Prove Function is Discontinuous Using Sequences

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• October 3rd 2007, 02:49 PM
tbyou87
Prove Function is Discontinuous Using Sequences
Hi I have to proove this function is discontinuous at x=0 using the formual definition of continuity:

{ 1 / squareroot(-x) x<0
f(x) = { 0 x = 0
{ 1 / squareroot(x) x > 0

Thanks a lot
• October 3rd 2007, 03:07 PM
Plato
Use this sequence: $a_n = \frac{{\left( { - 1} \right)^n }}{{n^2 }}$.
• October 3rd 2007, 03:31 PM
tbyou87
How did you find that sequence?
• October 3rd 2007, 03:52 PM
Plato
Quote:

Originally Posted by tbyou87
How did you find that sequence?

Thirty-five years of teaching the stuff.
You need not understand how I found it.
You must understand why it works as a counter-example.

Do you see that the sequence converges to zero?
But $\lim _{n \to \infty } f\left( {a_n } \right) = \infty$.
What does that contradict about continuity?
• October 3rd 2007, 05:28 PM
tbyou87
Yeah that makes sense. I don't know what I was thinking. I think in my head I wanted to do a proof as you would when you are proving that a function is continuous as opposed to discontinuous. But just to clarify my last question, in general it seems that any sequence of type (1/n^p) where p>0 or a^n where |a| < 1 are likely candidates for a contradiction.

Here's my complete answer by the way:
f(x0) = f(0) = 0
(xn) = 1/n^2 subset of dom(f) = R
(xn) -> 0 = x0 (can be easily proven using defenition of a limit)
lim f(xn) = lim 1/squareroot(1/n^2) = +infinity does not equal f(x0) = f(0) = 0.

You are basically looking for a sequence that goes to 0 but f(xn) goes to something other than 0 to prove discontinuity.

Thanks that was very helpful. I've got another question this time on proving that a function is continuous. Your help was much appreciated. Thanks