Is the writing of the proof ok? 
Problem. Let $f:(0,\infty)\to\mathbb{R}$. Prove that, $$\displaystyle\lim_{x\to\infty}f(x)=L\iff\displaystyle\lim_{x\to0 +}f\left(\dfrac{1}{x}\right)=L$$


My Solution.
Let us assume that $\displaystyle\lim_{x\to\infty}f(x)=L$. Then by definition, $$(\forall\varepsilon>0)(\exists M\in \mathbb{R}^{+})(x\ge M\implies |f(x)-L|<\varepsilon)$$which holds iff, $$(\forall\varepsilon>0)(\exists M\in \mathbb{R}^{+})\left(\dfrac{1}{x}\in\left(0,\dfrac{1}{M}\right)\implies \left|f\left(\dfrac{1}{x}\right)-L\right|<\varepsilon\right)$$

Is the writing of the proof ok? 
 A: Your proof is perfectly correct, and there is nothing else to justify. A more rigorous way would be
Let $\varepsilon>0$. Since $\lim_{x\to\infty }f(x)=L$,
$$\exists M>0: x>M\implies |f(x)-L|<\varepsilon.$$
Let $\delta=\frac{1}{M}$. Then, if $y=\frac{1}{x}$,
$$0<y<\delta\implies x>M\implies |f(x)-L|<\varepsilon\implies \left|f\left(\frac{1}{y}\right)-L\right|<\varepsilon.$$
The converse goes in the same way.
A: Another way to visualize this fact is by noticing that $\displaystyle \lim _{x \rightarrow \infty}f(x)=L$ iff the function $f$ can be extended to a continuous function $\overline{f}:(0,\infty] \rightarrow \mathbb{R}$ which evaluates $f(\infty)=L$. Now, taking the continuous functions
$$g: (0,\infty) \rightarrow (0,\infty)$$
$$x \mapsto \frac{1}{x}$$
and
$$\overline{g}: [0,\infty) \rightarrow (0,\infty]$$
$$x \mapsto \frac{1}{x},$$
we see that $f \circ g$ can be extended to a continuous function $\tilde{f} \circ \tilde{g}: [0,\infty) \rightarrow \mathbb{R}$. By a similar reason as before, we then have that $\displaystyle \lim_{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)=L$. The converse is similar.
