Prove that $|x^3|$ is continuous.

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.

• $|x^2+xy+y^2| \le |x+y|^2$ is not true in general – David Peterson Nov 16 '19 at 13:39
• Hint: the concatenation of continuous functions is continuous. Your function is the concatenation of which two functions? – Alexander Geldhof Nov 16 '19 at 13:45
• $x^2$ and $|x|$? – Elekhey Nov 16 '19 at 13:49
• Hint: You define $f(x)=x^3$ and $g(x)=|x|$ that are continuous. Then $g(f(x))=$? – Alex Pozo Nov 16 '19 at 13:49
• Ohhh I get it. Thank you! – Elekhey Nov 16 '19 at 13:51

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: $$h(x) = (f\circ g)(x)$$