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I need to prove that $\lim_{x \to 0}f(x^3)=\lim_{x \to 0}f(x)$. Then give an example of a function f for which $\lim_{x \to 0}f(x^2)$ exists but $\lim_{x \to 0}f(x)$ does not exist

Thank you in advance

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Looks like homework, you should tag it as such. What have you tried? Why could it make a difference to that statement whether the argument is $x^2$ or $x^3$? (You could have left calculus as a tag, too.) – K. Stm. Oct 4 '12 at 20:01
The limit of $f(x^3)$ and the limit of $f(x)$ both exist therefore the limit of the difference of the two functions exist as well – user43418 Oct 4 '12 at 20:03
@K.Stm. How about, "Looks like homework. If it is, you should tag it as such." – Graphth Oct 4 '12 at 20:27
@Graphtth Yes, that would be more correct and nice, but I think the condition is implied by common sense. I hope that wasn't too aggressive. – K. Stm. Oct 4 '12 at 20:31

2 Answers

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Hint: For the first part, do you know that $x\mapsto x^3$ is invertible on $\mathbb{R}$?
For the second part consider the function $$f(x)=\operatorname{sign}(x)=\left\{ \begin{array}{cc} 1 & x>0\\ 0 & x=0\\ -1 & x<1\end{array}\right.$$

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Hint: Can you show: If $g$ is continuous around $a$ and $g(a)=b$ and $\lim_{x\to b} f(x)$ exists, then $\lim_{x\to a} f(g(x))=\lim_{x\to b} f(x)$.

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Ok I proofed that. Could you help me now with the second part of the question – user43418 Oct 4 '12 at 22:23

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