Proving Discontinuity of f(x) = 1/(x^2)

In summary, the equation f(x) = 1/(x^2) is discontinuous at x = 0 because there exists an x such that |x-0| < delta but |1/(x^2)| >= Epsilon. However, there is a possible solution where ε > 0 such that 0 < |x| < δ and |1/x2 - L| < ε.
  • #1
ccsun
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Homework Statement


Prove that f(x) = 1/(x^2) is not continuous at x = 0 using the epsilon and delta definition of a limit


Homework Equations



definition of discontinuity
There exist epsilon > 0 such that for all delta > 0 there is an x such that |x-0| < delta but |1/(x^2)| >= Epsilon


The Attempt at a Solution



I don't know how to go from here. Like how do i prove it? I don't understand!
 
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  • #2


Start by assuming that [itex]|x| < \delta[/itex]. Consider how this piece of information affects your function [itex]1/x^2[/itex]. Can you derive an [itex]\epsilon[/itex] that is a function of delta such that [itex]|1/x^2| \geq \epsilon[/itex]?
 
  • #3


now, if 1/x2 IS continuous at 0,

[tex]\lim_{x \to 0} f(x) = f(0)[/tex] which is a problem, because f(0) doesn't exist.

but let's go even further, let's prove that there isn't ANY real number we can pick for f(0) that will make 1/x2 continuous at 0.

now, for this limit L to exist, we need to be able to find a δ > 0, so that:

0 < |x| < δ implies |1/x2 - L| < ε.

can L be 0? choose ε = 1. that would mean that we could find δ > 0 so that 0 < |x| < δ implies |1/x2| < 1. here, we have another problem.

|1/x2| = (1/|x|)2, so if 0 < |x| < δ, (1/|x|)2 > δ2. so no δ less than 1 will work. but if we choose δ ≥ 1, then (0,1) is a subinterval of (0,δ), and for 0 < |x| < 1, |1/x2| > 1.

since for the particular ε = 1, we can't find a δ, L can't be 0. so L > 0 (since 1/x2) (if L < 0, then on the interval (0,1], 1/x2 would have to take on the value 0 (by the intermediate value theorem) but if x ≠ 0, 1/x2 > 0).

so what happens if δ < 1/(√(2L))?

then 0 < |x| < 1/(√(2L)) means that |1/x2 - L| ≥ |1/x2| - |L|

= 1/|x|2 - L > 1/(1/(√(2L))2) - L = 2L - L = L

now if we choose 0 < ε < L (which we can, since L > 0, for example ε = L/2 would work fine), we have the same problem as before, no δ < 1/(√(2L)) will work, and any larger choice for δ will lead to |x| < 1/(√(2L)) for some x as well (larger δ's don't help).

so there is always some ε > 0 that doesn't have a δ, no matter what we choose for L. so such an L doesn't exist, we cannot define f(0) to be any real number in such a way as to make f(x) = 1/x2 continuous.
 
  • #4


Hello - where did the 1/sqrt(2L) come from? thank you.
 

1. What does it mean for a function to be discontinuous?

A function is considered discontinuous at a point if it either has a jump or a gap in its graph at that point. This means that the function is not continuous at that particular value of x.

2. How can I prove that a function is discontinuous?

To prove discontinuity of a function, you need to show that the function has a jump or a gap at a specific point. This can be done by evaluating the limit of the function at that point and showing that it does not exist or is not equal to the value of the function at that point.

3. What is the limit of f(x) = 1/(x^2) as x approaches 0?

The limit of f(x) = 1/(x^2) as x approaches 0 is undefined or does not exist. This is because as x gets closer and closer to 0, the value of the function approaches infinity.

4. Can a function be discontinuous at more than one point?

Yes, a function can be discontinuous at multiple points. This can happen if the function has jumps or gaps at multiple points or if the limit of the function does not exist at multiple points.

5. Can a function be discontinuous at all points?

Yes, there are functions that are discontinuous at every point in their domain. One example is the Dirichlet function which is defined as 1 for rational numbers and 0 for irrational numbers. This function is discontinuous at every point in its domain.

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