# Continuous function

how do i show $f(x)=\frac{x}{1-x^2}$ is a continuous function by means of an $\epsilon - \delta$ proof? oh and $x \in (-1,1)$

so far i have said:
let $\epsilon>0, \exists \delta>0 s.t. |x-x_0|< \delta$. now i need to show that $|f(x)-f(x_0)|< \epsilon$. yes?

can't do the rest of it though.......

## Answers and Replies

Mark44
Mentor
You can't just say that a delta exists, you need to show that one exists. Assume that you have some positive epsilon. Now work with your inequality involving f(x) and f(x_0) and epsilon, and manipulate this until you get an inequality with x and x_0.

its virtually impossible to factorise....

could i just say x is continuous and 1-x^2 is continuous and non zero so f is continuous?

Mark44
Mentor
It depends on how you are required to show continuity. If you have to give an $\epsilon - \delta$ proof, then that's what you need to do. OTOH, if you have some theorems about continuity you can use, then that would be easier. For example, polynomials are continuous everywhere, and quotients of polynomials are continuous on any interval whose points don't cause the denominator to vanish (become zero). Does that help?

HallsofIvy
Homework Helper
Since f(x)= x/(1- x^2),
$$f(x)- f(x_0)= \frac{x}{1- x^2}- \frac{x_0}{1- x_0^2}$$
$$= \frac{x(1-x_0^2)- x_0(1- x^2)}{(1- x^2)(1- x_0^2)}$$
$$= \frac{x- xx_0^2- x_0- x^2x_0}{(1- x^2)(1- x_0^2)}$$
$$= \frac{(x- x_0)- x_0x(x- x_0)}{(1- x^2)(1- x_0^2)}$$
$$= (x- x_0)\frac{1- x_0x}{(1- x^2)(1- x_0^2)}$$
Now you need to find a bound on
$$\frac{1- x_0x}{(1- x^2)(1- x_0^2)}$$

You will, of course, want to stay away form x or $x_0$ being 1 or -1 so you might do this: Let a be the smaller of $|1- x_0|$ and $|-1- x_0|[/tex] and require that [itex]|x- x_0|< a/2$.