# Math Help - Proving a Function is Continuous

1. ## Proving a Function is Continuous

Hi there,

The following is a suggested exercise I'm doing for my Analysis class.

Prove $f:x \rightarrow x^2$

Here's what I have, but it's straight from my class notes. I'm not sure what to do next.

f is continuous at a if:
$
\forall \epsilon > 0 \quad \exists \delta > 0 \medspace s.t.
$

$
\forall x \in A, \quad ||x-a||<\delta \implies ||f(x) - f(a)||< \epsilon
$

Any ideas?

2. Originally Posted by manygrams
Hi there,

The following is a suggested exercise I'm doing for my Analysis class.

Prove $f:x \rightarrow x^2$

Here's what I have, but it's straight from my class notes. I'm not sure what to do next.

f is continuous at a if:
$
\forall \epsilon > 0 \quad \exists \delta > 0 \medspace s.t.
$

$
\forall x \in A, \quad ||x-a||<\delta \implies ||f(x) - f(a)||< \epsilon
$

Any ideas?
Notice that

$|f(x)-f(a)|=|x^2-a^2|=|x-a||x+a|\le |x+a|\delta$

How can we control the factor $|x+a|$ ? can you think of a good upperbound?

3. What do you mean by control? And where did the less than or equal to |x+a|delta come from? Sorry, I'm really having trouble with all of this. Switched from Social Sciences into Mathematics and Economics... the questions are a lot different than what I'm used to.

4. I'm guessing |x-a| must be less than or equal to delta?

5. You shouldn't be "guessing" that, it is the definition of "limit"- besides it makes no sense until you have said what delta is! You want to have $|x^2- a^2|= |x+a||x- a|< \epsilon$ and you want to be able to guarentee that by choosing some $\delta$ so that "if $|x- a|< \delta$ then [tex]|x^2- a^2|< \epsilon[tex]".

As TheEmptySet told you, the crucial part is that |x+a|. You can always rewrite $|x-a||x+a|< \epsilon$ as $|x-a|< \frac{\epsilon}{|x+a|}$ and the only reason why you could not choose $\delta$ to be $\frac{\epsilon}{|x+a|}$ is that |x+a| is a function of x, not a constant.

So can you find some number, A, such that $\frac{\epsilon}{A}< \frac{\epsilon}{|x+a|}$? If so, then choosing $\delta= \frac{\epsilon}{A}$ will work:
If $|x- a|< \delta= \frac{\epsilon}{A}< \frac{\epsilon}{|x+a|}$, then, working backwards, we would have $|x- a||x+ a|= |x^2- a^2|< \epsilon$.

And how do we find that A? We want $\frac{\epsilon}{A}< \frac{\epsilon}{|x+ a|}$. Assuming, for the moment, that A is positive, that is the same as $A> |x+ a|$. Well, of course, there is no A for which that is true for all x but we are talking about x being close to a since this is a limit as x goes to a. Lets require that the distance from x to a be no greater than a/2: |x- a|< a/2. (I am assuming that a> 0 here. I will leave the cases a< 0 and a= 0 to you.) That is the same as -a/2< x- a< a/2 so, adding 2a to each part, 2a- a/2= 3a/2< x+ a< 5a/2. If we choose A= 5a/2, that will be true and so we will have $|x- a|< \frac{\epsilon}{A}= \frac{2\epsilon}{5a}$.

Normally, we are interested in the case that $\epsilon$ is very small number but there is no requirement that it be small. If we were just to choose $\delta= \frac{2\epsilon}{5a}$ then our requirement that |x-a|< a/2 might not be true. So what we do is choose $\delta$ to be the smaller of $\frac{2\epsilon}{5a}$ and $\frac{a}{2}$.

Again, this is for a> 0. You do the cases a< 0 and a= 0.