We have the function $f(x) = x\sin x$
and we want to show that $f(x)\to0$ for $x\to0$.
We want to find a $\delta$ where $\|x \sin x-0\|<\varepsilon$ and $\|x-0\|<\delta$
We know that $|\sin x| \leq |x|$
so therefore $x|\sin x| \leq x^2$
we now have the pattern $\|x-0\|^2$
and we can choose a $\delta = \sqrt{\varepsilon}$
Proof done.
Is this a valid proof? Did I make any mistakes? And if so, how do I fix it? Thanks in advance.