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I'm trying to prove that $$\lim_{x\to0}f(x) = \lim_{x\to0}f(x^3)$$ (The domain is not specified and neither the continuity, so it really is only about the limit of an arbitrary function)..

I'm guessing I need to simply use the definition. However I'm not sure where to start.

Here is what I had in mind, however I feel like it's not at all how it should be proved.

Proof

Let $\lim_{x\to0}f(x) = L$.

Then, we have that

$$ \forall \epsilon > 0, \exists \delta>0 \text{ such that whenever } 0<|x|< \delta \Rightarrow |f(x) - L| < \epsilon$$

Let $y = x^3$, then we want to show that $$|f(y) - L| < \epsilon \text{ whenever } 0<|x| = |y^{\frac{1}{3}}| < \delta$$

And here is where I get stuck. Is there any more efficient way to prove this?

Thank you!

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Asserting that $\lim_{x\to0}f(x)=L$ means, as you wrote, that$$(\forall\varepsilon>0)(\exists\delta>0):\lvert x\rvert<\delta\implies\bigl\lvert f(x)-L\bigr\rvert<\varepsilon.$$So, take $\delta^\star=\sqrt[3]\delta$ and then$$\lvert x\rvert<\delta^\star\implies\lvert x^3\rvert<\delta\implies\bigl\lvert f(x^3)-L\bigr\rvert<\varepsilon.$$In other words, $\lim_{x\to0}f(x^3)=L$. Can you do it in the opposite direction now?

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As an alternative we can proceed by composite function, assuming that $\lim_{x\to0}f(x)=L$ it means that

$$\forall\epsilon>0\quad \exists\delta>0\quad |x|<\delta\quad \bigl\lvert f(x)-L\bigr\rvert<\epsilon$$

Now consider $g(x)=x^3$ and we have that $\lim_{x\to0}g(x)=0$ that is

$$\forall\epsilon_1>0\quad \exists\delta_1>0\quad |x|<\delta_1\quad\bigl\lvert x^3\bigr\rvert<\epsilon_1$$

then if we assume $\delta=\epsilon_1$ we have that

$$\forall\epsilon>0\quad \exists\delta_1>0\quad |x|<\delta_1\quad \bigl\lvert f(g(x))-L\bigr\rvert<\epsilon$$

that is

$$\lim_{x\to0}f(g(x))=\lim_{x\to0}f(x^3)=L$$

and similarly for the opposite direction.

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