Question

can anyone help with epsilon-delta defn. of continuity?

Answer 1

i have to show that g(x)=sqrt(x) is continuous at 2, i am half way through but stuck.

Answer 2

So you want to show that for all epsilon > 0 there is a delta such that

\[ | x - 2 | < \delta \ \Rightarrow \ |\sqrt{x} - \sqrt{2}| < \epsilon \]

Where are you so far?

Answer 3

@Stiwan , I'm not sure what you are writing, but please don't write out an entire solution.

Answer 4

First off, I assume that your function is definded in the following way:\[g : \mathbb{R} \rightarrow \mathbb{R}^{+},\; g(x) = |\sqrt{x}| \] Because if you allowed negative function outputs, you would have 2 function values for any x and therefore the function would not be unambigous, hence by definition not a function.

Okay, Let's start with the proof. I'll make a case differentiation:

1) Let t be smaller than 2
\[|2 - t| = 2 - t = \sqrt{2}\cdot \sqrt{2} - \sqrt{t}\cdot \sqrt{t} = (\sqrt{2}+\sqrt{t})\cdot (\sqrt{2}-\sqrt{t})\]sqrt(2) > 1, sqrt(2) + sqrt(t) > 1, hence
\[\sqrt{2} - \sqrt{t} < \frac{2-t}{\sqrt{2}+\sqrt{t}} < 2 - t\] we know (it's easy to prove) that out of t < 2 follows sqrt{t} < sqrt{2}, hence
\[|\sqrt{2}-\sqrt{t}| = \sqrt{2} - \sqrt{t} < |2 - t|\] Now, if we set delta = epsilon, we can see that the definition of continuity is satisfied because
\[|2 - t| < \delta = \epsilon \Rightarrow |\sqrt{2}-\sqrt{t}| < \epsilon\]

Answer 5

@jamesj i have got to \[\frac{ \left| x-2 \right| }{ \left| \sqrt{x}+\sqrt{2} \right| }\]

Answer 6

The proof for t > 2 works in a very similar way

Answer 7

I'll let Stiwan now explain what to do, following the OpenStudy rule: You answer it, you explain it.

Answer 8

yeah, i could explain it, although I think it's already in my answer above

Answer 9

@jamesj stiwans answer although explains it, is not the way i am expected to do it. Based on what you said i guess you were going to do it the way i am looking for. could you help me out?

Answer 10

So what you want to do now is somehow bound the expression \( 1/| \sqrt{x} + \sqrt{2} | \) by a constant so you can say what a delta can be given an epsilon.

What can you say about that term?

Answer 11

you could say that \[\frac{ 1 }{ \left| \sqrt{x} +\sqrt{2}\right| } < \frac{ 1 }{ \left| \sqrt{x}\right| } \]

Answer 12

or can i say its less then 1/2 providing x<=2?

Answer 13

We want to get to a constant. In other words, suppose we can find a number K > 0 such that
\[ | \sqrt{x} - \sqrt{2} | = \frac{ |x - 2| }{|\sqrt{x} + \sqrt{2}|} < K |x - 2| \]
then we could argue that...

For all \( \epsilon > 0 \), choose \( \delta = \epsilon/K \) then
\[ |x-2| < \delta \ \Rightarrow \ \sqrt{x} - \sqrt{2} | < K |x - 2| < K \frac{\epsilon}{K} = \epsilon \]

So can you find such a number K? If so, you're done.

Answer 14

By the way, it doesn't have to be a tight bound K. Any bound K > 0 will do.

Answer 15

i'm still not sure

Answer 16

\[ \sqrt{2} > 1 \Rightarrow \sqrt{x} + \sqrt{2} > 1 \Rightarrow \frac{1}{\sqrt{x} + \sqrt{2}} < 1 \]
Hence \( K = 1 \) will do.

Answer 17

god i overcomplicated that in my head alot

Answer 18

Ok, do you have everything you need now?

Answer 19

yes thanks for the help.

Answer 20

Don't worry about struggling with these. Epsilon-delta proofs are subtle, and it took all of us a while to get used to them.