Ytukyg

I'm trying to prove that $$lim_{xto0}f(x) = lim_{xto0}f(x^3)$$ (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 $lim_{xto0}f(x) = L$.

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!

Asserting that $lim_{xto0}f(x)=L$ means, as you wrote, that$$(forallvarepsilon>0)(existsdelta>0):lvert xrvert<deltaimpliesbigllvert f(x)-Lbigrrvert<varepsilon.$$So, take $delta^star=sqrt[3]delta$ and then$$lvert xrvert<delta^starimplieslvert x^3rvert<deltaimpliesbigllvert f(x^3)-Lbigrrvert<varepsilon.$$In other words, $lim_{xto0}f(x^3)=L$. Can you do it in the opposite direction now?

As an alternative we can proceed by composite function, assuming that $lim_{xto0}f(x)=L$ it means that

$$forallepsilon>0quad existsdelta>0quad |x|<deltaquad bigllvert f(x)-Lbigrrvert<epsilon$$

Now consider $g(x)=x^3$ and we have that $lim_{xto0}g(x)=0$ that is

$$forallepsilon_1>0quad existsdelta_1>0quad |x|<delta_1quadbigllvert x^3bigrrvert<epsilon_1$$

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

$$forallepsilon>0quad existsdelta_1>0quad |x|<delta_1quad bigllvert f(g(x))-Lbigrrvert<epsilon$$

that is

$$lim_{xto0}f(g(x))=lim_{xto0}f(x^3)=L$$

and similarly for the opposite direction.