Given $f:\mathbb{R}\to\mathbb{R}$ and number $b$, prove the statements $\lim_{x\to 0}f(x)=b$ and $\lim_{x\to 0}f(x^3)=b$ are equivalent

Question

Given are $f: \mathbb{R} \to \mathbb{R}$ and a real number $b\in \mathbb{R}$. How can I prove that the statements $\lim_{x\to 0}f(x)=b$ and $\lim_{x\to 0}f(x^3)=b$ are equivalent?

So far I just have that I need to prove that if

1) $\lim_{x\rightarrow 0}f(x)=b$ then $\lim_{x\rightarrow 0}f(x^3)=b$

2) $\lim_{x\rightarrow 0}f(x^3)=b$ then $\lim_{x\rightarrow 0}f(x)=b$

And the definitions, so that if $\lim_{x\rightarrow 0}f(x)=b$ is true, then for $x\in\operatorname{dom}(f)$ and $|x-0|< \delta$ then $|f(x)-b|<\epsilon$

And if $\lim_{x\rightarrow 0}f(x^3)=b$ is true, then for each then for $x\in\operatorname{dom}(f)$ and $|x^3-0|< \delta $ then $|f(x^3)-b|<\epsilon$

Answer 1

This is a particular case of a more general situation: variable substitution in limits when the substitution is continuous and invertible.

Anyway, for your case it's not necessary to use the full strength of the argument.

Suppose $\lim_{x\to0}f(x)=b$. We want to show that also $\lim_{x\to0}f(x^3)=b$. So, let's take $\varepsilon>0$. We need to find $\delta>0$ such that $0<|x|<\delta$ implies $|f(x^3)-b|<\varepsilon$.

Start by choosing $\delta_0$ such that, for $0<|x|<\delta_0$, $|f(x)-b|<\varepsilon$. Take now $\delta=\sqrt[3]{\delta_0}$. Then $$ 0<|x|<\delta \implies 0<|x^3|<\delta_0 \implies |f(x^3)-b|<\varepsilon $$

Now do the converse.

Answer 2

$$\lim_{x\to 0}f(x) = b$$ means that for any $\epsilon>0$, we can find a $\delta > 0$ so that whenever $|x-0|<\delta$, $|f(x)-b|<\epsilon$.

If we take $t^3=x$, then whenever $|t-0|\leq \delta^{1/ 3}$ it follows that $|x-0|\leq\delta$ and $$|f(x)-b|=|f(t^3)-b|< \epsilon$$ i.e., $$\lim_{t\to 0}f(t^3) = b$$

Arguing the other way is pretty much the exact same.