Can someone explain mimits and how it relates to the formal limit definition?

mfk_1868
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i didnt understand delta definiton anyone can explain?
 
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Post it, and pinpoint what you don't understand.
 
i mean i didnt know it exactly but only thing i know it uses triangle inequality
 
You have a book, right?
In there is a definition of the limit.
Post that, and say what you didn't understand.
 
i have a book and at there there is epsilon definition. but teacher told a different method named delta using triangle inequality and i am asking this.i don't have any other info.
 
We say that a sequence of real numbers (s_n) converges to s (a real number) if given an \epsilon > 0 there exists a natural number N such that n \geq N implies that |s_n - s| < \epsilon. What don't you understand about this definition?
 
where did you used the triangle inequality i know the epsilon definiton and understood it but there is a delta definition.
 
Do you mean the limit of a function? If f is a real-valued function we say that \lim_{x \to a} f(x) = L, if given \epsilon > 0 there exists a \delta > 0 such that |x - a| < \delta imply that |f(x) - L| < \epsilon.

Is that the definition you don't understand? Are talking about metric spaces?

?
 
I've never heard of an epsilon definition of a limit, nor have I heard of a delta definition of a limit. I have, however, heard of the epsilon-delta definition of a limit, which is the standard definition. This definition does not make use of the triangle equality.
 
  • #10
cogito² said:
Do you mean the limit of a function? If f is a real-valued function we say that \lim_{x \to a} f(x) = L, if given \epsilon > 0 there exists a \delta > 0 such that |x - a| < \delta imply that |f(x) - L| < \epsilon.

Is that the definition you don't understand? Are talking about metric spaces?

?
yes can you show me a proving example. for example lim x->3 x^2/5 how to proof this.
 
  • #11
What do you mean by "proving example"??
This is from the DEFINITION of a limit ; not some alleged proposition of properties limits may have!
Do you understand that difference?
 
  • #12
for example there is a question that it gives you a limit equation then says prove this using delta method. I meaned this.
 
  • #13
What Is A Definition?
 
  • #14
So, you've some function and want to show f(x) tends to some given value L, say, as x tends to w? (using the full epsilon delta argument)

well, sorry, we can't do that without knowing what the question is. there is no method that works always, it depends on the question, though they often have the same underlying idea.
 
  • #15
it was a quiz question and teacher posted it on internet link is here http://www.fen.bilkent.edu.tr/~otekman/calc/quiz101.pdf first question number is 6 at the top of page.(look at the solution he uses triangle inequality)
 
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  • #16
he's perfectly entitled to use the triangle inequality. it's very useful.

i've not actually looked at the pdf but i can give you a proof of a property limits using the triangle inequality.

Suppose a(n) is a sequence tending to a, and b(n) tends to b, then a(n)+b(n) tends to a+b

proof:

consider

|a(n)+b(n)-a-b| = | a(n)-a + b(n)-b| <= |a(n)-a| + |b(n)-b| ***

by the triangle inequality.

Given e (epsilon) let N be chosen such that |a(n)-a| < e/2 and |b(n)-b| <e/2 for all n>N, which we may do since a(n) tends to a and b(n) tends to b.

then *** is less than or equal to e/2+e/2 =e, as we were required to show.

is that what you were thinking of?
 
  • #17
But look at problem 6 closely. It doesn't deal with "limits" at all.

The professor has defined a thing he calls "mimits" by reversing the roles of ε and δ

"The real number M is "mimit" of f at x0 if for every δ>0 there exist ε> 0 such that if 0<|x- x0|< δ then |f(x)- M|< ε"

Since in this case, you are given[\b] δ and asked to show that ε exists, finding ε is really just a matter of replacing x with x0+/- δ and calculating ε.

MFK 1868: We were having trouble understanding this because this is NOT a standard notation. Your professor was giving you a problem that requires you to look at the ideas from another point of view.
 

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