Finding f(x) and g(x) - limit

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December 17th 2010, 11:58 AM
Lafexlos

Finding f(x) and g(x) - limit

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

$\displaystyle \lim_{x \to a} f(x)$ DNE

$\displaystyle \lim_{x \to a} g(x)$ Converges

$\displaystyle \lim_{x \to a} f(x) - g(x)$ Converges

Find f(x) and g(x).
December 17th 2010, 12:03 PM
Also sprach Zarathustra

f(x)=1/x , g(x)=x , a=0
December 17th 2010, 12:19 PM
Lafexlos

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);

$\displaystyle \lim_{x \to a} f(x) - g(x) = \lim_{x \to a} \frac {1-(x-a)^2}{x-a}$

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 PM
chiph588@

Quote:

Originally Posted by Also sprach Zarathustra

f(x)=1/x , g(x)=x , a=0

I don't think this works...
December 17th 2010, 03:08 PM
snowtea

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 PM
zzzoak

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 AM
HallsofIvy

Quote:

Originally Posted by zzzoak

I don't know if it is possible.
Not nice example:

f(x):
1 for rational
0 for irrational

g(x):
1 for rational.

What is g for x irrational? If you mean that g is not defined for x irrational then $\lim_{x\to a} g(x)$ does not exist. In order that $\displaytype\lim_{x\to a} g(x)$ exist, you would have to have g(x)= 1 for x irrational also (at least close to a) and in that case, $\lim_{x\to a} f(x)- g(x)$ would not exist.

Quote:

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

Again, not defined for irrational x and so the limit would not exist.

There are no such f and g for the reason snowtea cited.
December 18th 2010, 04:32 AM
Lafexlos

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. =/