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Lim |f(x)| = +infty and f decreasing implies lim f(x) = +infty

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    If [tex]f[/tex] is defined on [tex](a,b)[/tex], [tex]\lim_{x \to a^{+}} |f(x)| = +\infty[/tex] and [tex]f[/tex] is decreasing on [tex](a,b)[/tex], show that [tex]\lim_{x \to a^{+}} f(x) = +\infty[/tex].

    2. Relevant equations



    3. The attempt at a solution
    If [tex]\lim_{x \to a^{+}} |f(x)| = +\infty[/tex], then [tex]\forall M > 0, \exists \delta > 0[/tex] such that [tex]\forall x \in (a,b)[/tex] and [tex]a < x < a + \delta[/tex], we have [tex]|f(x)| > M[/tex], which implies we either have [tex]f(x) > M[/tex] or [tex]f(x) < -M[/tex]. It suffices to show that [tex]f(x) < -M[/tex] does not hold.

    For contradiction, suppose that [tex]f(x) < -M[/tex] for [tex]x \in (a,b) \cap (a, a + \delta)[/tex]. Consider a sequence of real numbers [tex](x_n)[/tex] such that [tex]x_n \in (a,b) \cap (a, a + \delta)[/tex] and [tex]x_{n+1} > x_n[/tex] and [tex]\lim_{n \to \infty}x_n = a[/tex] for [tex]\forall n \in \mathbb{N}[/tex]. Since [tex]f[/tex] is decreasing, [tex]x_{n+1} > x_n[/tex] implies [tex]f(x_n) \geq f(x_{n+1})[/tex]. Thus, we have that [tex]-M > f(x_n) \geq f(x_{n+1})[/tex]....

    But this is the part where I got stuck..... any help appreciated! Thanks in advance!
     
  2. jcsd
  3. Sep 21, 2009 #2

    Office_Shredder

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    It's always entirely possible that for some x really close to a, f(x) is really large in magnitude and negative. The key is that once you have that, f(x) can never get smaller than that. So for example, if you pick M cleverly, like

    [tex]M=|f(\frac{a+b}{2})[/tex]|

    then you know that for x<(a+b)/2 f(x) cannot be less than -M
     
  4. Sep 21, 2009 #3
    Thanks for the reply! I just want to follow up on this.

    Suppose we have picked [tex]M=|f(\frac{a+b}{2})|[/tex], but since I want to show that [tex]f(x) < -M[/tex] cannot hold, I want to reach a contradiction somewhere. But I'm not seeing the contradiction. That is, suppose that we have [tex]f(x) < -M = -|f(\frac{a+b}{2})|[/tex] for [tex]x \in (a,b) \cap (a, a + \delta)[/tex]. But since [tex]f[/tex] is decreasing on [tex](a,b)[/tex], then we have for [tex]x < (a + b)/2, f(x) \geq f(\frac{a+b}{2})[/tex] (1)

    But even with this, how can I show that [tex]f[/tex] cannot be less than [tex]-M[/tex]? That is, since [tex]M = |f(\frac{a+b}{2})|[/tex] is in terms of absolute values, but expression (1) does not exactly involve [tex]M[/tex] (i.e. no absolute values).

    Any further help is appreciated! Thanks :)
     
  5. Sep 21, 2009 #4

    Office_Shredder

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    [tex]f(\frac{a+b}{2})[/tex]

    is either M or -M. I f(x)<-M, we have f(x)<M also (since M is positive). So regardless of whether [itex]f(\frac{a+b}{2}[/itex] is positive or negative, we have that f(x) is less than it.

    But if [itex] x< \frac{a+b}{2}[/itex] it must be [itex] f(x) >f( \frac{a+b}{2})[/itex] as f is decreasing
     
  6. Sep 21, 2009 #5
    Thanks for the quick reply! I get the contradiction part in your last sentence, but I'm a little fuzzy on the first sentence. Just to be absolutely clear, I'll just spell it all out. Hope I'm understanding you correctly!

    Let [tex]M > 0[/tex]. Then set [tex]M = |f(\frac{a+b}{2})|[/tex]. Then we must have that [tex]f(\frac{a+b}{2}) = M[/tex] (i.e. positive f) or [tex]f(\frac{a+b}{2}) = -M[/tex] (i.e. negative f). Now, we consider only the case [tex]f(x) < -M < 0[/tex]. Then, in this case, we must have the case of negative f. Hence, this would imply that [tex]f(x) < -M = -(-f(\frac{a+b}{2})) = f(\frac{a+b}{2})[/tex]. And from this, we use the fact that f is decreasing to reach the contradiction.

    I just want to be sure that this is how I'm interpreting the below:
    Thanks again for the help!
     
  7. Sep 21, 2009 #6
    Actually, just to be nitpicking a bit. I think just saying [tex]x < (a + b) / 2[/tex] is not sufficient. Since the statement [tex]f(x) < -M[/tex] only holds for [tex]x \in (a,b) \cap (a, a + \delta)[/tex], it is not necessarily true that [tex](a + b) / 2[/tex] falls into that range. Hence, I think it would be better if we write, for [tex]a < x < \min\{(a+b)/2, a + \delta\}[/tex], then since f is decreasing on [tex](a,b)[/tex], it follows that [tex]f(x) \geq f(\frac{a+b}{2})[/tex] --- contradiction to the above.
     
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