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Infinite limit as X tends to infinity

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    I am required to express in [tex]\varepsilon - \delta[/tex] way what I'm suppose to prove in case [tex]lim_\below{(x \rightarrow \infty)} f(x) = \infty[/tex]

    2. Relevant equations
    None.


    3. The attempt at a solution
    So first, intuitively I thought that what this means is that [tex]f(x)[/tex] is bigger than any arbitrary number when [tex]x[/tex] is bigger than any arbitrary number. So I attempted to combine the [tex]\varepsilon - \delta[/tex] definitions of when [tex]x[/tex] tends to infinity and when limit [tex]f(x)[/tex] tends to infinity.

    I came up with this:
    [tex]lim_\below{(x \rightarrow \infty)} f(x) = \infty[/tex] if for every [tex]M>0[/tex] there exists [tex]N>0[/tex] so that for every [tex]x>M[/tex], [tex]f(x)>N[/tex].

    I am unsure of whether it's the correct definition. Anyone can verify that?
     
  2. jcsd
  3. Mar 30, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi IdanH14! Welcome to PF! :smile:

    (have a delta: δ and an epsilon: ε and an infinity: ∞ :wink:)
    Yes :smile:, except it's the other way round …

    no matter how large N is, we can find an M above which f(x) > N. :wink:

    (you get the same result if you use 1/f, 1/δ, and 1/ε)
     
  4. Mar 30, 2010 #3
    Thanks! :)
    Let me summarize it to see if I got it. It should be
    For every N>0 there exists M>0, so that for every x>M, f(x)>N
    Right?
     
    Last edited: Mar 30, 2010
  5. Mar 30, 2010 #4

    tiny-tim

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    Right! :biggrin:
     
  6. Mar 30, 2010 #5
    I like you usage of icons. I think I'll adopt it. ;)

    Now, I'm trying to solve an exercise in my math book with this principle. Despite the fact I already learned infinite limits arithmetics, I'm required to prove that [tex]lim_\below{x\rightarrow \infty}) x cos\frac{1}{x} = \infty[/tex] in this cumbersome way. So I think I have a solution, but again, my insecurities creep in. :devil:

    So I noticed that the bigger x gets, the closer [tex]cos\frac{1}{x}[/tex] gets to 1. If I'll choose [tex]M=N[/tex] then, unless they [tex]N=\infty[/tex] I'll get something that's smaller than N, because X is multiplied by something which is close to 1, but doesn't equal 1. But if [tex]M=2N[/tex] then problem eliminated. Almost. :bugeye:

    Why almost? Because if [tex]\frac{1}{x}>1[/tex] then [tex]cos\frac{1}{x}[/tex] is potentially getting farther from 1. So, what I was thinking was to say [tex]M=max(1,2N)[/tex] and then problem solved.

    I hope I'm clear enough. Is this a valid proof? Did I find for every N>0 an M>0 that meet the requirements?

    Thanks :)
     
    Last edited: Mar 30, 2010
  7. Mar 30, 2010 #6

    tiny-tim

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    That's correct.

    I can work out why you're then using M = 2N (because of course M = N doesn't quite work), but you haven't actually specified why M = 2N does work. :smile:

    (for example, does it work for any N?)
     
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