Answer
$$\lim_{x \to 9}\sqrt{x}=3$$
Work Step by Step
We want to prove this limit by using $\varepsilon - \delta$ definition; that is, we must show that for each $\varepsilon >0$, there exists a $\delta >0$ such that $|\sqrt{x}-3|< \varepsilon$ whenever $|x-9|< \delta$.
Now, we have$$|\sqrt{x}-3|=\frac{|x-9|}{|\sqrt{x}+3|}.$$For all $x$ in the interval $(8, 10)$ we have$$|\sqrt{x}+3| > 5 \quad \Rightarrow \quad \frac{1}{|\sqrt{x}+3|} < \frac{1}{5}.$$So, letting $\delta $ be the minimum of $1$ and $5 \varepsilon$, we conclude that$$|x-9|< \delta = \min \{ 1, 5 \varepsilon \} \quad \Rightarrow \quad |\sqrt{x}-3|=\frac{|x-9|}{|\sqrt{x}+3|}< (5 \varepsilon )(\frac{1}{5}) = \varepsilon .$$Done.
