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Limits and derivative: is this proof accurate enough?

  1. Jan 8, 2013 #1
    1. The problem statement, all variables and given/known data

    f is differentiable in ##\mathbb{R^+}## and
    ##\displaystyle \lim_{x \to \infty} (f(x)+f'(x))=0##
    Prove that
    ##\displaystyle \lim_{x \to \infty}f(x)=0##

    3. The attempt at a solution

    I can split the limit in two:
    ##(\displaystyle \lim_{x \to \infty} f(x)+\displaystyle \lim_{x \to \infty} f'(x))=0##
    I consider the second one and say that, by definition of derivative I have:
    ##\displaystyle \lim_{x \to \infty} \displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}##
    As f is differentiable, then the second limit exists and is 0.
    So, i have ##\displaystyle \lim_{x \to \infty} 0 =0##
    And then, by hypothesis:
    ##\displaystyle \lim_{x \to \infty} (f(x)+0)=0##
    Are the passages logically correct?
    thank you in advance!
     
  2. jcsd
  3. Jan 8, 2013 #2

    micromass

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    No, you can't do that. You can only do that if you know that both limits exist. So:

    [tex]\lim_{x\rightarrow +\infty} f(x)+g(x) = \lim_{x\rightarrow +\infty} f(x) + \lim_{x\rightarrow +\infty} g(x)[/tex]

    is not true in general, but only if you know that the limits actually exist.

    As a counterexample:

    [tex]0=\lim_{x\rightarrow +\infty} x-x \neq \lim_{x\rightarrow +\infty} x + \lim_{x\rightarrow +\infty} -x[/tex]
     
  4. Jan 8, 2013 #3
    Thanks, you saved me from a major mistake!
    Maybe I should prove this way, then?:

    By definition of derivative:
    ##f(x)+f'(x)=f(x)+ \displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} ##=

    ## \displaystyle \lim_{h \to 0} f(x)-\frac{f(x)}{h}+\frac{f(x+h)}{h}## =
    ## \displaystyle \lim_{h \to 0} f(x)(1-\frac{1}{h})+\frac{f(x+h)}{h}##=

    and, by hypothesis:

    ##\displaystyle \lim_{x \to \infty} \displaystyle \lim_{h \to 0} f(x)(1-\frac{1}{h})+\frac{f(x+h)}{h}##= 0

    Say:

    ##\displaystyle \lim_{x \to \infty}f(x)=L## so that the expression above becomes:

    ##\displaystyle \lim_{h \to 0}L(1-\frac{1}{\epsilon})+\frac{L}{\epsilon}##=0

    ##\displaystyle \lim_{h \to 0} L=0## ##\Rightarrow## ##\displaystyle \lim_{x\to \infty }f(x)=L##=0
     
  5. Jan 8, 2013 #4

    micromass

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    Now you assume that

    [tex]\lim_{h\rightarrow 0} \lim_{x\rightarrow +\infty} f(x,h) = \lim_{x\rightarrow +\infty}\lim_{h\rightarrow 0} f(x,h).[/tex]

    This is also not true in general.
     
  6. Jan 8, 2013 #5
    oh.. I've run out of possible good ideas then, any hint?
     
  7. Jan 8, 2013 #6

    micromass

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    The intuition is this: if x is very large, then f(x) is very close to -f'(x). In particular, if f(x) is very large, then -f'(x) is very negative. And thus f(x) decreases very fast.

    Try to work with an epsilon-delta definition. What does it mean that f(x) does not tend to 0??
     
  8. Jan 8, 2013 #7
    can I simply say then, that:
    ##\forall \epsilon >0####\exists \delta>0## such that if ##0<|x-x_0|<\delta \Rightarrow |f(x)+f'(x)-0|<\epsilon##
    And so, being the function >0 because it is defined on ##\mathbb{R^+}:
    ##0<|f(x)|< \epsilon - f'(x)|## ##\Rightarrow## ##|f(x)|< \epsilon## and for the squeeze rule ##\lim f(x)=0##
     
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