Prove that $\lim_{x\to0}f(x)=\lim_{x\to0}f(x^3)$ I'm not really sure how to do this. Is $(1)$ the way to go?
$$\lim_{x\to0}f(x)-f(x^3)=0 \tag 1$$
What I tried to prove was $(2)$ with the stipulation that $x:=t^3$.
$$\lim_{x\to0}f(x) = L \iff \lim_{t\to0}f(t^3)=L \tag 1$$
Left to right: we know $|f(x)-L|<\epsilon$ when $0<|x|<\delta$. Let $\delta':=\min\{1/2, \delta^3\}$, so that if $|x|<\delta'$, then $|t|<\delta$. Then $|f(t^3)-L|<\epsilon$.
Right to left: we know $|f(t^3)-L|<\epsilon$ when $0<|t|<\delta$. Let $\delta':=\min\{1/2, \delta\}$, so that if $|t|<\delta'$, then $|x|<\delta$. Then $|f(x)-L|<\epsilon$.
Is this correct? Did I overcomplicate things?
 A: We can do it with sequences:
$$L:=\lim_{x \to 0} f(x)$$
This means that for all non-zero null-sequences $(x_n)$, we have that
$$f(x_n) \to L$$
Let $(x_n)$ be a given non-zero null-sequence. We want to prove that $$f(x_n^3) \to L$$
But since $y_n=x_n^3$ is a non-zero null-sequence as well, we have that
$$f(y_n) \to L$$
A: Here is what you can prove:

Let $f$ be a real valued function defined in some deleted neighborhood of $0$. Then the limiting behavior of $f(x^3)$ as $x\to 0$ is exactly the same as that of $f(x) $ as $x\to 0$.

The limiting behavior of a real valued function $f(x) $ defined in some deleted neighborhood of $a$ as $x\to a$ is one of the following types:


*

*$f(x) \to L$ as $x\to a$.

*$f(x) \to\infty $ as $x\to a$.

*$f(x) \to-\infty $ as $x\to a$.

*$f(x) $ oscillates finitely as $x\to a$.

*$f(x) $ oscillates infinitely as $x\to a$.


For your current problem if $f(x) $ exhibits one of above mentioned limiting behavior as $x\to 0$ then $f(x^3)$ also exhibits the same limiting behavior as $x\to 0$. And the above list is exclusive and exhaustive the converse also follows.
You can prove that

Corresponding to every deleted neighborhood $A$ of $0$ there is a deleted neighborhood $B$ of $0$ such that $f(A) =g(B) $ where $g(x) =f(x^3)$ and vice versa.

By definition of limit the result now follows. 
