# Proving $\lim_{x\to a}f(x)$ and $\lim_{h\to 0}f(a+h)$ are equivalent using $\epsilon$-$\delta$ argument

In Spivak's Calculus, he asks for a proof that $\lim_{x\to a}f(x)=\lim_{h\to 0}f(a+h)$.

He first shows that the existence of $\lim_{x\to a}f(x)$ implies the existence and equivalence of $\lim_{h\to0}f(a+h)$, and then he says the argument for the other direction is "similar," but I am having a hard time replicating it (I may be getting unnecessarily bogged down in notational issues). His proof of the first direction is essentially as follows:

(Spivak forward direction): Let $\ell=\lim_{x\to a}f(x)$ and define $g(h)=f(a+h)$. Then for every $\epsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-\ell|<\epsilon$. Now, if $0<|h|<\delta$, then $0<|(a+h)-a|<\delta$, so $|f(a+h)-\ell|<\epsilon$, which we can write as $|g(h)-\ell|<\epsilon$. Thus, $\lim_{h\to0}g(h)=\ell$, which can also be written $\lim_{h\to 0}f(a+h)=\ell$.

The same sort of argument shows that if $\lim_{h\to 0}f(a+h)=m$, then $\lim_{x\to a}f(x)=m$. So either limit exists if the other does, and in this case they are equal.

My attempt at other direction: Let $m=\lim_{h\to 0}f(a+h)$. Then for every $\epsilon>0$ there is a $\delta>0$ for all $h$ such that if $0<|h|<\delta$ then $|f(a+h)-m|<\epsilon$. Now, if $0<|x-a|<\delta$, then $|f(a+(x-a))-m|=|f(x)-m|<\epsilon$. Thus, $\lim_{x\to a}f(x)=m$.

What am I missing here? Is my proof okay? Why does Spivak use the function $g$ in the previous direction? Is it really necessary? What would such a $g$ be in the other direction?

• I wrote up a short proof here that does not make use of an auxiliary function $g$. Can anyone attest to its correctness or lack thereof? – interrogative Dec 10 '17 at 19:34

• In the classical $\varepsilon-\delta$ definition of limits, for educational purposes, one writes $\dots |f(x)-\ell|<\varepsilon$: there's only one variable $x$ inside the bracket. The author relabeled $f(a+h)$ as $g(h)$ so that
• readers understand it's a function of $h$ with $a$ fixed.
• the form $|g(h)-\ell|<\varepsilon$ looks like the book definition more like $|f(a+x)-\ell|<\varepsilon$. Though they both represent the same quantity, the author preferred the former for newcomers.
• Since the purpose of the auxiliary function is to simplify what's inside the bracket into one single variable, so that the relabeled function resembles symbolically a bit more to the classic $\varepsilon-\delta$ definition of limits. Therefore, I don't think changing the direction of the proof will change the way that the auxiliary function $g$ is defined. We still have $g(x)=f(a+x)$ in the other half of the proof. The difference is that in the forward direction of the proof, $g$ lives in "another side of the river", while in the you "see" $g$ at the very first moment.