# epsilon delta – Spivak Calculus Chapter 5 Problem 10 (b)

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:

1. $$Rightarrow$$“: Let $$limlimits_{x to 0} f(x)=m$$. Then for every $$varepsilon>0$$ there is a $$delta>0$$ such that, for all $$x$$, if $$0, then $$lvert f(x)-mrvert. Now let $$0 (which can als be written $$0), then $$lvert f(x-a)-mrvert. Thus $$limlimits_{x to a} f(x-a)=m$$.
2. $$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, then $$lvert f(x-a)-mrvert. Now let $$0. ($$0.) This inequality can be written $$0. It follows that $$lvert f((y+a)-a)-mrvert, which is $$lvert f(y)-mrvert. 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 $$0, then $$lvert f(x)-lrvert. Now, if $$0, then $$0, so $$lvert f(y+a)-lrvert. But this last inequality can be written $$lvert g(y)-lrvert. 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 $$lvert f(y+a)-lrvert be written $$lvert g(y)-lrvert? (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…