Spivak chapter 5 exercise 36 - prove $\lim_{x \to 0^-} f(1/x) = \lim_{x \to -\infty} f(x)$ There're similar questions already, but I somewhat struggled to apply their reasoning to this particular statement.
I wanted to ask if my proof is correct (and clarify several things which I've seemingly figured out when writing it here). Here's the proof:
$$\mbox{(1) }\lim_{x \to 0^-} f\left(\frac1x\right)=l \mbox{ if } \forall\varepsilon>0(\exists\delta>0(\forall x(0<0-x<\delta \longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$
$$\mbox{(2) }\lim_{x \to -\infty} f(x)=m \mbox{ if } \forall\varepsilon>0(\exists N(\forall x(x<N \longrightarrow|f(x)-m|<\varepsilon$$
We need to show that $l=m$.
from (1) we have
$$-\delta>x>0\longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$
let $g(x)=f\left(\frac1x\right)$, then we have
$$-\delta>x>0\longrightarrow|g\left(x\right)-l|<\varepsilon$$
in (2) we suppose that $x<N$. Hence $\frac1x>\frac1N$. Let $x'=\frac1x$. Furthermore, since $-\delta$ is negative and we can assume that $N$ is negative. (If it's not, we can take $N$ to be -1 or any other negative number, because if the conclusion holds $\forall x<N$, it surely holds for a subset $(-\infty,-1)$). We can take $-\delta=\frac1N$, which gives us
$$\frac1N=-\delta<x'<0$$ and this is the hypothesis from (1). Hence we conclude that
$$|g\left(x'\right)-l|<\varepsilon$$ or $$|f\left(\frac1{\frac1x}\right)-l|<\varepsilon$$ or $$|f\left(x\right)-l|<\varepsilon$$
To summarise, $x<N\longrightarrow|f\left(x\right)-l|<\varepsilon$. By the definition of the limit $$\lim_{x \to -\infty}f(x)=l$$ but also $$\lim_{x \to -\infty}f(x)=m$$ Therefore $l=m$.

The proof you give becomes a bit unwieldy and it does not make clear that $N$ must be negative.
I think it can be shortened as follows:

Assume $\lim_{x\to0^-} f(1/x)$ exists and equals $\ell$.  Then take any $\varepsilon > 0$.
Because the limit above exists, there exists $\delta > 0$ such that
$$\lvert f(1/x) - \ell\rvert \leqslant \varepsilon \text{ for all } -\delta \leqslant x < 0. \tag{1}\label{eq1}$$
Now take any $y \leqslant -1/\delta$, write $x=1/y$.  It follows $-\delta \leqslant x < 0$.  This satisifies the criterion in \eqref{eq1}, so $\lvert f(1/x)-\ell \rvert \leqslant \varepsilon$, which is the same as $\lvert f(y) - \ell\rvert \leqslant \varepsilon$.
Thus for every $\varepsilon > 0$ there exists $m$ (taking $m=1/\delta$) such that $\lvert f(y)-\ell\rvert \leqslant \varepsilon$ whenever $y \leqslant -m$. This provbes $\lim_{y\to-\infty}f(y)$ exists and equals $\ell$.
