# Trouble Understanding Difference in Epsilon-Delta Arguments: Why One Works But The Other Fails (Spivak Calculus Problem 5-10c)

In the problems for the limits chapter (5) of Spivak's Calculus, we are asked to prove: $$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} f(x^3)$$. The relevant to the question proof alternative is:

If we assume that $$\lim_{x \to 0} f(x^3)$$ exists, say $$\lim_{x \to 0} f(x^3) = m$$, then for all $$\epsilon > 0$$ there is a $$\delta > 0$$ such that if $$0 < |x| < \delta$$, then $$|f(x^3) - m| < \epsilon$$. Then if $$0 < |x| < \delta^3$$, we have $$0 < |\sqrt[3]{x}| < \delta$$, so $$|f((\sqrt[3]{x})^3) - m| < \epsilon$$, or $$|f(x) - m| < \epsilon$$. Thus $$\lim_{x \to 0} f(x) = m$$.

I thought I understood the proof for this, but then I saw the next question: "Give an example of a case when $$\lim\limits_{x \to 0} f(x^2)$$ exists, but $$\lim\limits_{x \to 0} f(x)$$ does not. (This is true, for example, if $$f(x) = 1,\text{for }x>0; -1, \text{for } x<0$$). The reason I now believe I do not understand the proof, is that I see no reason the proof cannot be repeated to show the same for $$x^2$$ as for $$x^3$$. Clearly though, from the counter-example above, the same result does not hold for $$x^2$$.

My question is, where in the logic of the proof Spivak gives, can the process not be followed to prove that if $$\lim\limits_{x \to 0} f(x^2) = l$$, then $$\lim\limits_{x \to 0} f(x) = l$$? As in, which step fails in the method when attempting the same proof with $$x^2$$ instead of $$x^3$$?

Any help is greatly appreciated!

It's true that $$(\sqrt[3]{x})^3=x$$, but $$\sqrt{x}$$ doesn't even make sense for $$x<0$$, and hence you can write $$(\sqrt x)^2=x$$ only for $$x>0$$.
For each $$x\in\Bbb R$$, $$\sqrt[3]x$$ exists (and this was used in the proof). But it is not true that every real number has a square root; only non-negative real numbers have it.