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For some analytic function $f(x)=\lim_{n\rightarrow\infty}\sum^n_{r=0}c_rx^r$, $$\lim_{x\rightarrow\infty}\frac{f(x)}{e^x}=\lim_{x\rightarrow\infty}\frac{\lim_{n\rightarrow\infty}\sum^n_{r=0}c_rx^r}{e^x}=\lim_{n\rightarrow\infty}\sum^n_{r=0}c_r\lim_{x\rightarrow\infty}\frac{x^r}{e^x}=\lim_{n\rightarrow\infty}\sum^n_{r=0}c_r(0)=\lim_{n\rightarrow\infty}0=0$$ But obviously this cannot be true? For say, $f(x)=e^x$ we would have $$\lim_{x\rightarrow\infty}\frac{e^x}{e^x}=1$$ and $$\lim_{x\rightarrow\infty}\frac{e^x}{e^x}=0$$ which would imply$$0=1$$

UPDATE: An answer has suggested that the two limits cannot be inverted. Is this true and why is this so? I would like a more complete explanation if this is the case. Also if this were the case, couldn't this modification to the question avoid that pitfall completely? $f(x)=\sum^\infty_{r=0}c_rx^r$, $$\lim_{x\rightarrow\infty}\frac{f(x)}{e^x}=\lim_{x\rightarrow\infty}\frac{\sum^\infty_{r=0}c_rx^r}{e^x}=\sum^\infty_{r=0}c_r\lim_{x\rightarrow\infty}\frac{x^r}{e^x}=\sum^\infty_{r=0}c_r(0)==0$$

Thanks in advance. Any other thoughts or comments welcome.

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2 Answers 2

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Hint: This is a perfect example of why you can't always switch limits!

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  • $\begingroup$ How about this: $f(x)=\sum^\infty_{r=0}c_rx^r$, $$\lim_{x\rightarrow\infty}\frac{f(x)}{e^x}=\lim_{x\rightarrow\infty}\frac{\sum^{\infty}_{r=0}c_rx^r}{e^x}=\sum^\infty_{r=0}c_r\lim_{x\rightarrow\infty}\frac{x^r}{e^x}=\sum^\infty_{r=0}c_r(0)=0$$ $\endgroup$
    – resgh
    Commented Feb 12, 2013 at 6:57
  • $\begingroup$ And why can't I switch the limits? Can you offer some explanation? $\endgroup$
    – resgh
    Commented Feb 12, 2013 at 7:00
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    $\begingroup$ @namehere: Your question gives an explanation: When you switch limits, you get different answers. Therefore if limits are to be switched, some justification must be given for why the switching is valid in a given case. $\endgroup$
    – Jonas Meyer
    Commented Feb 15, 2013 at 15:34
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An answer has suggested that the two limits cannot be inverted. Is this true and why is this so?

  1. "Is this true?" Yes, and your example shows why it is true. The switching of limits leads to $0=1$, and therefore it is wrong.

  2. "Why is this so?" Because the results of the limits in different orders is often different. You have given one example of this.

  3. "Also if this were the case, couldn't this modification to the question avoid that pitfall completely?...." No, that actually is not a modification, just a more concise notation with the exact same meaning. The value of an infinite series is, by definition, the limit of the sequence of partial sums. That is, $\sum\limits_{r=0}^\infty a_r$ means the same thing as $\lim\limits_{n\to\infty}\sum\limits_{r=0}^n a_r$, provided the latter exists. Thus, passing a limit through an infinite summation is a case of switching limits, and as your example shows, it is not always correct.

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  • $\begingroup$ Is there any criteria for when the switching of limits are the same? $\endgroup$
    – resgh
    Commented Feb 15, 2013 at 16:13
  • $\begingroup$ For limits of series of functions, uniform convergence is often a useful criterion. Abel's theorem gives a useful criterion for the case of power series that converge at the boundary of convergence: If $\sum\limits_{r=0}^\infty a_r$ converges, then $\sum\limits_{r=0}^\infty\lim\limits_{x\nearrow 1}a_rx^r=\lim\limits_{x\nearrow 1}\sum\limits_{r=0}^\infty a_rx^r$. $\endgroup$
    – Jonas Meyer
    Commented Feb 15, 2013 at 16:39

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