Prove if $\lim _{x\rightarrow0}(f(x))=L$ then $\lim _{x\rightarrow0}(f(x^3))=L$

Question

Prove if $\lim _{x\rightarrow0}(f(x))=L$ then $\lim _{x\rightarrow0}(f(x^3))=L$

Can i just say that because $f(0)=f(0^3)$ it is true?

Answer 1

Hint: let $y=x^3$ since $y \to 0$ as $x \to 0$ (at least as fast) we have

$$\lim_{y \to 0}f(y)$$

Now, how do you get $\varepsilon$ and $\delta$ given what you already know about $f(x)$'s limit.

Answer 2

Since the function has been assumed to have a limit at $x=0$, for every $\epsilon>0$, there exists $\delta$ such that $$ 0<|x|<\delta\implies |f(x)-L|<\epsilon. $$ Now ask yourself, how close do we have to make $x^3$ from $x=0$ for $f(x^3)$ to be from $L$? It has already been given to us by the epsilon-delta definition of a limit: $x^3$ must be within $\delta$ of $0$, so by extension $x$ must be within $\sqrt[3]{\delta}$ of $0$. Therefore, $$ 0<|x|<\sqrt[3]{\delta}\implies|f(x^3)-L|<\epsilon. $$ QED

Answer 3

Begining Hint:

Suppose $\lim_{x\to 0}f(x) = L$ then for every $\epsilon > 0$ there exists $\delta > 0$ such that

$$0 < |x| < \delta \ \ \ \Rightarrow \ \ \ |f(x) - L| < \epsilon$$

Now let $\epsilon > 0$ and $\delta = \ ?$ Think what delta needs to be set to so that $|f(x^3) - L| < \epsilon$