Calculus - Definition of the Limit - help Watch

username970964
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#1
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So the question is:

Use the definition of the limit to prove:
lim c = c
x→a

I've written, Given ε>0, δ=ε/0. If 0< 0< δ then IcI-c = Ic-cI = 0>0δ = 0(ε/0)=0

But it's supposed to =ε so what did I do wrong?

Thanks!
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Smaug123
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(Original post by Airess3)
So the question is:

Use the definition of the limit to prove:
lim c = c
x→a

I've written, Given ε>0, δ=ε/0. If 0< 0< δ then IcI-c = Ic-cI = 0>0δ = 0(ε/0)=0

But it's supposed to =ε so what did I do wrong?

Thanks!
You seem to have divided epsilon by zero? Also you have the line "if 0 &lt; 0 &lt; \delta", a condition which is never satisfied because 0 \geq 0.

Can you state the definition of "limit" for me first?

By the way, the pipe character | is much more comprehensible than I.
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username970964
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#3
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(Original post by Smaug123)
You seem to have divided epsilon by zero? Also you have the line "if 0 &lt; 0 &lt; \delta", a condition which is never satisfied because 0 \geq 0.

Can you state the definition of "limit" for me first?

By the way, the pipe character | is much more comprehensible than I.
Let f be a function defined by an open interval, containing 'a', except possibly not at 'a' itself. We say, the "limit as x approaches a is L", and write: lim(f(x))=L.
x→a
If for every ε greater than 0, there is delta greater than 0, such that if 0 is lesser than |x-a| lesser than delta then |f(x) -L| lesser than epsilon.
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Hasufel
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from your last bit, this is satisfied for every x , since:

|f (x)-L| = |c-c|=0&lt; \epsilon

take \delta= \epsilon
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username970964
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(Original post by Hasufel)
from your last bit, this is satisfied for every x , since:

|f (x)-L| = |c-c|=0&lt; \epsilon

take \delta= \epsilon
I understand it now, thanks!
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