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# Simple continuity question watch

1. Hello,

How do I prove that is continuous at x=1 by using the epsilon-delta method?

Working so far is:

(since we'll be close to x=1 so x+1 > 0)
and I don't know where to go from there.

If someone could give me a proof I'd appreciate it. I have no examples so seeing one would help. Thanks.
2. (1+h)^2 - 1 = 2h + h^2.
3. (Original post by Usar)
Hello,

How do I prove that is continuous at x=1 by using the epsilon-delta method?

Working so far is:

(since we'll be close to x=1 so x+1 > 0)
and I don't know where to go from there.

If someone could give me a proof I'd appreciate it. I have no examples so seeing one would help. Thanks.
I think you forgot the brackets around the x+1. Should be Now you need to find a delta such that < delta implies < epsilon
4. Ok, is this right: delta = min{epsilon/3, 1}?

Working is:
for some K.

To get K:
Assume , then , so K = 3.

Therefore .
5. Value for delta is right, but what you've got isn't really a proof that x^2 is cts. You've never even said what relevant "epsilon" has to what you're doing.

Ideally, a proof should start something like: "Take . Pick . Then ..." and conclude "... and so ".
6. Thanks. I shouldn't have said proof, just the method of getting a value for delta.
7. OK another question:
if for and for , how can I prove that it's discontinuous at x=2 using the epsilon-delta definition and proof by contradiction? I can't write |f(x) - f(2)| and substitute something in for f(x) cause f(x) is different either side of x=2.

Am I right in saying that I consider two cases: when x>2 and when x < 2 and find that they somehow lead to contradictions? (I've done this but am not sure if this is a valid method).
8. You only need to consider 1 case.

Big Spoiler
For any , what is a lower bound for ?
9. Oh so it leads to epsilon > (say) 3. I'm not sure how to spot "tricks" like that -- hope it comes with more familiarity. Thanks for the help.
10. Btw, why did you use delta/2 in that expression (as opposed to delta)?
11. Just in case you got confused because the normal formulation involves considering for , and isn't actually less than delta.

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