Understanding epsilon-delta def of limits

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SUMMARY

The discussion focuses on the epsilon-delta definition of limits, specifically addressing how to disprove a limit using this concept. The example provided illustrates that for the limit \(\lim_{x \to 1} x^2\), the assertion that it equals 2 is false. By analyzing the intervals defined by delta and calculating epsilon, it is shown that there exists a delta for which the condition does not hold, confirming that the limit is not equal to 2. Visual aids were referenced to enhance understanding of the concept.

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  • Understanding of limits in calculus
  • Familiarity with the epsilon-delta definition of limits
  • Basic graphing skills for functions
  • Knowledge of inequalities and their manipulation
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  • Study the formal epsilon-delta proof structure for limits
  • Practice disproving limits using various functions
  • Explore graphical representations of limits and continuity
  • Learn about the implications of limits in calculus, such as continuity and differentiability
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Students studying calculus, mathematics educators, and anyone seeking to deepen their understanding of limit concepts and epsilon-delta proofs.

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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?


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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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