For example to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{3})$
Assume $\lim_{x\to0}f(x) = l$ that is $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\forall x$: $0 < |x| < \delta_{1} \implies |f(x)-l|$
Then to show $\lim_{x\to0}f(x^{3}) = l$ which is by replacing $x$ with $x^{3}$ in the assumption: that is $0 < |x| < |x^{3}| < \delta_{2} \implies |f(x^{3})-l|$ for $x< 0$ and $x>1$ and
$0 < |x^{3}| < |x| < 1 \implies |f(x^{3})-l|$ for $0 < x < 1$
that is choosing $\delta_{2} = min(1, \delta_{1})$
thus showing: $\lim_{x\to0}f(x^{3}) = l$
Why would this argument not work if an attempt was made to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{2})$ which by my question here: Give an example about limits: $\lim\limits_{x\to0}f(x^{2})$ exists but $\lim\limits_{x\to0}f(x)$ does not. obviously would not work. Or is there something I overlooked in this proof?
Thanks.