Answer
$\lim_{x\to 4^-}\sqrt {4-x}=0$
Work Step by Step
Step 1. Given $\epsilon\gt0$ and $\sqrt {4-x}\lt\epsilon$, we have $0\lt4-x\lt\epsilon^2$ which gives $4-\epsilon^2\lt x\lt 4$
Step 2. The interval $I=(4-\delta, 4)$ leads to $4-\delta\lt x\lt4$. Comparing it with the results in step-1, we can set $\delta=\epsilon^2$ so that for $x\in I$, we have $\sqrt {4-x}\lt\epsilon$.
Step 3. We can identify this as the process of verifying the limt $\lim_{x\to 4^-}\sqrt {4-x}=0$
