Prove that if there is a number B such that |f(x)| <= B for all x doesnt equal 0, then limit x->0 xf(x) = 0
Where do I start?
I assume you proved that if $\displaystyle g(x)$ is bounded and $\displaystyle \lim_{x \to a}h(x) \to \infty$ or $\displaystyle \lim_{x \to a}h(x) \to 0$, then $\displaystyle \lim_{x \to a}h(x)g(x) = \lim_{x\to a}h(x)$...
If you have, simply use it and finish the proof. Otherwise, prove it and finish the proof! :P
since $\displaystyle \lim_{x\to0}x=0$ then for each $\displaystyle \epsilon>0$ we can let $\displaystyle |x|<\frac\epsilon{B}.$
on the other hand, all we need to prove is that $\displaystyle |xf(x)|<\epsilon,$ in efect $\displaystyle \left| xf(x) \right|=\left| x \right|\left| f(x) \right|<\frac{\epsilon }{B}\cdot B=\epsilon,$ and we're done.