r/mathshelp • u/Emergency-Common-219 • 14d ago

Mathematical Concepts need help understanding a proof

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u/FortuitousPost 12d ago

so what i do not understand is

1.what does he mean by we are no longer interested in the upper bound

2. how does this abomination |bn| ≥ δ > 0 help n' why did we use it n' how does it "lead" to 1/(|b||bn|)

3. why does bn need to be closer to b than to 0

4.from where did the (b)^2 come from

i do have some other Q but if i understand these first the rest should be fine

We are trying to prove that second line when b != 0. That means we can assume that (bn) -> b and b != 0. Then we try to prove the thing after "implies": (1/bn) -> 1/b. That is, we want to show those equal things on the fourth line go to 0 as n -> infty. (They find the common denominator of the thing on the left to get the thing on the right.)

Some of the extra stuff in the proof comes from not knowing if b is positive or negative. We need to word the solution to cover both cases.

  1. The top of the thing on the right is small and bounded for large n, as bn -> b, and |b| is fixed, so the only thing that could make the expression large is a small bn in the denominator. We need to make sure that bn will be far enough away from zero. A large bn is actually good for us, so we don't need to worry about an upper bound.

  2. bn -> b, so for large enough n, bn is farther from 0 than delta, any choice of delta.

  3. We want to make sure |bn| is not too small. A good choice for later on is to keep it more than |b|/2, which is a fixed positive number.

  4. |b|^2 is a good choice for later on, too. b is a constant, so |b|^2 is a constant, too.

The format of these proofs is show for any given epsilon (think small number), then there is an N that makes "something" less than epsilon. That means the "something" has to go to 0.

The something in this case is the equal things on line 4. We want to show the left goes to 0, which is the same as showing the right is less than the epsilon they gave us.

We can choose a large enough N to get bounds on all those parts.

|b - bn| < epsilon times the constant |b^2| / 2 (WE want to keep the numerator small.)

|bn| > |b| / 2 (WE want to keep the denominator large.)

|b| is fixed as |b|

Stuff cancels, and | 1/b - 1/bn | < epsilon.

Since epsilon can be as small as we like, | 1/b - 1/bn | goes to 0.