Question

prove or disprove
if \(\lim (g(x)) =\infty\) and \(\lim(f(x)) =\infty\)
then \( \lim (g(x).f(x)) =\infty \) as x goes to some value.

Answer 1

lim(g(x).f(x))=lim(g(x)).lim(f(x))=~×~=~
Am representing infinity by ~
Hence proved.

Answer 2

so ur saying infinite*infinite=infinite which is away impossible mathematically. you can't just touch infinity lol

Answer 3

Ha ha ....

Answer 4

unless u have a way to convince that was true.

Answer 5

It is.
Correct.

Answer 6

You can check...

Answer 7

although i'm going now. i shall see this next morning,

gn=)

Answer 8

Wait....

Answer 9

~×~=~
This is correct.

Answer 10

well @wolfram is not convincing enough if u think about that.

Answer 11

I can't quite remember how to prove these limit rules :)
But my book has a good explanation.. trying to find it >.< grr

Answer 12

In limits......it is correct ,it don't come under 7 indeterminate forms.

Answer 13

can't wait to see @zepdrix =D

Answer 14

well why it's correct is part of my question @Pawanyadav :)

Answer 15

I'm using that, as a result ,I can't prove that....

Answer 16

I prove the question ,that's final....

Answer 17

Grr I dunno ;c maybe I was mistaken..

Answer 18

Is anyone saying that
~×~=~ in limits is false.

Answer 19

in limits?!?!

Answer 20

No, that sounds right :)
I just can't figure out how to justify it hehe

Answer 21

so what is ~ + ~

using the new nomenclature?

Answer 22

That's what am saying, it is correct..
But @ikram002p is not ready to accept it

Answer 23

He was representing \(\rm \infty\cdot\infty\) by typing ~x~

Answer 24

@Irishboy123 ,it is already ~
About which new nomenclature are you talking

Answer 25

~+~=~
~×~=~
But. ~÷~ and ~-~ are indeterminate forms. ..do anyone have any problem with this.

Answer 26

@ikram002p as it is proved ,don't forget the bet now

Answer 27

Both f(x) and g(x) blow up to infinity with finite but possibly different x right? If both f(x) and g(x) only blow up at x to infinity, then take f(x)=g(x)=x. f(x).g(x)=x^2 and only blows up at infinity, which seems to contradict your requirement that f(x).g(x) blows up at some, presumably finite, value of x.

Answer 28

Do tan(x) and cot(x) satisfy your reqirement? Both blow up at different but finite x. The product of two is 1 and does not blow up.

Answer 29

Both the limits are infinity.
So, given any large positive number \(\epsilon\), we can find \(L\) and \(M\) such that \(f(x)\gt \epsilon\) and \(g(x)\gt \epsilon\) for all \(x\gt \max\{L,M\}\).


It follows \(f(x)*g(x)\gt \epsilon\) whenever \(x\gt \max\{L,M\}\). That is \(f(x)*g(x)\) arbitrarily increases for large values of \(x\). Same as saying \(\lim f(x)*g(x) = \infty \).

Answer 30

@thomas5267. here is only 0ne x
not two different x
both f(x) and g(x) tends to infinity at some but same value of x.

Answer 31

=)