Explain the delta-epsilon defintion of a limit.?

In summary: On the x,y domain graph, draw a line from x=0 to y=0. This is the domain graph. On the 1-D limit graph, draw a line from x=0 to the limit. This is the limit graph.If there is no region where the above exists, then the limit doesn't exist.
  • #36
why are you getting so annoyed with me, I am not doing this for homework. The reason I asked all these questioned was so that i could actually understand calculus. I asked that specific question because i thought it was an interesting scenario. I tried solving but could not, that why i asked you guys. Sorry if I offended anybody.
 
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  • #37
Mandelbroth said:
I think he's asking us to do an example so he can understand.

Thank you for writing that. I proposed this example to understand the concept about limits
 
  • #38
Mandelbroth said:
##\delta\in(0,\sqrt{2}-1)##.

What does this statement mean. Does that mean 0<delta<sqrt 2 - 1. If so if E>1 , how can delta be more than one number.
 
  • #39
global said:
What does this statement mean. Does that mean 0<delta<sqrt 2 - 1. If so if E>1 , how can delta be more than one number.
"There exists" does not mean "there exists one and only one."
 
  • #40
Why didn't you taken into account the interval (-√2,0) ? After all, in the interval (-√2,0), |f(x)-L|<E=1. Shouldn't the delta you choose take into account the intervals (-√2,0) and (0, √2). The delta you gave only applies for the x in (0, √2).
 
  • #41
global said:
Why didn't you taken into account the interval (-√2,0) ? After all, in the interval (-√2,0), |f(x)-L|<E=1. Shouldn't the delta you choose take into account the intervals (-√2,0) and (0, √2). The delta you gave only applies for the x in (0, √2).
If we extend ##x## into that interval, we have a bigger ##\delta##. We have to also remember that as ##\delta## gets bigger, so does the range of values for ##x##.

I originally didn't give much thought to this and, for some reason, thought we could just let ##\delta=1##. This doesn't work, of course, because this allows, for example, ##\frac{3}{2}## as a value of ##x##. Then, ##|\frac{9}{4}-1|=\frac{5}{4}\not< 1##. Thus, we don't satisfy our condition for ##\varepsilon##.

I'm off to bed. I'll answer any more questions in the morning.
 
  • #42
Mandelbroth said:
If we extend ##x## into that interval, we have a bigger ##\delta##. We have to also remember that as ##\delta## gets bigger, so does the range of values for ##x##.

I originally didn't give much thought to this and, for some reason, thought we could just let ##\delta=1##. This doesn't work, of course, because this allows, for example, ##\frac{3}{2}## as a value of ##x##. Then, ##|\frac{9}{4}-1|=\frac{5}{4}\not< 1##. Thus, we don't satisfy our condition for ##\varepsilon##.

I'm off to bed. I'll answer any more questions in the morning.

I don't get your previous example.

Why can't you let delta be bigger? If delta was bigger, delta would incorporate both intervals that |f(x)-L|<E=1. Also by picking a bigger delta one can also satify the proof by saying for this delta,
0<|x-c|<delta then |f(x)-L|<E=1. Can you show why picking such a delta would be wrong?
 
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  • #43


Explains it well. Khan Academy is great
 
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  • #44
global said:
I don't get your previous example.

Why can't you let delta be bigger? If delta was bigger, delta would incorporate both intervals that |f(x)-L|<E=1. Also by picking a bigger delta one can also satify the proof by saying for this delta,
0<|x-c|<delta then |f(x)-L|<E=1. Can you show why picking such a delta would be wrong?

Can you show why it would be correct/allowable?
 
  • #45
johnqwertyful said:


Explains it well. Khan Academy is great


i don't see how his examples applys anyway to my example or the question I just asked. Could you please just answer my question straight.
 
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  • #46
verty said:
Can you show why it would be correct/allowable?

I asked you a question first, so can you please answer it. The reason i asked this question is i don't see any reason why it wrong. If somebody could clearly show me why it is wrong, then i could understand why such a delta is not suitable.
 
  • #47
global said:
I don't get your previous example.

Why can't you let delta be bigger? If delta was bigger, delta would incorporate both intervals that |f(x)-L|<E=1. Also by picking a bigger delta one can also satify the proof by saying for this delta,
0<|x-c|<delta then |f(x)-L|<E=1. Can you show why picking such a delta would be wrong?

Suppose we consider ##x\in(-\sqrt{2},0)##. Clearly, we can have the case ##x=-1##. Then, we have ##0<|-1-1|=2<\delta##. If ##\delta>2##, we can also have ##x=2##, because ##0<|2-1|=1<2##. Then, we have ##|2^2-1|=3\not< \varepsilon=1##.
 

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