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!
Best Answer
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?