Question

Ms. Vic's Calculus Class (9:00) Join If Wanted!
Medal's Will Be Given

Answer 1

Session on Limits

Answer 2

Message Me If You Have A Question

Answer 3

Class Starts At 9:00

Answer 4

Ok Welcome To My Calculus Class Many People Have Been Having Trouble With.

Answer 5

If You Have A Question. or know the answer just type it out ok

Answer 6

So, todays lesson is on limits.
You have seen enough limits to be ready for a definition. It is true that we have
survived this far without one, and we could continue. But this seems a reasonable
time to define limits more carefully

Answer 7

The goal is to achieve rigor without rigor mortis.
First you should know that limits of Ay/Ax are by no means the only limits in
mathematics.

Answer 8

Here are five completely different examples. They involve n + a,not
Ax +0:

1.
a, = (n -3)/(n + 3) (for large n, ignore the 3's and find a, + 1)
2.
a, =)a,-, + 4 (start with any a, and always a, +8)
3.
an=probability of living to year n (unfortunately an +0)
4.
a, = fraction of zeros among the first n digits of n (an+h?)
5.
a, = .4, a2 = .49, a, = .493, .... No matter what the remaining decimals are, the
a's converge to a limit. Possibly a, +.493000 . . .,but not likely.

Answer 9

The problem is to say what the limit symbol + really means.
A good starting point is to ask about convergence to zero. When does a sequence
of positive numbers approach zero? What does it mean to write an +O? The numbers
a,, a,, a,, ..., must become "small," but that is too vague. We will propose four
definitions of convergence to zero, and I hope the right one will be clear.

Answer 10

All the numbers a, are below 10- lo. That may be enough for practical purposes,
but it certainly doesn't make the a, approach zero.
2. The sequence is getting closer to zero-each a,, is smaller than the preceding
a,. This test is met by 1.1, 1.01, 1.001, ... which converges to 1 instead of 0.
3. For any small number you think of, at least one of the an's is smaller. That pushes
something toward zero, but not necessarily the whole sequence. The condition would
be satisfied by 1, ),1, f, 1, i,. . . ,which does not approach zero.
4. For any small number you think of, the an's eventually go below that number and
stay below.

Answer 11

Does Anyone Know The True definition for limits out of these 4 choices?

Answer 12

anyone?

Answer 13

yes! that is correct!

Answer 14

I want to repeat that. To test for convergence to zero, start with a small number-
say 10-lo. The an's must go below that number. They may come back up and go
below again-the first million terms make absolutely no difference. Neither do the
next billion, but eventually all terms must go below lo-''. After waiting longer
(possibly a lot longer), all terms drop below The tail end of the sequence
decides everything

Answer 15

@andrejrusa12

Answer 16

Does this sequence \[10^{-3} , 10^{-2}, 10^{-6},10^{-5},10^{-9},10^{-8}\] aproach 0?

Answer 17

anyone?

Answer 18

The answer is Yes, These up and down numbers eventually stay below any

Answer 19

ok does
Does \[10^{-4 }, 10^{-4},10^{-4},10^{-6},10^{-8},10^{-10}\], ... approach zero?

Answer 20

anyone have a answer?

Answer 21

best guess?

Answer 22

No it does not reach 0

Answer 23

This sequence goes below but does not stay below

Answer 24

Any questions???

Answer 25

if there are no questions this class is done i will stay in for a while and answer any questions u might have

Answer 26

this class is now done and im closing the session