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idea of the proof of fact that $\lim_{x\to 0}f(x) = \lim_{x\to 0}f(x^3)$:

By delta-epsilon definition of limits, $$\forall \epsilon \gt0 \exists \delta \gt 0 \forall x:0 < |x| < \delta \implies |f(x)-l|<\epsilon, $$
where $ l = \lim_{x\to 0}f(x)$.

I also know $\forall \epsilon \exists \delta_1 >0$ such that $\forall x:0<|x|<\delta_1 \implies |x^3| < \epsilon$

Then if I can manage to show $0 < |x^3| < \delta$, then this would imply $|f(x^3)-l|<\epsilon$ (this is the part I was unsure about; why am I allowed to do this). Then I find the corresponding $\delta_1$ for $\epsilon = \delta$ to show that $\forall x:0<|x|<\delta_1 \implies |x^3| < \delta$, and the rest follows.

My big two questions are:

1) why am I allowed to "replace" $x$ with $x^3$?

2) why does this method not work in trying to prove $\lim_{x\to 0}f(x) = \lim_{x\to 0}f(x^2)$, because could I not just replace $x$ with $x^2$?

All help is appreciated!

Saran Wrap
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    Regarding your 2) we can say that $\lim_{x\to 0}f(x)=L\implies \lim_{x\to 0}f(x^2)=L$ but reverse implication does not work. – Paramanand Singh Nov 05 '19 at 02:59
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    In your above proof you deal with $\lim_{x\to 0}f(x)=L\implies \lim_{x\to 0}f(x^3)=L$. The reverse implication is true here but your proof does not deal with it. Thus your proof also works in exactly the same manner for $x^2$ for the forward implication (given in my previous comment). – Paramanand Singh Nov 05 '19 at 03:26

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