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?

Thanks for your help.

  • $\begingroup$ You have made the post wrong this time as well. $\endgroup$ – Akash Yadav Mar 26 at 13:25
  • $\begingroup$ Imagine if everyone made a title like yours. It'd be really difficult to find things! $\endgroup$ – Shaun Mar 26 at 13:27
  • $\begingroup$ What problem is it in Spivak? Are there other pre-requisites on $f$? Can you type the complete problem? $\endgroup$ – John Douma Mar 26 at 13:32
  • $\begingroup$ I have changed the title to your comfort $\endgroup$ – Julian Torres Mar 26 at 13:37
  • $\begingroup$ @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/… $\endgroup$ – 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 $f(x)=x$.

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  • $\begingroup$ Thanks for your help $\endgroup$ – Julian Torres Mar 26 at 14:36

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