Correct this improper definition of a limitby Easy_as_Pi Tags: correct, definition, improper, limit 

#1
Feb312, 06:49 AM

P: 31

1. The problem statement, all variables and given/known data
Eddy wrote on his midterm exam that the definition of the limit is the following: The sequence {an} converges to the real number L if there exists an N ∈ Natural numbers so that for every [itex]\epsilon[/itex] > 0 we have an − L < [itex]\epsilon[/itex] for all n > N. Show Eddy why he is wrong by demonstrating that if this were the definition of the limit then it would not be true that lim n→∞ 1/n = 0. (Hint: What does it mean if a − b < [itex]\epsilon[/itex] for every [itex]\epsilon[/itex] > 0?) 2. Relevant equations ab <ε means that ab < ε from the reverse triangle inequality 3. The attempt at a solution I know it has to do with the fact that the actual definition of a limit has "for every ε > 0, there exists an N [itex]\in[/itex] Natural numbers S.T. ...." so, Eddy reversed that part of the definition. I just haven't been able to quite see the difference of the two. A little push in the right direction would be greatly appreciated. I like figuring these out on my own, so no full on answers, please. 



#2
Feb312, 08:25 AM

Mentor
P: 16,703

So there exists a certain N.
Let [itex]\varepsilon = 1/N[/tex] and try to prove that 1/n does not converge to 0 with this definition. 



#3
Feb312, 08:51 AM

P: 31

I'm not quite sure I follow. I ended up answering it this way: Eddy's definition implies there is a single natural number, N, such that for all n>N 1/n< every epsilon greater than zero. Which is not true. For every epsilon you give me, I can find an N such that 1/n is less than that epsilon for all n>N, but if you pick a newer, smaller epsilon, my N has to be larger, and since the natural numbers are unbounded, we can do this forever. But, it's a different N for each new epsilon, not one single N like eddy implied. Does that make sense?




#4
Feb312, 11:58 AM

Emeritus
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P: 11,536

Correct this improper definition of a limit
That's right.



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