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Let $f$ be a function so that $\displaystyle\lim_{x\underset{>}{\to}0}f(x)$ exists and $\displaystyle\lim_{x\to\infty}f(x)$ exists.

Does the equation $\displaystyle\lim_{x\underset{>}{\to}0}f(x)=\displaystyle\lim_{x\to\infty}f(\frac{1}{x})$ hold for all such functions $f$?

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What is that ">" under arrows, Barto? –  Babak S. Oct 30 '12 at 13:12
    
What do you mean by $\lim\limits_{x\underset{>}{\to}0}f(x)$ is it one-sided limit $\lim\limits_{x\to0^+}f(x)$? –  Martin Sleziak Oct 30 '12 at 13:13
    
It means $x$ approaches $0$ with values larger then $0$. So it implies the limit of $f(x)$ as $x$ approaches from the 'right'. @MartinSleziak: yes. –  barto Oct 30 '12 at 13:14
    
This is the same problem as in this question: math.stackexchange.com/questions/225194/equivalence-of-2-limits –  Martin Sleziak Oct 30 '12 at 13:42

2 Answers 2

up vote 1 down vote accepted

Yes it does. And we needn't know about $\lim_{x\to\infty} f(x)$ as it is not important for the behavior of $f$ near $0$ (and as $x \mapsto \frac 1x$ maps $[n,\infty)$ to $(0, \frac 1n]$ the second limits also talks about the behavior of $f$ near 0 [and that's the proof]).

To be more exact, let $\epsilon > 0$. Then by existence of $a :=\lim_{x\to 0^+} f(x)$ there is an $\delta > 0$ such that $|f(x) - a| < \epsilon$ for $0 < x < \delta$. Let $M := \frac 1\delta$, for $x > M$ we have $0 < \frac 1x < \frac 1\delta$, hence \[ \left|f\left(\frac 1x\right) - a \right| < \epsilon \] This proves $\lim_{x\to\infty} f(\frac 1x) = a = \lim_{x\to 0^+} f(x)$.

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Hint: Starting with $\displaystyle\lim_{x\rightarrow\infty}f\left(\frac{1}{x}\right)$, can you make a change of variables to transform this limit into the form $\displaystyle\lim_{u\rightarrow 0^+}f(u)$? If so, then the answer to your question is yes (note that it doesn't matter whether the variable is called $u$ or $x$ or anything else).

I'm actually not sure what the assumption that $\displaystyle\lim_{x\to\infty}f(x)$ exists has to do with the question, because it doesn't seem that anywhere you need to deal with a limit as the argument of $f$ goes to $\infty$.

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Indeed the existence of $\displaystyle\lim_{x\to\infty}f(x)$ isn't necessary. –  barto Oct 30 '12 at 13:48

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