Proving limit by definition with absolute value

Question

This is what I have:

$\lim_{x \to 0} x \cdot |x|=0$

And I know: $$|x|=\begin{cases}x,&\text{if }|x|\ge 0\\-x,&\text{if }x <0\;.\end{cases}$$ Case 1:

$\lim_{x \to 0^+}=0$ $(0 < |x - 0| < \delta \Rightarrow |x-0|< \epsilon)$ since: $o < x-0< \epsilon \Rightarrow \fbox{$\delta=\epsilon$}$

So:

$\forall \epsilon > 0, \delta =\epsilon, 0<|x-0|< \delta \Rightarrow |f(x) - 0| < \epsilon $

Case 2: $\lim_{x \to 0^-} x \cdot |x|=0$

And I know: $\lim_{x \to 0^+}=0$ $(0 < |x - 0| < \delta \Rightarrow |-x-0|< \epsilon)$

And I'm assuming that (because adding or substracting zero is no relevant):

$(x+0 \Leftrightarrow -(x-0))$

And for absolute value property: $0 < |x-0| < \epsilon$ So: $\fbox{$\delta = \epsilon$}$

With this I know the limit is the same, now I do the same with $\lim {x \to 0} x =$ $(0 < |x - 0| < \delta \Rightarrow |x-0|< \epsilon)$ With this I know : So: $\fbox{$\delta = \epsilon$}$

This is by far the hardest in my book, doesnt contain the solution. Am I on the right track?

Answer 1

To start with, let's ignore any value of $x \ge 1$. We only need to consider function values close to $0$, so for our intents and purposes, $f(x) = |x|$. The only thing we need to be careful of is ensuring that $\delta \le 1$. That way, if $0 < |x - 0| < \delta$, then $|x| < 1 \implies x < 1$, and so $f(x) = |x|$ for all the $x$ we have to worry about.

So, we are showing $\lim_{x \to 0} x|x| = 0$. We have $|x|x| - 0| = |x|^2$, and this is less than $\varepsilon$ if and only if $|x| < \sqrt{\varepsilon}$. Thus, we can take $\delta = \min\{\sqrt{\varepsilon}, 1\}$, to ensure that $\delta \le 1$ as we needed above. We get, \begin{align*} 0 < |x - 0| < \min\{\sqrt{\varepsilon}, 1\} &\implies |x| < \sqrt{\varepsilon} \text{ and }|x| < 1 \\ &\implies |x|^2 < \varepsilon \text{ and } x < 1 \\ &\implies |x|x| - 0| < \varepsilon \text{ and } f(x) = |x| \\ &\implies |xf(x) - 0| < \varepsilon, \end{align*} completing the proof.

Answer 2

Sorry, but you're quite far from a proof. Using $\delta=\varepsilon$ is not sufficient and you're never actually using the function that you want to find the limit of.

There is no need to go to one-sided limits. Your purpose is, given $\varepsilon>0$, to find $\delta>0$ such that, for $0<|x|<\delta$, $$ \bigl| x|x|-0 \bigr|<\varepsilon $$ Now the key observation is that $\bigl|x|x|\bigr|=|x|^2$.

Hence you're done by taking $\delta=\sqrt{\varepsilon}$. Indeed, if $0<|x|<\sqrt{\varepsilon}$, then $0<|x|^2<\varepsilon$.