Spivak Caculus 4th edition wrong answer on chapter 5 (limits prove)?

Question

It's chapter 5 on page 109, since I suspect it's not right, I will write the whole question and its answer given by the book here:

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question 10 (b). Prove that: $$\lim_{x\to 0}f(x)=\lim_{x\to a}f(x-a)$$

The answer given on page 626 is:

Intuitively, making $x$ close to $a$ is the same as making $x-a$ close to $0$. Formally: Suppose that $\lim_{x\to a}f(x)=l$, and let $g(x)=f(x-a)$.Then for all $\epsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$. Now, if $0<|y|<\delta$, then $0<|(y+a)-a|<\delta$, so$|f(y+a)-l|<\epsilon$. But this last inequality can be written $|g(y)-l|<\epsilon$. So $\lim_{y\to 0}g(y)=l$. The argument in the reverse direction is similar.

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I think this answer only proved $$\lim_{x\to a}f(x)=\lim_{x\to 0}f(x+a)$$,not the given problem. If it's what I think, can you give me a formal proof please?

Answer 1

You are right. The argument starts by considering $\lim_{x\to a}f(x)$ which is not part of the equation. It should have started by defining $g(x) = f(x-a)$ and assuming $\lim_{x\to a}g(x) = l$ (i.e. by considering the RHS of the equation).

Then for all $\epsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|g(x)-l|<\epsilon$. Now, if $0<|y|<\delta$, then $0<|(y+a)-a|<\delta$, so $|g(y+a)-l|<\epsilon$. But this last inequality can be written as $|f(y)-l|<\epsilon$. So $\lim_{y\to 0}f(y)=l$.

Answer 2

You have proved that $g(x)$ converges to $l$ when $x$ goes to $0$. However, $g(x) = f(x-a)$. So you have correctly the limit $l$ of $f(x-a)$ when $x$ goes to $a$.

Also you've stated that the answer proved something else, but what was proved is strictly equivalent to the proposition that you want to prove. Make the change of variable $y = x+a$.

The LHS will become the RHS of what you want to prove, and the RHS will become the LHS of what you want to prove.