# Definition of continuity

1. Aug 24, 2009

### cd19

1. The problem statement, all variables and given/known data
Can anyone please explain to me 'informally' the definition of continuity and the conditions associated with. I can't grasp the concept. Any input would be much appreciated.

2. Relevant equations

The function f is undefined at c

The limit of does not exist as x approaches c

The value of the function and the value of the limit at c are different
[/quote]
I am not sure what you are asking. Each of the above is a way that a function may NOT be continuous at c. Basically, the ideas is that if you look at the value of f(x) at points very close to c, you will get values very close to f(c). All the rest is just saying what "very close" means. Of course, if f(c) does not exist, that is impossible. If there are points, say, a, b, "very close" to x= c so that f(a) and f(b) are not very close together then that is impossible- that's what "limit does not exist" means. And, if both limit and value are different, the function is not continuous.
3. The attempt at a solution

Last edited by a moderator: Aug 24, 2009
2. Aug 24, 2009

### tiny-tim

Hi cd19!

Informally? …

ok, take a function f(x,y) defined on the plane.

f is continuous at the point (3,7), say, if you get the same limit no matter which path you take towards (3,7).

For example, f(x,y) = xy/(x2+y2) is not continuous at the origin …

try approaching the origin along the four paths x = 0, y = 0, x = ay, x = ayn

3. Aug 24, 2009

### cd19

what i'm basically asking is what in the simplest terms is continuity, the way it is explained in my maths books i find very hard to understand. thanks for the help tiny tim but i still don't understand whats happening, junior freshman so this is all new to me!

4. Aug 24, 2009

### snipez90

For a function f that maps real numbers to real numbers we want to know what it means for f to be continuous at c. The essence of the definition is that no matter how small a neighborhood $$\left(f(c) - \varepsilon, f(c) + \varepsilon\right)$$ around f(c) (note: we take $$\varepsilon > 0$$; can you visualize this as an open interval about f(c) on the y-axis?), we can choose a sufficiently small neighborhood $$\left(c - \delta, c + \delta\right)$$ around c so that for any $$x \in \left(c - \delta, c + \delta\right),$$ f(x) is in $$\left(f(c) - \varepsilon, f(c) + \varepsilon\right)$$.

Note that informally, the previous paragraph just says that $$\varepsilon$$ is an arbitrarily small positive number that prescribes how close we want f(x) to be to f(c), and no matter how small $$\varepsilon$$ is, we can always choose $$\delta$$ so that if x is close enough to c, then f(x) will indeed be within the prescribed $$\varepsilon$$ of f(c).

5. Aug 24, 2009

### tiny-tim

"you get the same limit no matter which path you take" is in the simplest terms, isn't it?

what do you find difficult to understand about that?

(did you try the example i gave?)

6. Aug 24, 2009

### Elucidus

This example is a very rough intuitive way of looking at continuity of a real-valued function of one real variable.

Intuitive idea of continuity: A funtion f(x) is continuous at x = c if the curve y = f(x) is unbroken at (c, f(c)).

For example f(x) = x + 1 is continuous at x = -1 since y = x + 1 is a line in the plane (and is unbroken everwhere along its length).

But g(x) = (x2 - 1)/(x - 1) equals f(x) everywhere except at x = -1 where it is undefined. The graph of y = g(x) looks like y = f(x) except that it has a puncture (a hole in the graph) at (-1, 0). So the curve is broken when x = -1 and the function g(x) is discontinuous at x = -1.

Another example: Consider

$$h(x) = \left\{ {\begin{array}{cc} 0 & x < -1 \\ x^2 & -1 \leq x \leq 0 \\ \ln(x)+1 & 0 < x < 1 \\ \frac{1}{x} & x \geq 1 \end{array}}$$

Investigating the graph of y = h(x) you should discover that h is discontinuous at x = -1 and 0, but is continuos at x = 1.

I will mention in caution that this sense of "unbrokenness" is only an intuitive aid. There are functions that are continuous at certain points but not unbroken, but these functions are usually edge cases and rarely show up in day-to-day work.

Technically a function f(x) is continuous at x = c if all of the following three things are true:

(1) $\lim_{x \rightarrow c} f(x)$ exists.

(2) $f(c)$ is defined.

(3) and $\lim_{x \rightarrow c}f(x) = f(c)$.

--Elucidus

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7. Aug 24, 2009

### mg0stisha

The easiest way I've ever heard it is that a function is continuous if you can draw it's graph without lifting your pencil.

8. Aug 24, 2009

### VietDao29

Maybe he's asking about one-variable function. Junior freshman hasn't covered multi-variable yet, so your example is a little bit over his head, methinks. :)

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@OP: You can have a look http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html" [Broken] to get a closer view about epsilon-delta definition of limit. Roughly speaking, if you have: $$\lim_{x \rightarrow \alpha} f(x) = L$$, it means, no matter how "close" you want the value of f(x) and L to be (i.e $$\forall \epsilon > 0, |f(x) - L| < \epsilon$$), there always exists a neighbor of $$\alpha$$ (the region arround $$\alpha$$) $$\alpha$$ excluded (i.e $$\exists \delta > 0, 0 < |x - \alpha| < \delta$$), on which region, the value of f(x) is that "close" to L, i.e:
$$\forall \epsilon > 0, \exists \delta > 0 : 0 < |x - \alpha| < \delta \Rightarrow |f(x) - L| < \epsilon$$.

The definition of the continuity of a function is somewhat similar to the above idea. f is continuous at x0 iff the limit of f as x tends to x0 is f(x0), i.e f will tend to f(x0) as x tends to x0.

Last edited by a moderator: May 4, 2017
9. Aug 24, 2009

### mg0stisha

In addition to VietDao29's post, here's a video tutorial on continuity (There's also other video tutorials for all limits and derivatives. They're extremely good.)

http://www.calculus-help.com/funstuff/tutorials/limits/limit05.html [Broken]

Last edited by a moderator: May 4, 2017
10. Aug 25, 2009

### tiny-tim

Hi VietDao29!
ah … good point … I didn't think of that.

in that case, i'd go with mg0stisha's idea …