Proving continuity of a absolute value function 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?)
 A: 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 $h\left(x\right)=|x|$?
A: Note that you can write ( by the triangular inequality)  $$|(x+h)|x+h| - x|x|| = |(x+h)|x+h| - x|x+h| + x|x+h| - x|x|| \leq (|(x+h)|x+h| - x|x+h||) + (|x|x+h| - x|x||) = |x+h|\cdot|h|+|x|\cdot||x+h|-|x|| \leq |x+h|\cdot|h|+|x|\cdot|h| \leq (|x|+|h|)\cdot|h|+|x|\cdot|h|$$
but $h \rightarrow 0 \ $ and so if $|h| < \delta$ then $$|(x+h)|x+h| - x|x||\leq (|x|+\delta)\cdot\delta+|x|\cdot\delta$$ and so if you want that $|(x+h)|x+h| - x|x|| < \epsilon$ you have to solve the equation in $\delta $ $$(|x|+\delta)\cdot\delta+|x|\cdot\delta < \epsilon$$
A: I wouldn't go for epsilon-delta. Since the two functions have limits everywhere, their product has the product of the two limits as a limit itself everywhere, and since the limits are equal to the values of the functions, the product of the limits is equal to the product of the values of the functions. The proof is very basic, you can check it on Spivak or actually most acclaimed calculus textbooks and/or notes anytime you like :)
