When proving that $$\lim_{x \to 0} f(x)=\lim_{x \to 0}f(x^3)$$ does it suffice to show that as $x \to 0, x^3 \to 0$ ? If so then by the epsilon-delta proof we have:$\lim_{x \to 0} x^3=0$
$|x^3-0|=|x|^3<ε$ $\implies$ $|x|<\frac {ε}{x^2}$. Suppose we want $|x-0|<1 \iff 0<x^2<1 \iff \frac{1}{x^2}>1$.
Is this so far correct? If so then how do we proceed to find the proper δ?
Yes, it would suffice to show that $\lim_{x\to 0} x^3 = 0$ and use a theorem about the limit of the composition of continuous functions. A $\delta-\epsilon$ proof would go something like this:
Suppose $\lim_{x\to 0} f(x) = L$ where $L$ is finite. Let $\epsilon > 0$. Then by the definition of the limit, there is a $\hat\delta > 0$ such that for all $x$ satisfying $|x| < \hat\delta$, we have $|f(x) - L|<\epsilon$. Set $\delta = (\hat\delta)^{1/3} > 0$. Then if $|x| < \delta = \hat\delta^{1/3}$, $|x^3|<\hat\delta$, so $|f(x^3) - L| < \epsilon$, hence $\lim_{x\to 0} f(x^3) = L$.