# Thread: Continuous and discontinuous functions

1. ## Continuous and discontinuous functions

Hello,

I'm having trouble understanding continuous and discontinuous functions. I understand the basic concept of continuous functions are functions where small changes in input result in small changes in output, but isn't that the case with all functions? I understand them as far as graphing goes, but I'm not fully understanding the overall concept.

Why is y=1/x discontinuous but y=x*2 continuous?

Can I get a simple explanation?

Thanks.

2. Originally Posted by SeanC
Hello,

I'm having trouble understanding continuous and discontinuous functions. I understand the basic concept of continuous functions are functions where small changes in input result in small changes in output, but isn't that the case with all functions? I understand them as far as graphing goes, but I'm not fully understanding the overall concept.

Why is y=1/x discontinuous but y=x*2 continuous?

Can I get a simple explanation?

Thanks.
Look at a graph of y=1/x near x=0, you will see that it goes to +infty as you
approch 0 from above and -infty as you approach 0 from below, so near 0
a small change in x cn result in a large change in y, hence this is
discontinuous at x=0.

For you other example if you change x to x+e, y changes from x*2 to
x*2+e*2, and as e is small so is e*2.

Note by convention in ASCII (plain text) * denotes multiplication. If you
meant x squared you should use x^2. In this case y changes from x^2 to
(x+e)^2 = x^2 + 2e x + e^2, and if e is small enough 2e x+e^2 is also
small.

RonL

3. The actual definition of a continous function is as follows. Vageuly as I remember it from 1st and 2nd year undergrad.

a function f(x) is continous at a point a if

$\lim_{\epsilon \rightarrow 0} f(a-\epsilon)=\lim_{\epsilon \rightarrow 0}f(a+\epsilon)$.

A function is continous everwhere if it is continous for all a. So take the function
$f(x)=\frac{1}{x}$

at x=0.

$\lim_{\epsilon \rightarrow 0} (\frac{1}{-\epsilon}) \rightarrow -\infty$

$\lim_{\epsilon \rightarrow 0}(\frac{1}{\epsilon}) \rightarrow +\infty$

The two aren't equal, so the function is dicontinous at x=0

The simple definition is...

A function is continous if you can draw it WITHOUT lifting pen off paper.