In the proof of the weak maximum principle of the diffusion equation, I came across the part that says:
Let $\epsilon$ be a positive constant and let $v(x, t)=u(x, t)+\epsilon x^{2} .$ Our goal is to show that $v(x, t) \leq M+\epsilon l^{2}$ throughout $R .$ Once this is accomplished, we'll have $u(x, t) \leq M+\epsilon\left(l^{2}-x^{2}\right) .$ This conclusion is true for any $\epsilon>0$ Therefore, $u(x, t) \leq M$ throughout $R,$ which is what we are trying to prove.
Why does showing $u(x,t)\leq M + \epsilon \cdot l^2 $ finish our proof that $u(x,t)\leq M $? Surely, $\epsilon>0$ and $l^2-x^2 >0$, but why would $u(x,t)\leq M $ hold?
Intuitively, I think it's sending epsilon to zero, but I don't know the detailed process of it.
Suppose $a \leq b + \epsilon$ for all $\epsilon > 0$. Equivalently, $a < b + \epsilon$ for all $\epsilon > 0$. If $a>b$, then $a-b >0$, so $a < b + (a - b) = a$, contradiction. Hence, $a \leq b$.