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I'm trying to prove that if $\lim_{x\to0}f(x)=L$, then $\lim_{x\to0}f(x^3)=L$. This is what i have but i'm not sure if its right.

Proof:

Since $\lim_{x\to0}f(x)=L$, there exists $\delta>0$ such that $0<|x|<\delta \implies |f(x)-L|<\epsilon$. But this means

$0<|x^3|<\delta \implies |f(x^3)-L|<\epsilon$, so $0<|x|<\frac{\delta}{|x^2|} \implies |f(x^3)-L|<\epsilon$. Thus, $\lim_{x\to0}f(x^3)=L$.

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    $\begingroup$ But I'm saying if $ 0<|x^3|<\delta$ then $|f(x^3)-L|<\epsilon$ $\endgroup$ – chrismc Jul 31 '16 at 23:23
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Since $\displaystyle \lim_{x \to 0} f(x) = L$, this means Given $\epsilon > 0$, there exists an $\delta > 0$ such that:

$\forall x : 0 < |x| < \delta \implies |f(x)-L| < \epsilon$. But with this $\delta > 0$, choose $\delta' = \sqrt[3]{\delta}$ then we have:

$\forall x: 0 < |x| < \delta' \implies |x^3| < \delta'^3 = \delta $.

Combining these $2$ statements we have at once:

$\forall x: 0 < |x| < \delta' \implies |f(x^3)- L| < \epsilon$.

This means $\displaystyle \lim_{x \to 0} f(x^3) = L$

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