How do you prove that $\lim_{x\rightarrow0^-} f(x) = \lim_{x\rightarrow -\infty} f(\frac{1}{x})$

Question

Given $$\lim_{x\rightarrow0^-} f(x) = \lim_{x\rightarrow -\infty} f(\frac{1}{x})$$ I need to prove it with a simple proof using $\epsilon-\delta$. So, here is what I ve got so far:

Given $\lim_{x\to 0-} f(x) = L$ if, for every $\varepsilon>0$ there is a $\delta>0$ such that for all $|x-0|<\delta \implies$ $|f(x)-L|<\varepsilon$.

We have that $\lim_{y\to -\infty} g(y) = M$ if, for every $\varepsilon>0$ there is a $N$ such that for all $y<N \implies$ $|g(y)-M|<\varepsilon$.

Taking functions such that $g(y)=f(1/y)$.

Consider the limit $$\lim_{x\to 0^{-}} f(x) = L \exists$$ , then for any $\varepsilon>0$ find a $\delta>0$, so that $|f(x)-L|<\varepsilon$ for $|x|<\delta$. Let $N=\frac{1}{\delta}$.

So, if $y<N$ then $\frac{1}{y}\in (0,\delta)$ so $$|g(y)-L| = |f(\frac{1}{y})-L| < \varepsilon$$

So, $\lim_{x\to 0-} f(x) = L$ then $\lim_{y\to -\infty} f(1/y) = L$, too.

$\lim_{y\to -\infty} g(y) = L$, then, given $\varepsilon>0$ and $N \in \mathbb{R}$, we can take $\delta>0 = \frac{1}{N}$ if $N<0$ and $\delta = 1$ if $N\leq 0$.

Such that $N'=\max(N,1)$.

So, we can conclude $$\lim_{x\rightarrow0^-} f(x) = \lim_{x\rightarrow -\infty} f(\frac{1}{x})$$ Suggestions will be welcome. Thanks.

Answer 1

Your general idea is correct, but it could be much cleaner. Also, in your $\delta$ claims, you are using absolute values and aren't avoiding the $x=0$ case (you should have $-\delta<x<0$, not $|x|<\delta$). All you have to do is:

Let $\epsilon>0$. Then there exists $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $-\delta<x<0$ $\iff$ there exists $M:=1/\delta$ such that $|f(1/x)-L|<\epsilon$ whenever $x<-M$.

Answer 2

It's pretty simple.

Suppose $\lim_{x\to0^-}f(x)=L$, finite. You want to prove that $\lim_{x\to-\infty}f(1/x)=L$.

Let $\varepsilon>0$. Then, by assumption, there exists $\delta>0$ such that, for $-\delta<x<0$, we have $|f(x)-L|<\varepsilon$.

Set $K=1/\delta$. Then, if $x<-K$, we have $-\delta<1/x<0$, hence $|f(1/x)-L|<\varepsilon$.

Now do similarly for $L=\pm\infty$.

Finally, adapt the proof to show the converse.