# Delta - epsilon proof

1. Apr 21, 2004

### gimpy

Ok well i did problems like this before but now im having trouble with this one for some reason.

Let $$f(x) = \frac{1}{\sqrt{x}}$$. Give a $$\delta$$ - $$\epsilon$$ proof that $$f(x)$$ has a limit as $$x \rightarrow 4$$.

So the defn of a limit is
$$\forall \epsilon > 0 \exists \delta > 0$$ such that whenever $$0 < |x - 4| < \delta$$ then $$|f(x) - l| < \epsilon$$
Assuming the limit we are trying to prove is $$l$$.

So i know i somehow have to turn $$|f(x) - l| < \epsilon$$ into something with $$x - 4 < ...$$ and that will prove that the limit exists. Am i correct? Am i on the right track? can i assume that $$l = \frac{1}{2}$$ since $$\frac{1}{\sqrt{4}}$$?

2. Apr 22, 2004

### arildno

1. There exists no general technique to find a limit.
What the d,e-business is about in this case (I think), is to investigate whether an arbitrarily chosen number l is, in fact, the limit value.
So, it is perfectly legitimate to investigate whether the number 1/2 is the limit value or not.

3. Apr 22, 2004

### gnome

Well, I tried to do this, but I sort of got stuck at the end, so if anyone can check & let us know where I've gone wrong and how to complete it correctly, I'd appreciate it.

I started with the "guess" that the limit is 1/2, and therefore we need to show that
$$\left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| < \epsilon \: \text{whenever}\: 0 < |x-4| < \delta$$
If I combine the fractions and require δ < 1, then 3 < x < 5 and √x > 1, so
$$\left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| = \left| \frac{2 - \sqrt{x}}{2\sqrt(x)}\right| < \left| \frac{2 - \sqrt{x}}{2}\right| < \epsilon$$

$$-\epsilon \;< \; \frac{2 - \sqrt{x}}{2} \;<\; \epsilon$$

$$-2\epsilon \;< \; 2 - \sqrt{x} \;<\; 2\epsilon$$

$$-2\epsilon -2 \;<\; - \sqrt{x} \;<\; 2\epsilon -2$$

$$4\epsilon^2 +8\epsilon +4 \;>\; x \;>\; 4 - 8\epsilon + 4\epsilon^2$$

$$4\epsilon^2 + 8\epsilon \;>\; x-4\; >\; 4\epsilon^2 - 8\epsilon$$

$$4\epsilon^2 - 8\epsilon \text{ is no good, since for small values of }\epsilon \text { that will be negative,}$$
$$\text{ so does this mean that for any value of }\epsilon \text{ , choosing}$$
$$\delta \;<\; 4\epsilon^2 + 8\epsilon \:\text{satisfies the proof?}$$

Edit: well, the more I look at it, the more convinced I am that this is wrong, so if anyone can help, please do.

Last edited: Apr 22, 2004
4. Apr 22, 2004

### matt grime

|2- sqrt(x)| < (2+sqrt(x))(2+sqrt(x)) = |4-x| < d

that any help?

5. Apr 22, 2004

### gnome

No, I still don't see it.
I assume you meant to write |2- sqrt(x)| < (2-sqrt(x))(2+sqrt(x)) = |4-x| < d

But we need
$$\left| \frac{2 - \sqrt{x}}{2\sqrt{x}}\right| < \epsilon$$

whenever
$$\left|2-\sqrt{x}\right| \left|2+\sqrt{x}\right| = \left|4-x\right| < \delta$$

and I'm still not seeing how to relate ε to δ

Presumably we need
$$\delta = ({2+\sqrt{x})\epsilon}$$
but I don't know how to accomplish that.

6. Apr 22, 2004

### Hurkyl

Staff Emeritus
I bet you know something smaller than (2 + &radic;x)&epsilon; that doesn't depend on x!

Last edited: Apr 22, 2004
7. Apr 22, 2004

### gnome

I should probably stick to my own homework.

That was all wrong. Still working on it.

Last edited: Apr 22, 2004
8. Apr 22, 2004

### gnome

OK, now I let δ = 3ε, and keep the restriction 3<x<5 which is reasonable since we're interested in values of x close to 4.

So if
$$0 < |x-4| < \delta$$
then
$$0 < |\sqrt{x}-2|\times(\sqrt{x}+2) < 3\epsilon$$
Now divide by 3
$$0 < |\sqrt{x}-2|\times \frac{(\sqrt{x}+2)}{3} < \epsilon$$
And since √x is definitely bigger than 1, we know that
$$\frac{(\sqrt{x}+2)}{3} > 1$$
so
$$\left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| = \left| \frac{2 - \sqrt{x}}{2\sqrt{x}}\right| < \left|{2 - \sqrt{x}}\right| < |\sqrt{x} - 2|\times \frac{(\sqrt{x}+2)}{3} < \epsilon$$

Is that it? Did I spend all this time just to find that?

Now it looks right to me. Is anything missing?

Is there some obvious clue that I should have seen that would have told me to try δ = 3ε?

Thanks for your help, Matt & Hurkyl. Please let me know if this still needs more work.

(Edited to correct typos.)

Last edited: Apr 23, 2004
9. Apr 23, 2004

### Hurkyl

Staff Emeritus
Aside from the typo, looks right.

Incidentally, if you wanted to sew up the loose end, the trick to saying:

"OK, now I let &delta; = 3&epsilon;, and keep the restriction 3<x<5 which is reasonable since we're interested in values of x close to 4."

is "let &delta; be the smaller of 3&epsilon; and 1"

So that way you have both |x-4| < 3&epsilon; and |x-4| < 1