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Limits of functions at infinity

  1. Oct 4, 2009 #1
    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?
     
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  3. Oct 4, 2009 #2

    arildno

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    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.
     
  4. Oct 5, 2009 #3

    HallsofIvy

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    "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 [itex]\lim_{x\to a} f(x)= \infty[/itex] and [itex]\lim_{x\to a} g(x)= \infty[/itex]. Then, given any Y> 0, there exist [itex]\delta_1> 0[/itex] such that if [itex]|x-a|< \delta_1[/itex] then f(x)> Y and there exist [itex]\delta_2< 0[/itex] such that if [itex]|x- a|< \delta_2[/itex] then g(x)> 0. Take [itex]\delta[/itex] to be the smaller of [itex]\delta_1[/itex] and [itex]\delta_2[/itex] so that if [itex]|x-a|< \delta[/itex] both [itex]|x-a|< \delta_1[/itex] and [itex]|x-a|< \delta_2[/itex] 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 [itex]\delta[/itex] such that if [itex]|x-a|< \delta[/itex], [itex]f(x)> Y[/itex] and [itex]g(x)> 1[/itex]. 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.
     
  5. Sep 2, 2010 #4
    Could someone delineate/prove HallsofIvy's last comment regarding f-g and f/g in more detail?
     
  6. Sep 2, 2010 #5
    f=g so f-g=0 and f/g=1...
     
  7. Sep 3, 2010 #6

    HallsofIvy

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    Excellent dilineation!
     
  8. Sep 7, 2010 #7
    Very Kind of you PF..
     
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