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 $limlimits_{x to 0} f(x)=limlimits_{x to a} f(x-a)$.

My attempt:

- “$Rightarrow$“: Let $limlimits_{x to 0} f(x)=m$. Then for every $varepsilon>0$ there is a $delta>0$ such that, for all $x$, if $0<lvert x-0rvert<delta$, then $lvert f(x)-mrvert<varepsilon$. Now let $0<lvert (x-a)-0rvert<delta$ (which can als be written $0<lvert x-arvert<delta$), then $lvert f(x-a)-mrvert<varepsilon$. Thus $limlimits_{x to a} f(x-a)=m$.
- “$Leftarrow$“: Let $limlimits_{x to a} f(x-a)=m$. Then for every $varepsilon>0$ there is a $delta>0$ such that, for all $x$, if $0<lvert x-arvert<delta$, then $lvert f(x-a)-mrvert<varepsilon$. Now let $0<lvert yrvert<delta$. ($0<lvert y-0rvert<delta$.) This inequality can be written $0<lvert (y+a)-arvert<delta$. It follows that $lvert f((y+a)-a)-mrvert<varepsilon$, which is $lvert f(y)-mrvert<varepsilon$. Thus $limlimits_{y to 0} f(y)=m$ (which is equivalent to $limlimits_{x to 0} f(x)=m$).

Spivak’s solution:

Suppose that $limlimits_{x to a} f(x)=l$, and let $g(x)=f(x-a)$. Then for all $varepsilon>0$ there is a $delta>0$ such that, for all $x$, if $0<lvert x-arvert<delta$, then $lvert f(x)-lrvert<varepsilon$. Now, if $0<lvert yrvert<delta$, then $0<lvert (y+a)-arvert<delta$, so $lvert f(y+a)-lrvert<varepsilon$. But this last inequality can be written $lvert g(y)-lrvert<varepsilon$. So $limlimits_{y to 0} g(x)=l$. The argument in the reverse direction is similar.

My troubles:

Why does he start with $limlimits_{x to a} f(x)=l$? Why can $lvert f(y+a)-lrvert<varepsilon$ be written $lvert g(y)-lrvert<varepsilon$? (If $g(x)=f(x-a)$, then shouldn’t $f(y+a)=g(y+2a)$?) It looks like he is trying to prove $limlimits_{x to a} f(x)=limlimits_{x to 0} f(x-a)$, which is very weird…