studies support thread

[kryss]

1. work out how many moles of lead that is and then how much the equivalent number of moles of lead dioxide weighs.

2a) similar but you need to use coulombs to work out how many molecules are produced. and factor in that this is only 75% of the actual figure.

[kryss]

http://homepage.smc.edu/kennedy_john/LimitProofTechnique.pdf
though it does sound like a textbook.

[Hikaru]

proving limits hmmm
well, there's the cool way where you use already known limits to proof other limits, and there's the way that goes by the definition, that is, show that there is a delta that brings x close enough to A so that f(x) will be close enough to L.
I haven't done proving by definition for a long time, and I can't quite remember the trick, but it should be pretty simple and straight forward.
for example:

lim of 2x where x->0

0<|x-0|<d ---> |2x-0|<e

what needs to happen: |2x|<e
and so, it is sufficient to have |x|<e/2, and so we choose a d <= e/2,
and then you can clearly see the condition is true:

0< | x - 0 | < d < e/2 -----> 2 | x - 0 | < e ---> | 2x - 0 | < e

I do remember that for all polynomial function f(x), lim f(x) where x->a will be f(a) because (f(x) - a) is divided by (x - a)
(because of polynomial division)

[Hikaru]

well, d is supposed to be delta and e is supposed to be epsilon.

erm... lemmy remember...

you need to alter f(x) - L so it will look like something you can work with. I think.
1/x is a bit troublesome, but it's fairly easy with polynoms.

OK
say
0 < | x - a | < d ---> | 1/x - 1/a | < e

| 1/x - 1/a | =  | ( x - a ) /  ( x * a ) | = | x - a | / |x * a|   <   d / | x * a |  #

now, we need to put some limit on x, so that | 1 / x * a | won't go to infinity.
that's why we have d.
0 < | x - a | < d  ---> -d < x - a < d <---> -d + a < x < d + a (*)
I want | d - a | not to "be close" to 0.
for example, if a is 2, I will choose a d not bigger than 1, so 1 < x < 3

in general, I will choose a d that is no bigger than a/2
however, if d < a/2, we can be assured that according to (*) above,
-a/2 + a < x < a/2 + a <--->  a/2 < x < a*3/2
<--->  |a * 2/3 | < |1/x| < |2/a|
(the absolute value is just convenient, otherwise I would have to make a different case for negative a)

going back to #:
we showed that
| 1/x - 1/a | <   d / | x * a |

and I say, 1/|x| < 2/|a|
that's why, |1/x*a| < |2/a*a|

and we get that d/| x * a | < 2d / |a|^2
and so, | 1/x - 1/a | < 2d / a^2
if 2d/a^2 were to be smaller than e, it will be just great, and so, we choose a d that follows this condition: 2d/a^2 < e
==> d < e/2 * a^2

so, d < min { a/2, e/2 * a^2 } makes the condition true. :O



thas is the draft, you use it to find d. then in the final answer you need to write something like this:
"Let e be a real number bigger than 0.  (you can use mathematical signs, I just don't have any here)
 we will choose d < min { a/2, e/2 * a^2 }.

 if d < a/2, then
 0 < | x - a | < d < a/2 ---> -a/2 < x - a < a/2 <---> a/2 < x < a3/2
 and in absolute value, we will get that
 | 2/3a | < | 1/x  | < | 2/a | ---> 0 =/= | 1 / x | < | 2 / a |  (*)

 now let's look at the expression | 1/x - 1/a |.
 | 1/x - 1/a | = | a - x | / | x * a | < 2d / |a * a|  < 2 (e/2 * a^2) / a^2 = e
                                       |x-a|<d & (*)      d<(e/2)*a^2

 and so we proved that for any Real e greater than 0, | 1/x - 1/a | < e if | x - a | < d, QED."
 (you can phrase that last sentence however you want)

[kryss]

just find lots of examples and do them. that's the only way to learn it properly.

[Hikaru]

well, it's not that difficult, you "just" need is to limit | F(x) - L | using delta :S
a more video like example would be better I think.

[Hikaru]

Well, you start with limiting (selecting) delta with epsilon, eg: d < 2e or d = 2e etc etc.
That way, | x - a | is somehow limited by epsilon, and the proof ends when you show that | f(x) - L | is also limited by epsilon.
it doesn't exactly matter how it is limited, as long as the limitation aspires to 0.

for example,
if you show that | f(x) - L | < 3e, that's just as good.
or more generally, if | f(x) - L | < g(e) ----------> 0
                                                       e--->0
then f(x) ------> L
              x->a

it's the sandwich law
0 <= | f(x) - L | < e -------> 0

( if  h(x) <=  f(x) <= g(x) g(x) --------> L, and h(x) -------> L then  f(x) ------> L   )
(                                       x->a                        x->a                        x->a        )