I am trying to solve this question $\lim_{x \to 9} \sqrt{x-5} = 2$. If you could tell me if I am going about this the right way I'd appreciate it.
Given $\epsilon > 0$, Let $\delta =min(9,\epsilon)$
Given $0<|x-9|<\delta$
$|f(x)-L| = |\sqrt{x-5}-2| =\left|(\sqrt{x-5}-2) \cdot \frac{\sqrt{x-5}+2}{\sqrt{x-5}+2}\right| = \left|\frac{x-9}{\sqrt{x-5}+2}\right|< \frac{\delta}{1}=\epsilon$
Note: $\frac{1}{\sqrt{x-5}+2}$ is always less than $\frac{1}{1}$
Does this make sense and is there a better way than doing this? Whenever I see a limit proof question with a root like this, my go-to approach is to multiply by the conjugate, is that the best approach?
Thanks