Question:
Prove that:
$\underset{x\rightarrow11^{-}}{\lim}\sqrt{x-2}=3$]
Answer:
Basically need to prove that : $\forall\varepsilon>0\,\,\exists\delta>0\forall x\in\mathbb{R}\quad-\delta<x-11<0\rightarrow\left|\sqrt{x-2}-3\right|<\varepsilon$
Perpwork: Let $\varepsilon>0$ be given. $$\begin{array}{cc} & \left|\sqrt{x-2}-3\right|<\varepsilon\\ \iff & -\varepsilon<\sqrt{x-2}-3<\varepsilon\\ \iff & 3-\varepsilon<\sqrt{x-2}<3+\varepsilon\\ \iff & \left(3-\varepsilon\right)^{2}<|x-2|<\left(3+\varepsilon\right)^{2}\\ \\ \end{array}$$
Given that $ \forall\varepsilon>0\quad\left(3-\varepsilon\right)^{2}\geq0$
$$\begin{array}{cc} & \left(3-\varepsilon\right)^{2}<|x-2|<\left(3+\varepsilon\right)^{2}\\ \rightarrow & \left(3-\varepsilon\right)^{2}<x-2<\left(3+\varepsilon\right)^{2}\\ \iff & \left(3-\varepsilon\right)^{2}+2<x<\left(3+\varepsilon\right)^{2}+2\\ \iff & \left(3-\varepsilon\right)^{2}-9<x-11<\left(3+\varepsilon\right)^{2}-9 \end{array}$$
Set $\delta<9-\left(3-\varepsilon\right)^{2}$
$$\begin{array}{ccccc} & -\delta<x-11<0\\ \iff & \left(3-\varepsilon\right)^{2}-9<x-11 & \rightarrow & \left(3-\varepsilon\right)^{2}<|x-2|<\left(3+\varepsilon\right)^{2}\\ & & \rightarrow & -\varepsilon<\sqrt{x-2}-3<\varepsilon\\ & & \rightarrow & \left|\sqrt{x-2}-3\right|<\varepsilon \end{array}$$ Q.E.D
Is this a legit proof ?