Hi again, after a loong brake. Hope, i posted this question to right forum(calculus).

Question is;

DNE

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Converges

Find f(x) and g(x).

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- December 17th 2010, 11:58 AMLafexlosFinding f(x) and g(x) - limitHi again, after a loong brake. Hope, i posted this question to right forum(calculus).

Question is;

DNE

Converges

Converges

Find f(x) and g(x).

- December 17th 2010, 12:03 PMAlso sprach Zarathustra
f(x)=1/x , g(x)=x , a=0

- December 17th 2010, 12:19 PMLafexlos
i think "a" should remain as itself, and that way your answer converts to this.

f(x)=1/(x-a) and g(x)=(x-a)

when you appy this to f(x)-g(x);

i couldn't understand how it's converges.

btw, my head is really full today, cuz of exams. so if it's answer is too easy sorry for bothering.

edit: and sorry again for my bad grammer. =) - December 17th 2010, 12:30 PMchiph588@
- December 17th 2010, 03:08 PMsnowtea
I don't think it is possible.

If the limit of f(x) - g(x) and g(x) exists, then the limit of f(x) = (f(x) - g(x)) + g(x) exists. - December 17th 2010, 03:34 PMzzzoak
I don't know if it is possible.

Not nice example:

f(x):

1 for rational

0 for irrational

g(x):

1 for rational.

And

f(x)-g(x) for rational. - December 18th 2010, 03:17 AMHallsofIvy
What is g for x irrational? If you mean that g is not defined for x irrational then does not exist. In order that exist, you would have to have g(x)= 1 for x irrational also (at least close to a) and in that case, would not exist.

Quote:

And

f(x)-g(x) for rational.

There are no such f and g for the reason snowtea cited. - December 18th 2010, 04:32 AMLafexlos
Thanks guys. Really helped.

My friend -who asked question- missed some information. After i get all information, if i still cant solve, i'll ask again with no missing info.

Sorry for taking your time. =/