Quick Question about continuity at a point

[tylerc1991]

Homework Statement

I have always been comfortable with proving continuity of a function on an interval, but I have been running into problems proving that a function is continuous at a point in it's domain. For example:

Prove [itex]f(x) = x^2[/itex] is continuous at [itex]x = 7[/itex].

Homework Equations

We will be using the delta epsilon definition of continuity here.

The Attempt at a Solution

Let [itex]f(x) = x^2[/itex] and [itex]\varepsilon > 0[/itex].

Choose [itex]\delta[/itex]= ________ (usually we choose [itex]\delta[/itex] last, so I am just leaving it blank right now).

Now, if [itex]|x - y| = |7 - y| = |y - 7| < \delta[/itex], then

[itex]|f(x) - f(y)| = |49 - y^2| = |y^2 - 49| = |y + 7||y - 7|.[/itex]

This is where it gets a little awkward for me. I know that I may say [itex]|y - 7| < \delta[/itex], but what do I do with the [itex]|y + 7|[/itex]? Could I say that [itex]|y + 7| < \delta + 14[/itex]? Then I would have to choose a [itex]\delta[/itex] such that [itex]\delta (\delta + 14) = \varepsilon[/itex].

Thank you for your help anyone!

 

[Stephen Tashi]

There is probably a way to write the proof using mostly references to absoulte values. However, it is useful to know how to "grunge it out" when no elegant way comes to mind.

When you have to get down and dirty, it is best to write things like [itex] |y-7| < \delta [/itex] in the equivalent form of:

eq 1. [itex] 7 - \delta < y < 7 + \delta [/itex]

(For simplicity I'll label them "equations" but the they actually are inequalities.)

To square eq. 1 and keep the inequality marks pointed the same way, we must make sure that all the terms are positive. We can make [itex] 7 - \delta > 0 [/itex] by chosing [itex] \delta < 7 [/itex], so remember this condition. Squaring eq 1., we get:

eq. 2. [itex] 49 - 14 \delta + \delta^2 < y^2 < 49 + 14\delta + \delta^2 [/itex]

To get the functions of [itex] \delta [/itex] to be closer to [itex] y^2 [/itex] than [itex] \epsilon [/itex] we need eq. 3 and eq. 4 to hold:

eq 3. [itex] 49 - \epsilon < 49 - 14 \delta + \delta^2 [/itex]

eq. 4. [itex] 49 + 14\delta + \delta^2 < 49 + \epsilon [/itex]

Thos equations simplify to eq 5. and eq 6. respectively:

eq 5. [itex] -\epsilon < -14 \delta + \delta^2 [/itex]

eq 6. [itex] 14 \delta +\delta^2 < \epsilon [/itex]

Mutliplying eq 5. by -1 and reversing the inequality sign gives:

eq 7. [itex] 14 \delta - \delta^2 < \epsilon [/itex]

If eq. 6 holds then eq 7 would also, so we only worry about eq 6.

Rather than worry about solving quadratic equations, it's simpler to take advantage of the fact that we are dealing with inequalities and trying to make [itex] \delta [/itex] small.

So add the condition [itex] 0 < \delta < 1 [/itex] so that we can say [itex] \delta^2 < \delta [/itex]

This and eq 6. imply that we want:

eq 8. [itex] 0 < 14 \delta + \delta^2 < 14\delta + \delta < \epsilon [/itex]

eq 9. [itex] 15 \delta < \epsilon [/itex]

So this imples we want:

eq 10. [itex] \delta < \frac {\epsilon}{15} [/itex]

We can satisfy eq 10. by setting [itex] \delta [/itex] equal to various things, for example [itex] \delta = (0.5)\frac{\epsilon}{15} [/itex] or [itex]\delta = \frac{\epsilon}{16} [/itex] etc.

We have to remember the previous assumptions we made on [itex] \delta [/itex].
To incorporate all of them , it is sufficient to say:

eq 11. Let [itex] \delta = min\{ \frac{\epsilon}{16}, 1.0 \} [/itex]


To have a real proof you have to go through the reasoning in reverse order.