Do Functions Tending to Infinity Always Result in Infinity When Combined?

[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?

 

[arildno]

To help you a bit with the first one.

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.

 

[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 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?

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.

Similarly f(x)*g(x) would also tend to infinity wouldn't it?

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).

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?

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.

 

[flux_factor]

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

 

[woodyallen1]

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

 

[HallsofIvy]

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

Excellent dilineation!

 

[woodyallen1]

Very Kind of you PF..