# Equivalence of limits, Spivak Chapter 5 Number 10b — $\lim\limits_{x \to a} f(x)$ $\iff$ $\lim\limits_{x \to 0} f(x-a)$

Good day,

I want to apologize because I made this post wrong last time. I am going to ask you again. I hope this time will be better.

I am trying to solve an exercise from the Spivak book, but I think there's an error in the theorem.

We are asked to prove the following:

$$\lim\limits_{x \to a} f(x)$$ $$\iff$$ $$\lim\limits_{x \to 0} f(x-a)$$

To test the theorem I take two examples:

First: $$f(x)=x^2$$ and $$a=2$$

$$\lim\limits_{x \to a} f(x)$$ $$\iff$$ $$\lim\limits_{x \to 2} x^2=4$$

$$\lim\limits_{x \to 0} f(x-a)$$ $$\iff$$ $$\lim\limits_{x \to 0} (x-2)^2=4$$

Perfect, this work for this function.

Second: $$f(x)=x^3$$ and $$a=2$$

$$\lim\limits_{x \to a} f(x)$$ $$\iff$$ $$\lim\limits_{x \to 2} x^3=8$$

$$\lim\limits_{x \to 0} f(x-a)$$ $$\iff$$ $$\lim\limits_{x \to 0} (x-2)^3=-8$$

So in the second example the theorem is not true, and I think there is an error in the book.

Could anyone tell me if there is an error in the analysis I made?

• What problem is it in Spivak? Are there other pre-requisites on $f$? Can you type the complete problem? – John Douma Mar 26 at 13:32
• @JulianTorres I don't see that problem as $10b$ in the version of Spivak I found online but problem $9$ uses $l$ instead of $a$ and has it outside the function, i.e. $f(x)-a$, not $f(x-a)$. See problem $9$ of chapter $5$ of ia801906.us.archive.org/29/items/Calculus_643/… – John Douma Mar 26 at 18:05
It should be $$f(x+a)$$. You are right, you found a counterexample in the other case. You could have used $$f(x)=x$$.