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.


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!


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?


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


and similarly for the opposite direction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.