Real Analysis

>>9129960
>How the hell does one prove (iii)?
There is no point in asking here.
You could find the same information on a book.
You should really try and prove it yourself.

>>9129960
[math] \frac{1}{a_n} - \frac{1}{a} = \frac{a - a_n}{a a_n} [/math]

Given that [math] a_n [/math] converges then you can find some k such that [math] k < |a a_n| [/math] and then [math] |\frac{a - a_n}{a a_n}| < |\frac{a - a_n}{k}| = \frac{1}{|k|}|a-a_n| [/math]

and you know that goes to zero.

>>9129962
No, it's not in the book.

And trust me, I tried.

>>9129968
Magnitude of a*an will be different for all n, for all you know a could be converging to an as an oscillating sequence from both sides of 1.

>>9129974
>a could be converging to an
an to a*

>>9129970
>No, it's not in the book.
>THE

>And trust me, I tried.
The post your math.

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>>9129960
>can't take the limit of a continuous, invertible function

Might as well drop out now OP. This is bread and butter stuff.

>>9129990
It's actually something from the lecture notes that I couldn't find in any of the texts I was referring to.

Will post where I got later, but I don't think it's relevant to the solution.

>>9130003
...I'm not quite sure you understood the question?

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>>9130008

>>9130014
Didn't bother to write something more formal since this wasn't getting anywhere, there's absolutely no constraint on the sequence an being limited to all ans being above 1 or below - 1.

>>9129974
For every epsilon > 0 there is a N such that
|a a_n| < |a|^2 + epsilon |a|
for all n > N.
Since it's true for every epsilon you can take for example epsilon = 1 then
|a a_n| < |a|^2 + |a|
so you can define k := |a|^2 + |a|

>>9130018
That k would be more than abs(a*an)

>>9130030
Sorry, you're completely right.

There is an N such that for all n>N
|a_n - a| < |a|/2

So for such n
|a| <= |a - a_n| + |a_n| < |a|/2 + |a_n|
Subtract |a|/2 to get
|a_n| > |a|/2
Now you can take k := (|a|^2)/2