Is f(x) = 1/√x Continuous as x Approaches 4?

[gimpy]

Ok well i did problems like this before but now I am 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}}?

 

[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.

 

[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| &lt; \left| \frac{2 - \sqrt{x}}{2}\right| &lt; \epsilon

-\epsilon \;&lt; \; \frac{2 - \sqrt{x}}{2} \;&lt;\; \epsilon

-2\epsilon \;&lt; \; 2 - \sqrt{x} \;&lt;\; 2\epsilon

-2\epsilon -2 \;&lt;\; - \sqrt{x} \;&lt;\; 2\epsilon -2

4\epsilon^2 +8\epsilon +4 \;&gt;\; x \;&gt;\; 4 - 8\epsilon + 4\epsilon^2

4\epsilon^2 + 8\epsilon \;&gt;\; x-4\; &gt;\; 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 \;&lt;\; 4\epsilon^2 + 8\epsilon \:\text{satisfies the proof?}

I really don't feel good about this conclusion and I'm getting confused over the directions of the inequalities, so someone please help.

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

 

[matt grime]

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

that any help?

 

[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| &lt; \epsilon

whenever
\left|2-\sqrt{x}\right| \left|2+\sqrt{x}\right| = \left|4-x\right| &lt; \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.

 

[Hurkyl]

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

 

[gnome]

I should probably stick to my own homework.

That was all wrong. Still working on it.

 

[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 &lt; |x-4| &lt; \delta
then
0 &lt; |\sqrt{x}-2|\times(\sqrt{x}+2) &lt; 3\epsilon
Now divide by 3
0 &lt; |\sqrt{x}-2|\times \frac{(\sqrt{x}+2)}{3} &lt; \epsilon
And since √x is definitely bigger than 1, we know that
\frac{(\sqrt{x}+2)}{3} &gt; 1
so
\left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| = \left| \frac{2 - \sqrt{x}}{2\sqrt{x}}\right| &lt; \left|{2 - \sqrt{x}}\right| &lt; |\sqrt{x} - 2|\times \frac{(\sqrt{x}+2)}{3} &lt; \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.)

 

[Hurkyl]

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