# Show x^2 is continuous.

1. Jul 14, 2010

### Daveyboy

1. The problem statement, all variables and given/known data
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.

2. Jul 14, 2010

### Hurkyl

Staff Emeritus
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.

3. Jul 14, 2010

### Tobias Funke

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

4. Jul 15, 2010

### HallsofIvy

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

You might start by factoring $|x^2- a^2|< |x-a||x+ a|$.

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

Last edited by a moderator: Jul 17, 2010
5. Jul 16, 2010

### Daveyboy

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.

6. Jul 17, 2010

### HallsofIvy

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

Now you can say $|(x-a)(x+a)|= |x-a||x+a|< (a+\delta)|x- a|$.

Last edited by a moderator: Jul 17, 2010
7. Jul 17, 2010

### Staff: Mentor

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.

8. Jul 17, 2010

### HallsofIvy

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

9. Jul 17, 2010

### Staff: Mentor

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

10. Jul 17, 2010

### 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

11. Jul 18, 2010

### Daveyboy

shouldn't we conclude that

$|(x-a)(x+a)|= |x-a||x+a|< (2a+\delta)|x- a|$
?

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
$(2a+\delta)|x- a|<\epsilon$

for delta would that give me a correctly bounded delta?

12. Jul 18, 2010

### Hurkyl

Staff Emeritus
I think you have a typo there....

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