Understanding epsilon-delta def of limits

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Homework Help Overview

The discussion revolves around the epsilon-delta definition of limits in calculus, specifically focusing on understanding how to use this definition to disprove a limit, such as showing that \(\lim_{x \to 1} x^2 \neq 2\).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the epsilon-delta definition and how it can be applied to demonstrate that a limit does not exist. Participants explore the implications of varying delta and epsilon values in relation to the limit.

Discussion Status

Some participants provide graphical representations to aid understanding, while others question how to formally structure an epsilon-delta proof based on the initial discussions. There is an ongoing exploration of the conditions under which the limit can be disproven.

Contextual Notes

Participants note the importance of specific delta and epsilon values in the context of the limit, and there is a mention of the need for clarity in the definitions and conditions being discussed.

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Homework Statement


I'm having trouble conceptually understanding the epsilon-delta definition of limits. How do you use it to disprove a limit? For example, how would you use it to show that \lim_{x \to 1} x^2 \neq 2?


Homework Equations





The Attempt at a Solution

 
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imagine you've got the function x plotted in a graph(you can actually draw it in a piece of paper). and you want to show that <br /> \lim_{x \to 1} x \neq 2. First, set an interval ]1-delta,1+delta[ in x axes, see the image of the 1+delta; epsilon will be=|f(1+delta)-f(1)|. (in general, it will be the max{|f(1+delta)-f(1)|,|f(1-delta)-f(1)|}, in this case, the function is similar to the right and left of f(1), so, only one part is needed because both are equal).
Now, you see that if the delta is very big, say 4, then epsilon=4 rigth?! so, (the second part of the condition, the epsilon part) is |2-f(1)|=2-1=1<epsilon: the condition is verified.(2 because you are testing the condition on the point 2 you are asking: "is it really true that the condition is verified to EVERY delta?")

Note however that if you put delta lower, say 0.5, epsilon is 0.5 and |2-f(1)|=|2-1|>epsilon:. Exists a delta that don't verify the condition, that implies that isn't true that for every delta, exists..bla bla bla...so, the limit isn't 2.

Hope this helps solve your problem...
and hope i didn't make any mistake...^_^
 
Last edited:
Perhaps in this way
can Be clear idea

[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/6225e0aea0.jpg[/url][/PLAIN]
 
m_s_a said:
Perhaps in this way
can Be clear idea

[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/6225e0aea0.jpg[/url][/PLAIN]

but how do you put that into an epsilon-delta proof?
 
[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/df5040ccd8.jpg[/url][/PLAIN]
[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/4c7f5603e7.jpg[/url][/PLAIN]

No requirement that the value of f(x0)

The smaller values for delta & epsilon Leads to ...what?
 

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