Proving Continuity of x^2 using Delta-Epsilon Argument

[Daveyboy]

Homework Statement

Show x^2 is continuous, on all reals, using a delta/epsilon argument.

Let E>0. I want to find a D s.t. whenever d(x,y)<D d(f(x),f(y))<E.
WLOG let x>y
|x^2-y^2|=x^2-y^2=(x-y)(x+y)<D(x+y)

I am trying to bound x+y, but can't figure out how.

 

[Hurkyl]

One problem is that you have the epsilon-delta statement wrong. What you wrote is the definition of "uniformly continuous", and squaring is not uniformly continuous.

Uniform continuity requires a delta that works for all values of y. Continuity only requires that, for each value of y, there exists a delta. If it helps, you should think of the delta you are choosing as a function of y.

 

[Tobias Funke]

I am trying to bound x+y, but can't figure out how.

Maybe writing x+y=x-y+2y will help?

 

[HallsofIvy]

You need to show: given any real number, a, then for any \epsilon&gt; 0, there exist \delta&gt; 0 such that if |x- a|&lt; \delta then |x^2- a^2|&lt; \epsilon.

You might start by factoring |x^2- a^2|&lt; |x-a||x+ a|.

Now, if x is close to a, so x- a is close to 0, how large can x+ a be?

 

[Daveyboy]

Now, if x is close to a, so x- a is close to 0, how large can x+ a be?

then x+a approaches 2a...


I feel like I should just take delta = sqrt(epsilon), and I'm fairly confident any delta less than that will suffice. I do not really understand how to show that though.

 

[HallsofIvy]

Saying x+ a "approaches" 2a is not enough. If |x- a|&lt; \delta then -\delta&lt; x- a&lt; \delta so, adding 2a to each part, 2a- \delta&lt; x+ a&lt; 2a+ \delta.

Now you can say |(x-a)(x+a)|= |x-a||x+a|&lt; (a+\delta)|x- a|.

 

[Mark44]

Fixed your LaTeX.

Saying x+ a "approaches" 2a is not enough. If |x- a|&lt; \delta then -\delta&lt; x- a&lt; \delta so, adding 2a to each part, 2a- \delta&lt; x+ a&lt; 2a+ \delta.

Now you can say |(x-a)(x+a)|= |x-a||x+a|&lt; (a+\delta)|x- a|.

Tip: Make sure that LaTeX expressions start and end with the same type of tag. [ tex] ... [ /tex] and [ itex] ... [ /itex]. You sometimes start the expression with [ itex] and end with [ /math]. I'm not sure that this Web site can render [ math] ... [ /math] expressions, but I am sure that you can't mix them.

 

[HallsofIvy]

Thanks, Mark44. I just need to learn to check my responses before going on!

 

[Mark44]

Thanks, Mark44. I just need to learn to check my responses before going on!

That's advice I'm trying to give myself, too.

 

[Telemachus]

How do this demonstrate the continuity? I mean, what you're saying is the same than \lim_{x \to c}{f(x)} = f(c), but in delta epsilon notation? I don't get it :P

 

[Daveyboy]

shouldn't we conclude that

<br /> |(x-a)(x+a)|= |x-a||x+a|&lt; (2a+\delta)|x- a|<br />
?

I still want to clearly define delta as a function of x, epsilon, and a (even though a is a constant)

I see the trick that was used to bound |x^2-a^2| and that was neat, but now I am confused.

If I solved
<br /> (2a+\delta)|x- a|&lt;\epsilon<br />

for delta would that give me a correctly bounded delta?

 

[Hurkyl]

shouldn't we conclude that

<br /> |(x-a)(x+a)|= |x-a||x+a|&lt; (2a+\delta)|x- a|<br />
?

I think you have a typo there...

I still want to clearly define delta as a function of x, epsilon, and a (even though a is a constant)

No, you want it as a function of epsilon and a. (Check your quantifiers -- delta is "chosen" before x enters the picture)