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

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. 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.

Cheers,
Pascal

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

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

meaning

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

meaning

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

meaning

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

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 $$\lim_{x \in \mathbb{R}^+ \rightarrow 0} = \infty$$ ?