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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!

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    Add more context, please. What do we know about $f$? – 2017-02-07
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    @theSongbird that's the only information provided, apart from that the domain is R. – 2017-02-07
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    Possible duplicate of [Prove that if one of $\lim_{x \rightarrow 0 }{f(x)}$ and $\lim_{x \rightarrow 0}{f(x^3)}$ exists, then the other one also exists.](http://math.stackexchange.com/questions/332859/prove-that-if-one-of-lim-x-rightarrow-0-fx-and-lim-x-rightarrow-0) See also http://math.stackexchange.com/questions/627946/how-to-prove-that-lim-x-to-0fx-l-is-equivalent-to-lim-x-to-0fx3-l ; http://math.stackexchange.com/questions/928725/prove-that-the-lim-x-rightarrow-0fx-b-is-equivalent-to-the-lim-x-rig ; http://math.stackexchange.com/q/211066/147873 – 2017-02-07

4 Answers 4

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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 $x\mapsto x^3$).

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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.

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    How would you piece those two things together though? – 2017-02-08
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    It's pretty straightforward that those two statements imply that for any $\varepsilon>0$ there exists $\zeta>0$ such that $|f(x)-A|<\varepsilon$ whenever $|x|<\zeta$, which is what you want to prove. – 2017-02-09
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    Just look closely at them. – 2017-02-09
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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 $$lim_{x\rightarrow 0}x=A$$

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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 $|f(x) - A| < \varepsilon$.