# Limits of functions at infinity

1. Oct 4, 2009

### Juggler123

Suppose that the two functions f(x) and g(x) both tend to infinity then surely f(x) + g(x) also tends to infinity? How can you prove this though? Similarly f(x)*g(x) would also tend to infinity wouldn't it? f(x) - g(x) and f(x)/g(x) wouldn't tend to anything though surely since infinity minus infinity and infinity over infinity are both undefined. Can anyone help me?

2. Oct 4, 2009

### arildno

Saying that g(x) tends to infinity certainly implies the less strict interpretation, namely that for x>X, we have g(x)>N, where X and N are both some positive numbers.

Thus, for all x>X, we have the inequality

f(x)+g(x)>f(x)+N>f(x)

Thus, if f(x) is tending to infinity, then surely, f+g also.

3. Oct 5, 2009

### HallsofIvy

"limit of functions at infinity" implies you are talking about the limit as x goes to infinity but you are actually asking about infinite limits.
Suppose $\lim_{x\to a} f(x)= \infty$ and $\lim_{x\to a} g(x)= \infty$. Then, given any Y> 0, there exist $\delta_1> 0$ such that if $|x-a|< \delta_1$ then f(x)> Y and there exist $\delta_2< 0$ such that if $|x- a|< \delta_2$ then g(x)> 0. Take $\delta$ to be the smaller of $\delta_1$ and $\delta_2$ so that if $|x-a|< \delta$ both $|x-a|< \delta_1$ and $|x-a|< \delta_2$ are true. Then f(x)> Y and g(x)> 0 so f(x)+ g(x)> Y+ 0.

Similar proof. Given any real number, Y, there exist real number $\delta$ such that if $|x-a|< \delta$, $f(x)> Y$ and $g(x)> 1$. Then f(x)g(x)> (Y)(1).

Let f(x)= g(x)= 1/(x-a). Then f and g both tend to infinity as x goes to a but f- g and f/g tend to 0 and 1 respectively.

4. Sep 2, 2010

### flux_factor

Could someone delineate/prove HallsofIvy's last comment regarding f-g and f/g in more detail?

5. Sep 2, 2010

### woodyallen1

f=g so f-g=0 and f/g=1...

6. Sep 3, 2010

### HallsofIvy

Excellent dilineation!

7. Sep 7, 2010

### woodyallen1

Very Kind of you PF..