prove that $\lim_{x\to 0 } f(x)=l$ ,then there is a number $\lambda >0$ and a number $M$ such that $|f(x)|<M$ if $|x-a|<\lambda$
my attempt:
$\lim_{x\to 0 } f(x)=l$ means $\forall \epsilon>0, \exists \lambda >0 $: if $|x-a|<\lambda$ then $|f(x)-l|<\epsilon$ or $ -|l|-\epsilon\leq l -\epsilon<f(x)<\epsilon+l\leq \epsilon+|l|$
so it suffice to take $M=\epsilon + |l|$
so i think i proved that if $\lim_{x\to 0 } f(x)=l$ then there is a number $\lambda $ and number $M$ such that $|f(x)|<M$
-does my proof is correct ,and if yes ,do i can consider it as a rigorous proof ?