Suppose $\,f:\mathbb{R}\rightarrow\mathbb{R},\,\lim\limits_{x\downarrow0}\,f(x) = c \in \mathbb{R}\cup\{\pm\infty\}$. I'd like to know if the follwing statement is true.

$$\lim\limits_{x\downarrow0}\,f(x) = \lim\limits_{x\rightarrow\infty}\,f(\frac1x)$$

My intuition about this is that for any $x$ in $\mathbb{R}^*_+$ there is a $y$ in $\mathbb{R}^*_+$ so that $\frac1y<x$. So it's like following the reversed graph in the reverse direction.

If the statement is true could you either:

  • proof it
  • explain your intuition why this is true

     ... and also would this be the same for $\lim\limits_{x\uparrow0}\,f(x) = \lim\limits_{x\rightarrow-\infty}\,f(\frac1x)$?

If the statement is false could you either:

  • hint me a counter-example
  • disproof the statement

     ... and also is there a way to correct the statement? E.g. constraining $\,f$ to be conitious.



say $c$ is a finite real number then $\lim\limits_{x\downarrow0}\,f(x) = c $ means :

$\forall \epsilon > 0, \exists \delta > 0\, \text{such that } \, |x | < \delta \implies |f(x)|<\epsilon $


$\forall \epsilon > 0, \exists \delta > 0\, \text{such that } \, |\frac1x | < \delta \implies |f(\frac1x)|<\epsilon $


$\forall \epsilon > 0, \exists \delta' = \frac{1}{\delta} > 0\, \text{such that } \, |x | > \delta' \implies |f(\frac1x)|<\epsilon $


$\lim\limits_{x\rightarrow\infty}\,f(\frac1x) = c$

if $c$ is $\pm \infty$ it's the same reasoning


It is true.

First assume that $\lim_{x \in \mathbb{R}^+ \rightarrow 0} = K$ for some finite $K$.

Then for all $\epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| \leq \epsilon$ for all positive $x'' \le x'$. This implies that $|f(\frac{1}{y}) - K| \le \epsilon$ for all $y \geq \frac{1}{x'}$. Which implies $\lim_{y \rightarrow \infty} f(\frac{1}{y}) = K$.

Can you handle the case where $\lim_{x \in \mathbb{R}^+ \rightarrow 0} = \infty$ ?


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