I am unsure about how to solve this problem and unfortunately I do not understand the given solution. The problem goes as follows (Spivak, Calculus, 3rd edition, pg. 107, Problem 10-(b)):
Prove that $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to a} f(x-a)$.
My attempt:
- "$\Rightarrow$": Let $\lim\limits_{x \to 0} f(x)=m$. Then for every $\varepsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<\lvert x-0\rvert<\delta$, then $\lvert f(x)-m\rvert<\varepsilon$. Now let $0<\lvert (x-a)-0\rvert<\delta$ (which can als be written $0<\lvert x-a\rvert<\delta$), then $\lvert f(x-a)-m\rvert<\varepsilon$. Thus $\lim\limits_{x \to a} f(x-a)=m$.
- "$\Leftarrow$": Let $\lim\limits_{x \to a} f(x-a)=m$. Then for every $\varepsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<\lvert x-a\rvert<\delta$, then $\lvert f(x-a)-m\rvert<\varepsilon$. Now let $0<\lvert y\rvert<\delta$. ($0<\lvert y-0\rvert<\delta$.) This inequality can be written $0<\lvert (y+a)-a\rvert<\delta$. It follows that $\lvert f((y+a)-a)-m\rvert<\varepsilon$, which is $\lvert f(y)-m\rvert<\varepsilon$. Thus $\lim\limits_{y \to 0} f(y)=m$ (which is equivalent to $\lim\limits_{x \to 0} f(x)=m$).
Spivak's solution:
Suppose that $\lim\limits_{x \to a} f(x)=l$, and let $g(x)=f(x-a)$. Then for all $\varepsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<\lvert x-a\rvert<\delta$, then $\lvert f(x)-l\rvert<\varepsilon$. Now, if $0<\lvert y\rvert<\delta$, then $0<\lvert (y+a)-a\rvert<\delta$, so $\lvert f(y+a)-l\rvert<\varepsilon$. But this last inequality can be written $\lvert g(y)-l\rvert<\varepsilon$. So $\lim\limits_{y \to 0} g(x)=l$. The argument in the reverse direction is similar.
My troubles:
Why does he start with $\lim\limits_{x \to a} f(x)=l$? Why can $\lvert f(y+a)-l\rvert<\varepsilon$ be written $\lvert g(y)-l\rvert<\varepsilon$? (If $g(x)=f(x-a)$, then shouldn't $f(y+a)=g(y+2a)$?) It looks like he is trying to prove $\lim\limits_{x \to a} f(x)=\lim\limits_{x \to 0} f(x-a)$, which is very weird…
Best Answer
Indeed, Spivak makes a mistake in his argument: the inequality $$ |f(y+a)-l|<\epsilon $$ can not be written as $$ |g(y)-l|<\epsilon $$ The definition of $g$ is $g(y)=f(y-a)$, not $g(y)=f(y+a)$.
This error is also in the fourth edition of his book.
Once you understand the underlying ideas and can write down a formal proof correctly, the "solution" is not so important. In textbooks, authors occasionally make mistakes; sometimes, when they make an argument $X$ to show statement $Y$, they may really mean to write argument $X'$.
The spirit of the proof for this problem is in Spivak's comment in his solution:
Your attempt is correct. I may change your wording slightly though.
Suppose $\lim_{x\to 0}f(x)=m$. We want to show that $\lim_{x\to a}f(x-a)=m$. Let $\epsilon>0$. There exists $\delta>0$ such that $$ \textrm{for every $x$ with } 0<|x-0|<\delta,\quad|f(x)-m|<\epsilon. \tag{1} $$ If $0<|y-a|=|(y-a)-0|<\delta$, then (1) implies that $|f(y-a)-m|<\epsilon$. Thus $ \lim_{y\to a}f(y-a)=m, $ i.e. $ \lim_{x\to a}f(x-a)=m. $
Conversely, suppose $ \lim_{x\to a}f(x-a)=m. $ We want to show that $\lim_{x\to 0}f(x)=m$. Let $\epsilon>0$. There exists $\delta>0$ such that $$ \textrm{for every $x$ with } 0<|x-a|<\delta,\quad|f(x-a)-m|<\epsilon. \tag{2} $$ If $0<|y|=|(y+a)-a|<\delta$, then (2) implies that $|f(y)-m|=|f((y+a)-a)-m|<\epsilon$. Hence $\lim_{y\to 0}f(y)=m$.