# Thread: Limits As X approaches infinity

1. ## Limits As X approaches infinity

Suppose that f and g are real functions such that $f(x),g(x)\to\infty\$, as $x\to\infty\$.

Prove that $f(x)+g(x)\to\infty\$ as $x\to\infty\$ approaches infinity.

What (if anything) can be said in general about the difference $f(x)-g(x)$, the product $f(x)g(x)$ and the quotient $\frac{f(x)}{g(x)}$ as $x\to\infty\$??

Justify any assertions that you make.

2. Originally Posted by I Congruent
Suppose that f and g are real functions such that $f(x),g(x)\to\infty\$, as $x\to\infty\$.

Prove that $f(x)+g(x)\to\infty\$ as $x\to\infty\$ approaches infinity.

What (if anything) can be said in general about the difference f(x) - g(x), the product f(x)g(x) and the quotient f(x)/g(x) as z (this should be x right?) approaches infinity?

Justify any assertions that you make.

anyways, it would be better if you show us your own proof..

3. I was thinking that since $f(x)\to\infty\$ it means that for any arbitary $A>0$, there exsists a $x>k$ such that $f(x)>A$, similarly since $g(x)\to\infty\$, it means that for any arbitary $B>0$, there exsists a $x>h$ such that $g(x)>B$.

But i dont really know were to go from here.

4. Originally Posted by I Congruent
I was thinking that since $f(x)\to\infty\$ it means that for any arbitary $A>0$, there exsists a $x>k$ such that $f(x)>A$, similarly since $g(x)\to\infty\$, it means that for any arbitary $B>0$, there exsists a $x>h$ such that $g(x)>B$.

But i dont really know were to go from here.
So if we let $C=\min(A,B)$ and $j=\max(k,h)$, we have :
$A \ge C > 0$ and $B \ge C > 0$

And we have $k \ge j$ and $h \ge j$

So for both arbitrary A>0 and B>0, we have an arbitrary $C>0$, such that for all x > j (that is > k and > h) we have f(x)>A and f(x)>B, that is f(x)>C

the red parts give you the definition of $f(x)+g(x) \to \infty$

Note : this is not really a formal proof, it's just giving you the idea.

5. Thanks Moo, i can understand and see why it works but i still have know idea how to prove it .

6. I still need help

7. Originally Posted by I Congruent
I still need help
$B > 0\, \Rightarrow \,\left( {\exists n_f } \right)\left[ {x \geqslant n_f \, \Rightarrow \,f(x) \geqslant \frac{B}{2}} \right]$
$B > 0\, \Rightarrow \,\left( {\exists n_g } \right)\left[ {x \geqslant n_g \, \Rightarrow \,g(x) \geqslant \frac{B}{2}} \right]$

$x \geqslant n_f + n_g \, \Rightarrow \,\left( {f + g} \right)(x) \geqslant B$

8. Do we have to do all that?

Surely:

$f(x)-g(x)=\infty-\infty=0$

$f(x)g(x)={\infty}^2=\infty$

$\frac{f(x)}{g(x)}=1$

Or should I leave this thread to the big boys and girls? (lol!)

9. Originally Posted by Showcase_22
Or should I leave this thread to the big boys and girls?
NO! You should learn how to do these before posting.
$\lim _{n \to \infty } n^2 = \infty \,,\,\lim _{n \to \infty } n = \infty \mbox{ but }\lim _{n \to \infty } \left( {n^2 - n} \right) = \infty$

10. Can we really state that $\infty-\infty=0$ ? No, since, $\infty$ is not a number.

11. Oh right, sorry.

I was just wondering if there was a faster way of doing it, but I was mistaken

12. in short, given the situation, the cases

$f(x) - g(x)$and $\frac{f(x)}{g(x)}$ are INCONCLUSIVE as it is..

(but there are many ways to find the limit, if any, or conclude that the limit does not exist.)

13. I don't think they are inconclusive i can only think of 3 possible outcomes for $f(x) - g(x)$ and 3 possible outcomes of $\frac{f(x)}{g(x)}$.

14. yes! true! that is why they are inconclusive because you HAVE 3 POSSIBLE outcomes AS IT IS.. (actually, there are only two.. either the limit does or does not exist..)

but if you manipulate it (i mean if you are given with specific functions), you will find THE ONLY and the RIGHT outcome.

15. By 3 outcomes i ment $f(x) - g(x)$ is either going to tend to 0, $-\infty$, or $\infty$ as $x\to\infty$.

and $\frac{f(x)}{g(x)}$ is either going to tend to 0, 1, or $\infty$ as $x\to\infty$.