Epsilon-Delta Proof for Discontinuous Functions

__THE QUESTION:__

Suppose we have f:R --> R defined by the piecewise function:

f(x) = 1/(x^2), when x ≠ 0

f(x) = 7, when x = 0.

Prove that f is not continuous at x = 0 using an epsilon-delta proof.

__WHAT I'VE DONE:__

I know by definition: "a function f is not continuous at a point a if we can find an ε > 0 such that for every δ > 0, we can find a real x such that if |x-a| < δ then |f(x)-f(a)| ≥ ε".

I started by fixing ε = 8 and then left δ > 0 arbitrary. I now need to find x that satisfies the condition: |x - 0| = |x| < δ implies |1/(x^2) - 7| ≥ 8.

From this point on, I'm not quite sure what to do but I've tried playing around and used |1/(x^2) - 7| ≥ 8 to say:

|1/(x^2) - 7| ≤ |1/(x^2)| + |-7| (using triangle inequality)

|1/(x^2) - 7| ≤ 1/(x^2) + 7

So

1/(x^2) + 7 ≥ 8

1/(x^2) ≥ 1

x^2 ≤ 1 which implies -1 ≤ x ≤ 1

But I haven't done anything with δ here so I'm not convinced that this is the right x that I'm supposed to find. Have I gone about it the right way like this and need something more? Or is this completely wrong? If so, please show me what the right way to go about this is. Thank you!

Re: Epsilon-Delta Proof for Discontinuous Functions

Hi,

You're trying to prove the wrong thing!

A function f is discontinuous at a provided there is $\displaystyle \epsilon>0$ such that for any $\displaystyle \delta>0$ there is some x such that $\displaystyle |x-a|<\delta$ __and__ $\displaystyle |f(x)-f(a)|\ge\epsilon$.