Prove that $\lim_{x \to 5} \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}=2$ using $\delta-\epsilon$ method.

Question

I need to prove that Prove that $\lim_{x \to 5} \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}=2$ using $\delta-\epsilon$ method. So far, what I have is $\forall \epsilon >0$, there exists a $\delta>0$ such that if $0<|x-5|<\delta$, then $\left| \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}-2 \right| = \left| \frac{-2x+10+\sqrt{x-3}-\sqrt{7-x}}{x-4} \right|$. At this point, I don't know how to deal with the two radicals in the numerator.

Answer 1

HINT:

$$\sqrt{x-3}-\sqrt{7-x}=\frac{2(x-5)}{\sqrt{x-3}+\sqrt{7-x}}$$

so that we have

$$\left|\frac{(-2x+10)+\sqrt{x-3}-\sqrt{7-x}}{x-4}\right|=2\frac{|x-5|}{|x-4|}\left|1-\frac{1}{\sqrt{x-3}+\sqrt{7-x}}\right|$$