Questions on Spivak's proof that making x close to a is the same as making $x - a$ close to 0.

Question

1. On my first attempt, I started with the left side of the question, $\lim\limits_{x \rightarrow 0} f(x)$. I defined it as $L$. I was befuddled after 20 minutes.

On my second attempt, I defined the right side of the question, $\lim\limits_{x \rightarrow a} f(x-a)$. Again, I got befuddled for another 20 mins.

I didn't commence with $\lim\limits_{x \rightarrow a} f(x) = L$ because it doesn't appear in the question. Isn't it more natural to start with what appears in the question, $\lim\limits_{x \rightarrow 0} f(x)$ or $\lim\limits_{x \rightarrow a} f(x-a)$? How do I do this?

2. I know $|y|$ means absolute value of y, but what does $y$ represent here? $y$ doesn't appear in the question.

3. Spivak "let $g(x) = f(x - a)$". Then $g(y) = f(y - a)$. Not $f(y + a)$?

Spivak, Calculus 2008 4 edn. His website's errata lists no errata for these pages. I haven't MathJax'ed this to avoid typos.

Answer 1

Let's go through the proof again, a bit more carefully.

Suppose ${\lim_{x\to 0}f(x)=L}$. We will show ${\lim_{x\to a}f(x-a)}$ also ${=L}$.

By definition of the first limit, ${\forall\ \epsilon>0\ \exists\ \delta>0\ |\ |x|<\delta\Rightarrow |f(x)-L|<\epsilon}$.

Let ${y=x+a}$. Notice that ${|x|=|(x+a)-a|=|y-a|}$. So we overall now have, with our new variable $y$ defined, that equivalently ${\forall\ \epsilon>0\ \exists\ \delta>0\ |\ |y-a|<\delta\Rightarrow|f(y-a)-L|<\epsilon}$ (all we have done is write the first limit definition with $y$ instead of $x$ being used). This is equivalent to saying $$ \lim_{y\to a}f(y-a)=L $$ which is what we wanted to show.

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More generally, if you have some function ${c(x)}$ continuous at ${x=a}$, you have that $$ \lim_{x\to c(a)}f(x)=\lim_{x\to a}f(c(x)) $$