.e. deleted neighbourhood)
– user
Commented Sep 6, 2022 at 22:42
• The definition of the limit involves $$\color{red}{0 < } ~|x - a| < \delta,$$ rather than $$|x - a| < \delta.$$ Commented Sep 6, 2022 at 23:58
• @user2661923 This seems in direct contradiction with most definitions of limits I've seen in calculus courses. It's a matter of convention of course, but the convention I've most often seen is that the definition involves $|x-a| < \delta$, not $0 < |x-a| < \delta$; with this convention, you can still exclude point $a$, but you need to be explicit: $$\lim_{\substack{x \to a \\ x \neq a}} f(x)$$ Which convention is "better" is a discussion for matheducators.stackexchange.com rather than for math.stackexchange, but the only answer for OP's question is "what definition were you given?"
– Stef
Commented Feb 14 at 13:07
• @Stef $~|x-a| = 0 \iff x=a.~$ If you don't (somehow) prevent $~x=a,~$ then you blur the distinction between the definition of a limit, and the definition of a function being continuous at a specific point. The limit of a function as $~x \to a~$ can exist, without this limit equaling $~f(a)~$ (i.e. without the function being continuous at $~x=a~$). Commented Feb 14 at 15:10
• @user2661923 For the first half of your comment: I don't blur anything. In fact, I make things more explicit. If I need to exclude $a$, then I can exclude $a$ explicitly. Which many math teachers will argue is better for the students.
– Stef
Commented Feb 14 at 15:16

No.

$$f(x)= \begin{cases} 2, x=0\\ x, x \ne0 \\ \end{cases}$$

$$f(0)=2$$, but $$\lim_{x\to 0}f(x)=\lim_{x\to 0}x=0$$.

You can have a function with a hole in it, for example:

In fact, the function may not even be defined at $$a$$. One example is $$f(x) = \begin{cases} -x + 4 & \text{if x \neq 0}. \end{cases}$$ Then $$\lim\limits_{x \to 0} f(x) = 4$$, but $$f(0)$$ doesn't exist. I could also define $$f(0)$$ to be anything, say, $$5$$. Then the limit exists and is equal to $$4$$, but differs from $$f(0)$$. In either of these cases, $$f(x)$$ is not continuous at $$x = 0$$.

• @TedShifrin Yes, this was a silly typo. Thank you for pointing this out! Commented Sep 7, 2022 at 0:08

This confusion often occurs because most, if not all, of the functions people are acquainted with in elementary calculus are continuous. In other words, they do have the property that if $$f$$ is defined at $$a$$, then $$\lim_{x \to a}f(x)$$ exists and is equal to $$f(a)$$. But not every function is like this. Consider, for example, the sign function or signum function, written as $$\DeclareMathOperator{\sgn}{sgn}\sgn$$: $$\sgn(x)= \begin{cases} 1 & \text{if x>0} \, ,\\ 0 & \text{if x=0} \, ,\\ -1 & \text{if x<0} \, . \end{cases}$$ It's called the sign function because it tells us whether a real number is positive, negative, or zero. Note that $$\sgn$$ is defined at $$0$$, but not continuous at $$0$$. It is continuous at all other points.