Proof $\lim_{x \rightarrow \infty}f(x)=\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})$ I want to prove that $\lim_{x \rightarrow \infty}f(x)=\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})$.

Proof:
Suppose that $\lim_{x \rightarrow \infty}f(x)$ exists and we denote $L:=\lim_{x \rightarrow \infty}f(x)$.
Then we have that $\forall \epsilon> 0, \exists N>0 : \text{if }x>N \Rightarrow |f(x)-L|< \epsilon  $.
We want to prove that $\forall \epsilon'>0, \exists \delta'>0 : \text{if } 0<x<\delta \Rightarrow |f(\frac{1}{x})-L|< \epsilon' $.
If we take $\epsilon'=\epsilon$ we have $N_{\epsilon}>0 : |f(x)-L|<\epsilon$ if $x>N_{\epsilon}$.
But we have that if $a < b$ (with $a,b \neq 0$) then $\frac{1}{b}<\frac{1}{a}$ so if we take $\delta'=\frac{1}{N_{\epsilon}}$ we have that if $0<x < \delta'$ then $\frac{1}{x}>\frac{1}{\delta'}=N_{\epsilon}$ so $|f(\frac{1}{x})-L|<\epsilon$.
How this is verified $\forall \epsilon'>0$ we have that $\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})=L=\lim_{x \rightarrow \infty}f(x)$.\
The Next Case:
Suppose that $\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})$ exists and we denote $L:=\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})$.
Then we have that $\forall \epsilon'> 0, \exists \delta'>0 : \text{if }0<x<\delta' \Rightarrow |f(\frac{1}{x})-L|< \epsilon'  $.
We want to prove that $\forall \epsilon>0, \exists N>0 : \text{if } x>N \Rightarrow |f(x)-L|< \epsilon $.
If we take $\epsilon'=\epsilon$ we have that exists $\delta'>0 : |f(\frac{1}{x})-L| < \epsilon$ if $0 < x<\delta'$.
But if we take $N_{\epsilon}=\frac{1}{\delta'}$ we have that if $x>N_{\epsilon}$ then $\frac{1}{x}<\frac{1}{N_{\epsilon}}=\delta'$ so $|f(\frac{1}{\frac{1}{x}})-L|< \epsilon$ but $f(\frac{1}{\frac{1}{x}})=f(x)$ so we have $|f(x)-L|<\epsilon$.
How this is verified $\forall \epsilon>0$ we have that $\lim_{x \rightarrow \infty}f(x)=L=\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})$.

Is this proof correct? I want to do it in the most rigorous way possible
 A: I have not checked each detail, but the idea seems correct. The underlying principle is:
$\lim_{x\to 0,x>0}g(x) = L$ iff for each $\epsilon>0$ there is a $\delta$ so that $g((0,\delta))\subset B_\epsilon(L)$. $\lim_{x\to\infty} h(x) = L$ iff for each $\epsilon>0$ we have a $\rho$ so that $h((\rho,\infty))\subset B_\epsilon(L)$.
Now the basic idea is that the transformation $x\to 1/x$ maps each $(0,\delta)$ bijectively to some $(\rho,\infty)$ with $\rho=1/\delta$.
Note that an even simpler proof can be done using this characterisation of limit: We say $\lim_{x\to x_0} f(x) = L$ iff for any sequence $x_n$ with $x_n\to x_0$ we have $f(x)\to L$ ($x_0$ can be $\pm\infty$). It is clear that this is equivalent to the other idea of convergence:
Trivially if $\lim_{x\to x_0}f(x)=L$ by the first characterisation then also $\lim_{n\to \infty} f(x_n)=L$ for any $x_n\to x_0$. On the other hand if the first characterisation is not met one can easily construct a sequence $x_n\to x_0$ so that $|f(x_n)-L|\geq \epsilon$ for some $\epsilon>0$.
So if we use this characterisation your property follows directly from
$$ x_n \to 0 \Leftrightarrow 1/x_n \to \infty$$
for $x_n>0$.
