# Is there another way to prove this?

A question has been provided where - $L, M \in \mathbb R$ and $L\le M+\epsilon$ for every $\epsilon >0$ and I have to prove that $L\le M$ also holds.

Well , I only used the condition that $M+\epsilon$ is approximately equal to $M$ when $M \gg \epsilon$ . Hence $L\le M$ for an infinitesimal value of $\epsilon$.

Is this the correct way to prove it and if there any other way to prove it because I am satisfied by my way of proving this mathematically.

Suppose $L>M$. Then $L=M+\epsilon$ for some $\epsilon>0$. Then $L$ can not be smaller than $M+\epsilon/2$ which is a contradiction.