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# analysis watch

1. A bit confused about using sequences for limits of functions.

I understand the equivalence of f(x) tending to q as x tends to p and f(p(n)) twending to q as p(n) tends to p.

However, when proving things like

(f+g)(x) tends to a + b as x tends to b if f tends to a and g tends to b

The book says these imediately follow from the earlier results about sequences. I don't see how.

ntuitively I understand that f(p(n)) and g(q(n)) say are both sequences (in the loose sense) tending to a and b so by AOL this is ok, but not quite sure how to really formalise it.

I don't have a problem individually proving the above - it is just the claim that all results follow immediately I find a bit confusing.
2. (Original post by DeanK22)
A bit confused about using sequences for limits of functions.

I understand the equivalence of f(x) tending to q as x tends to p and f(p(n)) twending to q as p(n) tends to p.
I'm not clear what you're trying to say here, but with at least one plausible interpretation, what you are saying is false.

Consider f(x) = sin(pi/x). Then f(x) doesn't tend to 0, but if we take p(n) = 1/n, then p(n)->0 and f(p(n))->0.
3. (Original post by DFranklin)
I'm not clear what you're trying to say here, but with at least one plausible interpretation, what you are saying is false.

Consider f(x) = sin(pi/x). Then f(x) doesn't tend to 0, but if we take p(n) = 1/n, then p(n)->0 and f(p(n))->0.
I left out the for all sequences tending to p but npt equal to p [as it had little to do with the question].

I guess I should really out everything in;

f(x) tends to q as x tends to p iff lim n tends to infinty f(p(n)) for every sequence p(n) in the domain of f, p(n) =/= p, and p(n) tends to p. I understand this bit.

suppose f(x) tends to a and g(x) tends to b as x tends to p say. If x tends to p, then;

(f+g)(x) tends to a + b

(fg)(x) tends to ab

(f/g)(x) tends to a/b with b =/= 0

The book then says that these immediately follow from the ana;afous results for sequences - like if a(n) tends to a and b(n) tends to b then (a(n) + b(n)) tends to a + b, etc.

I don't see how they imediately follow.
4. Well, to show f+g tends to a + b, pick an arbitrary sequence p(n)->p (with p(n) never equalling p).
Then you know f(p(n))->a, and g(p(n))->b. But (f+g)(p(n)) = f(p(n))+g(p(n)) -> a+b by the analagous result for sequences.
Since this holds for all sequences p(n), we're done.
5. (Original post by DFranklin)
Well, to show f+g tends to a + b, pick an arbitrary sequence p(n)->p (with p(n) never equalling p).
Then you know f(p(n))->a, and g(p(n))->b. But (f+g)(p(n)) = f(p(n))+g(p(n)) -> a+b by the analagous result for sequences.
Since this holds for all sequences p(n), we're done.

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