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 $|f(x^3)-b|<\epsilon$

  • 1
    $\begingroup$ Try using Cauchy criteria $\endgroup$
    – Itay4
    Apr 26, 2017 at 8:16
  • $\begingroup$ @Itay4 Thanks! In my case I would want to show that $|f(x)-f(x^3)| < \epsilon$ right? $\endgroup$
    – Fruitdruif
    Apr 26, 2017 at 8:37
  • 1
    $\begingroup$ Suppose $\lim_{x\to 0}f(x)=b:$ $|f(x^3)-b|=|f(x^3)-f(x)+f(x)-b|\leq|f(x^3)-f(x)|+|f(x)-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ $\endgroup$
    – Itay4
    Apr 26, 2017 at 8:41
  • $\begingroup$ @Itay4 don't understand why $|f(x^3)-f(x)|+|f(x)-b|< \frac{\epsilon}{2}+\frac{\epsilon}{2}$. We only know that $|f(x)-b|< \epsilon$ right? Sorry for all the (probably very easy) questions, I'm new to proving limits $\endgroup$
    – Fruitdruif
    Apr 26, 2017 at 8:48
  • $\begingroup$ @Itay4 Why send the OP estimating $|f(x^3)-f(x)|$? This is certainly not the simplest approach. $\endgroup$
    – Did
    May 21, 2017 at 10:02

2 Answers 2


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.


$$\lim_{x\to 0}f(x) = b$$ 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.

  • $\begingroup$ Shouldn't $lim_{t \rightarrow 0} f(t^3)$ equal $b$ instead of $0$? $\endgroup$
    – Fruitdruif
    Apr 26, 2017 at 8:53
  • $\begingroup$ Yep, fixed it. Thanks $\endgroup$
    – adfriedman
    Apr 26, 2017 at 16:08

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