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For example to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{3})$

Assume $\lim_{x\to0}f(x) = l$ that is $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\forall x$: $0 < |x| < \delta_{1} \implies |f(x)-l|$

Then to show $\lim_{x\to0}f(x^{3}) = l$ which is by replacing $x$ with $x^{3}$ in the assumption: that is $0 < |x| < |x^{3}| < \delta_{2} \implies |f(x^{3})-l|$ for $x< 0$ and $x>1$ and

$0 < |x^{3}| < |x| < 1 \implies |f(x^{3})-l|$ for $0 < x < 1$

that is choosing $\delta_{2} = min(1, \delta_{1})$

thus showing: $\lim_{x\to0}f(x^{3}) = l$

Why would this argument not work if an attempt was made to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{2})$ which by my question here: Give an example about limits: obviously would not work. Or is there something I overlooked in this proof?


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$x^2$ isn't invertible around $0$, while $x^3$ is. That matters if the left-sided limit of $f(x)$ doesn't exist, while the right-sided one does. In that case, $\lim_{x\rightarrow 0}f(x^2)$ may exist even if $\lim_{x\rightarrow 0}f(x)$ doesn't. One could conjecture though that if both limits exists, they must be equal.

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That conjecture would be true as $\displaystyle\lim_{x\to 0}f(x^2) = \lim_{x\to 0^+}f(x) = \lim_{x\to 0}f(x)$ under the assumption that the last limit exists. – Michael Albanese Oct 11 '12 at 13:05
So assuming both limit exists: this argument can be used to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{n})$ for all n? – mathnoob Oct 11 '12 at 13:46
@mathnoob Yes. If you know that $\lim_{x\rightarrow 0}f(x)$ exists, then $\lim_{x\rightarrow 0}f(x^n)$ exists for all $n$ and is equal to $\lim_{x\rightarrow 0}f(x)$ . – fgp Oct 11 '12 at 13:49

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