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 $l=m$.
 A: 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 $\ell$.
A: WA Don nailed it, but I feel compelled to add an answer.
I'm going through Spivak myself and when first encountering problems like this, I tried approaches very similar to super.t's.
It took a while to absorb what was going on. Initially I found it a bit confusing trying to keep straight which conditions are fixed, which can vary, and what implies what.
Here I'll rework the problem, annotating with a few comments. The aim is to hopefully help clarify the thinking behind the approach. Apologies for stating things that may seem obvious to you, and generally for over-explaining. Hopefully this is of some use to someone, someday.
Show that
$$\lim_{x \to 0^-} f\left(\frac1x\right)= \lim_{x \to -\infty} f(x)$$
We can see intuitively that this makes sense: as $x$ gets very small negative on the left hand side, $1/x$ becomes very large negative, and the 2 sides both look like $f$ of some very large negative number.
Let's assume that the first limit exists, and is equal to $\ell$
$$\lim_{x \to 0^-} f\left(\frac1x\right)= \ell$$
If this is true, then from the definition of the limit at $0$ from below, we have: for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$, if
$$0 < -x < \delta, \text{then} \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$
or, multiplying the $\delta$ expression by $-1$, we have for all $x$, if
$$0 > x > -\delta, \text{then} \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$
Let's pause for a moment and consider what this says. Here, $x$ is just some number. This says that, if we have some number between $0$ and $-\delta$, then $f(\frac{1}{\text{number}})$ will be "sufficiently close" to $\ell$.
Let's try sticking $1/y$ in as "number". The motivation for this choice is that we have an expression involving $\left\vert f\left(\frac{1}{x}\right) - \ell \right\vert$ and we'd like to instead have an expression for $f(x)$. Sticking $1/y$ into the above expression for $x$ we have
$$\text{If } 0 > 1/y > -\delta \text{ then } \left\vert f\left(\frac{1}{(1/y)}\right) - \ell \right\vert < \varepsilon$$
or
$$\left\vert f(y) - \ell \right\vert < \varepsilon$$
Again, the original $x$ expression was true for all numbers $x$ that satisfied the $\delta$ condition on the left hand side, so this new expression for $f(y)$ is true for all numbers $1/y$ that satisfy $0 > 1/y > -\delta$.
Now if $0 > 1/y > -\delta$, what does this tell us about $y$?
Multiplying by $-1$ to make everything nonegative, to avoid any sign confusion:
$$0< -1/y < \delta$$
Inverting
$$0<1/\delta<-y$$
Finally, multiplying again by $-1$ we have
$$0 > -1/\delta > y$$
Let's summarize what this all implies:
We know that if the first limit exists, then for any $\varepsilon >0$ there exists some $\delta>0$ such that for all $y$ if
$$y < -1/\delta \text{ then } \left\vert f(y) - \ell \right\vert < \varepsilon$$
Or, substituting $N = -1/\delta$ we have, for any $\varepsilon >0$ there exists some $N<0$ such that for all $y$, if
$$y < N \text{ then } \left\vert f(y) - \ell \right\vert < \varepsilon$$
Another way of writing this is:
$$\lim_{y \to -\infty} f(y) = \ell$$
Thus, if the first limit exists, then so too does the second, and they are equal.
A key point here is the equivalence of the conditions $y < -1/\delta < 0$ and $0 > 1/y > -\delta$. The first implies the second and vice versa.
Finally to complete the proof, we can begin with the second limit and similarly show that if it exists, then so too does the first, and they are equal.
The steps will be very similar to what we've already done.
Alternate justification for substituting in $y = 1/x$
Going back to the beginning, we see that the first limit involves $f(1/x)$. What do the $\delta$-restrictions tell us about $1/x$?
If the first limit is $\ell$ then, for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$ if
$$0 < -x < \delta, \text{then} \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$
If
$$0 < -x < \delta$$
then
$$0 < 1/\delta <\frac{1}{-x}$$ or
$$0 > -1/\delta >\frac{1}{x}$$
Thus, for any $\varepsilon > 0$ there exists some $N = -1/\delta < 0$ such that for all $x$, if
$$\frac{1}{x}< N \text{ then } \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$
We can see here we're almost to the second limit. We just need to replace the number $\frac{1}{x}$ with $x$ (or $y$, or any other name).
