Prove that $\lim_{x→0} f(x) = A$. Suppose that $\lim_{x→0} f(x^3) = A$, where $A$ is a constant number. Prove
that $\lim_{x→0} f(x) = A$.
The domain of f is R. But that is the only information provided.
I am completely lost on how to approach this. Any help would be greatly appreciated!
 A: Suppose that $\lim_{x\to0}f(x)\neq A$ to get a contradiction. Then there exist a sequence $x_n$ converging to $0$ as $n\to\infty$ and such that $f(x_n)\not\to A$ as $n\to\infty$. Letting $y_n=x_n^{1/3}$ this can be reworded into $\lim_{n\to\infty}f(y_n^3)\neq A$, which imply that $\lim_{x\to0}f(x^3)\neq A$ since $y_n\to0$ as $n\to\infty$ (by the continuity of $x\mapsto x^3$).
A: Let $y=x^{\frac{1}{3}}$, we have that $f(y^3)=f(x)$.
Since $\lim_{y\rightarrow 0} f(y^3)=A$ we have that, for every $\varepsilon>0$ there exists $\delta=\delta(\varepsilon)$ such that, whenever $|y|<\delta$, $|f(x)-A|=|f(y^3)-A|<\varepsilon$. 
Now note that function $x\mapsto x^{\frac{1}{3}}$ is continuous, so for any $\delta>0$ there exists $\zeta=\zeta(\delta)$ such that $|y|=|x^{\frac{1}{3}}|<\delta$ whenever it holds that $|x|<\zeta$. 
Now piece these two things together.
A: Let $y=x^{1/3}$. We have:
$$lim_{x\rightarrow 0}x^3=lim_{y\rightarrow 0}y=A$$
Knowing that $f(x),f(y)\in\mathbb{R}$ and $y=x^{1/3}$, by swapping y with x we have $$lim_{x\rightarrow 0}x=A$$
A: Let $\varepsilon > 0$. By definition, exists $\tilde{\delta} > 0$ such that if $x \in \mathbb{R}$, with $|x| < \tilde{\delta}$, then $|f(x) - A| < \varepsilon$. Let $\delta = {\tilde{\delta}}^{\frac{1}{3}}$. Then $\delta > 0$. Further, if $x \in \mathbb{R}$, with $|x| < \delta$, then $|f(x) - A| < \varepsilon$.
