Show that if $f : (a,\infty)\to \mathbb R$ is such that $\lim_{x\to \infty}xf(x)=L$ where $ L \in \mathbb R$, then $\lim_{x\to \infty}f(x)=0$
Let $\epsilon>0$,$\exists K(\epsilon)>0$: $\forall x:|x|>K(\epsilon)\implies \left|\left|xf(x)\right|-\left|L\right|\right| \leq|xf(x)-L|<\epsilon$
$\epsilon>0$,$\exists K(\epsilon)>0$: $\forall x:|x|>K(\epsilon)\implies \frac{|L|-\epsilon}{|x|}<|f(x)|<\frac{|L|+\epsilon}{|x|}$
Choose $M=\max\left\{\frac{|L|+\epsilon}{\epsilon}, \, K(\epsilon)\right\}\implies \forall x>M ,|f(x)|<\epsilon$
Am I correct? Can I choose $M$ like this?