$\epsilon-\delta$ proof for $ \lim _ {x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}=6. $

Question

I want to prove that

$$ \lim _ {x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}=6. $$

Here is my approach:

For all $\epsilon >0 $ Let $\delta = \epsilon$, then whenever we have that $|x-9|<\delta$, we know that :

$$\left|\frac{x-9}{\sqrt x-3}-6\right|=\left|\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}-6\right|=\left|\sqrt x+3-6\right|=|\sqrt x-3|. $$ Now notice that $$|\sqrt x-3|< |\sqrt x-3|\cdot |\sqrt x+3|=|x-9|<\delta=\epsilon. \square$$

Is my proof sufficient - I'm not sure about the final step, is it okay what I do here?

Answer 1

Yes, but you can also solve the simplified limit $$\lim_{x \rightarrow 9} \sqrt{x} +3 =6$$