Is tan(x) continuous when x = pi/2?

1. Sep 5, 2014

johann1301

Is the tangents function tan(x) continuous when x = 90 degrees or x = pi/2?

2. Sep 5, 2014

da_nang

No, it isn't.

3. Sep 5, 2014

johann1301

because it isn't defined at that point?

4. Sep 5, 2014

Staff: Mentor

Correct. The domain of tan(x) doesn't include odd multiples of π/2.

5. Sep 5, 2014

johann1301

is this true for all functions? If a function f(x) isn't defined at some point x=a, then f(x) isn't continuous at the point a?

6. Sep 5, 2014

Staff: Mentor

Yes, if the function isn't defined at a point, then it's not continuous there. When we talk about continuity, we are necessarily talking about its domain of definition.

7. Sep 5, 2014

Thanks!

8. Sep 5, 2014

HallsofIvy

Do you not know the definition of "continuous" at a give point? It is
"A function, f, is continuous at x= a if and only if these three conditions are satisified:
1) f(a) exists (f is defined at x= a)
2) $\lim_{x\to a} f(x)$ exists
3) $\lim_{x\to a} f(x)= f(a)$

Since (3) certainly implies that the left and right sides of the equation exist, often we just state (3) alone. But it is part of the definition of "continuous" that f is defined at x= a.

9. Sep 5, 2014

johann1301

Ok, this makes sense, but what about epsilon and delta argumentation? Is that another definition?

10. Sep 5, 2014

caveman1917

Yes, you can define continuity in several different ways that are (mostly) equivalent. None of them has a function being continuous at a point in which it doesn't exist.

11. Sep 8, 2014

willem2

No. That's just used in the definition of a limit, wich is used in the definition of a continuous function.

12. Sep 8, 2014

caveman1917

That's not necessarily correct, the Weierstrass definition can be used on its own without requiring you to define limits. But even if you define both they need not necessarily correspond, some authors will for instance define them slightly different and then give the "$\lim_{x \to c} f(x) = f(c)$ iff f continuous at c" as a theorem for limit points only (at least my analysis textbook did so). At the end of the day it doesn't make much difference of course, but it's still something to keep in mind.

Last edited: Sep 8, 2014
13. Sep 8, 2014

caveman1917

As it turns out wikipedia (see article on "continuous function") falls for such an issue, probably due to using several different textbooks by different authors together. The Weierstrass definition has $|x-p| < \delta$, the definition for a limit has $0 < |x-p| < \delta$.

The difference is small but relevant, define $f \subseteq \{ p \} \times \mathbb{R} = \{ (p, f(p)) \}$. Then the limit at p doesn't exist yet f is continuous at p.

14. Sep 8, 2014

DrewD

Unless I misunderstand what you are saying, this is not an error. For a continuous function $|x-c|<\delta$ is used because the $\epsilon$-$\delta$ relationship must hold when $x=c$. For a limit $0<|x-c|<\delta$ because the value of $f(c)$ is unimportant (not even required to exist).

If the limit doesn't exist, then there is some $\epsilon>0$ such that no corresponding $\delta$ can be found. If this is the case, it will still be the case when the restriction that $0<|x-c|$ is included. Of course $\delta>0$. If not, then every point of a function would be continuous.

15. Sep 8, 2014

caveman1917

It is not an error by itself, but using these ways of defining it means that $\lim_{x \to c} f(x) = f(c)$ iff f continuous at c is not a theorem unless restricted to limit points.

The function provided above was intended as a counter-example. It is continuous since for all strictly positive $\epsilon$ there exists a strictly positive $\delta$ such that for all $x \in \{ p \}$ it is true that
$|x - p| < \delta \Rightarrow |f(x) - f(p)| < \epsilon$ which, since the only element of $\{ p \}$ is p itself boils down to
$0 < \delta \Rightarrow 0 < \epsilon$ which is true by definition of $\epsilon, \delta$.

That the function does not have a limit at p can be seen by that, as you say, it doesn't depend on f(p) but f(p) is all you have to base it on. Or more formally, no (non-empty) punctured neighbourhoods of p exist.

16. Sep 9, 2014

DrewD

That's certainly correct. I wasn't paying enough attention. Thank you for pointing out my mistake.

The theorem could be stated as $f:U\rightarrow\mathbb{R}$ with $U$ open instead of referring directly to limit points (I think open sets would save the day).

17. Sep 9, 2014

caveman1917

You're welcome, but i wouldn't really call it a mistake. I was just trying to point out that the answer to the question whether they are different definitions is indeed yes, they are almost equivalent but not completely. In the end, as long as you're aware of the precise consequences, it's mostly personal preference as to how exactly you define continuity.

But is $\{p\}$ an open set? It seems impossible for it not to be. Are you thinking of an open interval (of real numbers)? Or are you thinking of some other larger space in which you consider $\{ p \}$ not to be an open set?

Last edited: Sep 9, 2014