How can i prove the function $f: x \mapsto x|x|$ is continuous over $\mathbb{R}$ using epsilon-delta definition.
I've tried:
Given a certain $\epsilon$ we want to prove that there exists a $\delta$ such that $|x-x_0|< \delta \Rightarrow |f(x)-f(x_0)|< \epsilon$.
Starting from the RHS, i get: $$|x|x|-x_0|x_0|| \leq |x|x|| + |x_0|x_0||$$ Implies $$|x|x|-x_0|x_0|| \leq x^2 + x_0^2$$
But then, how do I transform this into $|x-x_0|$ or should I use another method?
(Could I state that $x$ is continuous, and so is $|x|$. And the product of 2 continuous functions is continuous?)
Yes, you can use that reasoning, that the product of two continuous functions is continuous (and $\rm{id}$ and $|\cdot|$ are, indeed, both continuous), to show that $f$ is continuous. Proving from the $\varepsilon$-$\delta$ definition is still a good exercise.
Hint: How do you prove that the product of two continuous functions $g,h$ is continuous? Can you rewrite the proof in terms of $g\left(x\right)=x$ and $h\left(x\right)=|x|$?