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A question about the existence of a limit...

by Artusartos
Tags: existence, limit
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Artusartos
#1
Dec16-12, 01:25 PM
P: 247
1. The problem statement, all variables and given/known data

For question 20.18 in this link:

http://people.ischool.berkeley.edu/~...04hw7sum06.pdf

I understand how they got the value 3/2 for the limit, but I don't see where they proved the existence of that limit...because the question is not just asking us to determine the value, right? It's also asking us to prove that the limit exists...

2. Relevant equations



3. The attempt at a solution
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LCKurtz
#2
Dec16-12, 04:10 PM
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Did you miss the part about "using the limit laws" near the end? For example, the limit of a quotient is the quotient of the limits if the limit of the denominator isn't zero, etc...
Artusartos
#3
Dec17-12, 05:41 AM
P: 247
Quote Quote by LCKurtz View Post
Did you miss the part about "using the limit laws" near the end? For example, the limit of a quotient is the quotient of the limits if the limit of the denominator isn't zero, etc...
Thanks...

This is what I understood, but I'm not sure if it is correct...

Since the function they are giving us is continuous, we know that for any sequence x_n in the domain that converges to 0, [tex]f(x_n) \rightarrow f(0)[/tex]. Is that the proof?

lurflurf
#4
Dec17-12, 06:05 AM
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A question about the existence of a limit...

Yes, in the initial form the function is undefined when x=0. The new form is a continuous function so the value when x=0 is the same as the limit, which is the same as the original limit. It is also of interest to notice that the limit is a Newton quotient so its value is recognized as 3sqrt'(1).
Artusartos
#5
Dec17-12, 06:06 AM
P: 247
Quote Quote by lurflurf View Post
Yes, in the initial form the function is undefined when x=0. The new form is a continuous function so the value when x=0 is the same as the limit, which is the same as the original limit. It is also of interest to notice that the limit is a Newton quotient so its value is recognized as 3sqrt'(1).
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HallsofIvy
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Dec17-12, 06:34 AM
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Quote Quote by Artusartos View Post
Thanks...

This is what I understood, but I'm not sure if it is correct...

Since the function they are giving us is continuous, we know that for any sequence x_n in the domain that converges to 0, [tex]f(x_n) \rightarrow f(0)[/tex]. Is that the proof?
Actually, the function they give us is NOT continuous because it is not defined at x= 0. Of course, IF the limit exists and we redefine the function to have that value at x= 0 then it is continuous at x= 0. The "limit theorem" used here is "if the sequence [itex]\lim_{n\to\infty}f(x_n)= L[/itex] for every sequence [itex]x_n[/itex] that converges to [itex]x_0[/itex], then [itex]\lim_{x\to x_0} f(x)[/itex] exists and is equal to [itex]L[/itex]".
lurflurf
#7
Dec17-12, 09:51 PM
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I think the explanation in the link is very poor.
It is an often used method that is confusing
asked to find
lim f
we find a continuous function g so that f=g on the domain of f then
lim f=lim g=g(a)
where a is not in the domain of f

When poorly explained, limits seem like a function evaluation with some extra steps.


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