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I want to do it with the epsilon-delta definition. So for $\forall x\in D(f)$ and $\forall \epsilon >0$ $\exists \delta>0$ $\forall y$ such $|y-x|<\delta$ $\implies$ $|f(y)-f(x)|<\epsilon$. Let $|y-x|<\delta$ , and $|y^3-x^3|=|y-x||y^2+yx+x^2|\leq|y-x||y+x|^2$ $\implies \delta|y-x|^2 =\epsilon$, and here im stuck.

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  • $\begingroup$ $|x^2+xy+y^2| \le |x+y|^2$ is not true in general $\endgroup$ – David Peterson Nov 16 '19 at 13:39
  • $\begingroup$ Hint: the concatenation of continuous functions is continuous. Your function is the concatenation of which two functions? $\endgroup$ – Alexander Geldhof Nov 16 '19 at 13:45
  • $\begingroup$ $x^2$ and $|x|$? $\endgroup$ – Elekhey Nov 16 '19 at 13:49
  • $\begingroup$ Hint: You define $f(x)=x^3$ and $g(x)=|x|$ that are continuous. Then $g(f(x))=$? $\endgroup$ – Alex Pozo Nov 16 '19 at 13:49
  • $\begingroup$ Ohhh I get it. Thank you! $\endgroup$ – Elekhey Nov 16 '19 at 13:51
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We know that $g(x) = x^{3}$ is continuous. Let us show that the function $f(x) = |x|$ is continuous at an arbitrary point $a \in \mathbb{R}$. To do this, let $\delta = \epsilon$ and suppose $|x-a| \le \epsilon$. Then, because of the triangle inequality, we have: $$||x|-|a||\le |x-a|\le \epsilon$$ Which proves $f$ is continuous at $a$. Now, we know that the composite of continuous functions is continuous, so if $h(x) = |x^{3}|$ then: $$h(x) = (f\circ g)(x)$$

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