# Prove that a function is continuous on an interval

1. Dec 4, 2012

1. The problem statement, all variables and given/known data
I have to prove that $\sqrt{x}$ is continuous on the interval [1,$\infty$).

2. The attempt at a solution
Throughout the school semester I believed that to show that a function is continuous everywhere all I need to do was show that $\lim\limits_{h\rightarrow 0}f(x+h)-f(x)=0$ and I never thought much about it. It never really came up as a problem on any of the homeworks or exams so I never had much problem with it. I am now doing review problems for the final and I realized that f(x)=1/x is a clear counterexample to what I stated above. The limit as h approaches 0 of f(x+h)-f(x) is 0 but 1/x is not continuous everywhere. Since this is the case, how do I show that a function is continuous everywhere within an interval?

Last edited: Dec 4, 2012
2. Dec 4, 2012

### Zondrina

To show it is continuous, you could go back to your first principles and use an epsilon delta argument.

Also given the interval, you can use the definition of the derivative, it's up to you.

Last edited: Dec 4, 2012
3. Dec 4, 2012

Say I didn't know that $\sqrt{x}$ was differentiable. How would I use epsilon delta to prove that it was differentiable everywhere in that interval? The only epsilon delta proof for continuity that I know only proves that it is continuous at one point.

I read somewhere that I could choose my delta to not only depend on epsilon but also to depend on x. Is this correct? If this is, would it be right if I said

Choose $\delta=\epsilon(|\sqrt{x}+\sqrt{a}|)$. Then $|x-a|<\delta\Rightarrow|x-a|<\epsilon(|\sqrt{x}+\sqrt{a}|)\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}=|\sqrt{x}-\sqrt{a}|<\epsilon$.

Last edited: Dec 4, 2012
4. Dec 4, 2012

### LCKurtz

No, that isn't right. When you are discussing continuity at the point $x=a$, the $\delta$ will normally depend on $\epsilon$ and $a$. If it doesn't depend on $a$ you have uniform continuity. So in your example, $\delta$ can depend on the $a$ but not $x$. But you have$$|\sqrt x -\sqrt a| = \frac {|x-a|}{|\sqrt x + \sqrt a|}$$Given that you have $x\ge 1$ can you find an overestimate of the right side that doesn't involve $x$ in the denominator, then choose $\delta$ to make it work?

5. Dec 4, 2012

I'm looking at $|x-a|<\delta$ and I'm thinking that the way I would find a bound for for $|\sqrt{x}|$ is to first make $\delta_{1}=1$ so that I have $|x-a|<1\Rightarrow -1<x-a<1\Rightarrow a-1<x\Rightarrow \sqrt{a-1}+\sqrt{a}<\sqrt{x}+\sqrt{a}\Rightarrow \frac{1}{\sqrt{x}+\sqrt{a}}<\frac{1}{\sqrt{a-1}+\sqrt{a}}$ and since these numbers are positive $\frac{1}{|\sqrt{x}+\sqrt{a}|}<\frac{1}{|\sqrt{a-1}+\sqrt{a}|}$
What if I then chose $\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))$?

Then my epsilon delta argument would be:
Choose $\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))$. Then
$|x-a|<\delta\Rightarrow |x-a|<\epsilon(|\sqrt{a-1}+\sqrt{a}|)\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon$.

Also while we are talking about uniform continuity, if delta doesnt depend on $a$ for uniform continuity then what does delta depend on? Does it depend on x and y in that case? Like say I wanted to prove that $\sqrt{x}$ was uniformly continuous on [1,$\infty$) and I have $|x-y|<\delta$, can I then choose a $\delta$ that depends on $x$ and $y$ to show that $|\sqrt{x}-\sqrt{y}|<\epsilon$?

6. Dec 4, 2012

### LCKurtz

That just makes my head hurt. Why don't you just use my suggestion, since $x\ge 1$:$$\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}$$and use that to figure out $\delta$?
No. In that case $\delta$ depends only on $\epsilon$, not $x$ or $y$. Look at my suggestion above and see if you can also overestimate getting rid of the $a$ in the denominator to get a $\delta$ that depends only on $\epsilon$.
No.

Last edited: Dec 4, 2012
7. Dec 4, 2012

OOOOOOOOOOOhhhhhhhh. So since $\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}$ we can choose $\delta=\epsilon|1+\sqrt{a}|$ and then get

$|x-a|<\epsilon|1+\sqrt{a}|\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<\epsilon|1+\sqrt{a}|\frac{1}{|1+\sqrt{a}|}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon$. Third times a charm?

Hmm. Thanks. This helps to clarify a lot! I will do this part on my own time. As long as I get the top part, it seems like this part shouldn't be too much harder to do.

8. Dec 4, 2012

### LCKurtz

Good, now you are getting somewhere. Your $\delta$ depends on $\epsilon$ and $a$, so you have continuity for each $a$. Now, can you find a $\delta$ that doesn't depend on $a$ either? If it depended only on $\epsilon$ you would have shown uniform continuity.

9. Dec 5, 2012

As per your advice I noted that the smallest value of $a$ would also be 1 on the interval so choosing $\delta=2\epsilon$ I would have:

$|x-a|<\delta\Rightarrow |x-a|<2\epsilon\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<2\epsilon\frac{1}{2}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon$.

Thank you! This has clarified a lot for me!

10. Dec 5, 2012

### LCKurtz

OK, you have all the pieces figured out, but you could present it a little more smoothly. Start with
$$|\sqrt x - \sqrt a| = \frac{|x-a|}{\sqrt x + \sqrt a}\le \frac{|x-a|}{2}$$since $x\ge 1$ and $a\ge 1$. (This is really the "exploratory argument" where you are figuring out what $\delta$ might work.)

Suppose $\epsilon>0$. Let $\delta=2\epsilon$. Then if $|x-a|<\delta = 2\epsilon$ we have$$|\sqrt x - \sqrt a| <\frac{2\epsilon} 2=\epsilon$$