Proving $\lim_{x\to 4} \left(\frac{\sqrt {2x-1}}{\sqrt {x-3}}\right) = \sqrt 7$ using $ \varepsilon - \delta$ 
Prove that $$\lim_{x\to 4} \left(\frac{\sqrt {2x-1}}{\sqrt {x-3}}\right) = \sqrt 7$$ using $\varepsilon - \delta$.

We find $\delta$ such that
$0<|x-4| <\delta$
$$\left|\frac{\sqrt {2x-1}}{\sqrt {x-3}}-\sqrt 7\right|= \left|\frac{\sqrt {2x-1}-\sqrt{7x-21}}{\sqrt {x-3}}\right|$$
I know that we should get to $|x-4|$ but i dont know how 

We want to show that $\lim\limits_{x\to 4} \dfrac{\sqrt{2x-1}}{\sqrt{x-3}}=\sqrt{7}$. By the $\epsilon-\delta$ limit definition, this means $\forall \epsilon >0,\exists \delta >0 \space(0<|x-4|<\delta \Rightarrow \left |\dfrac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right |<\epsilon)$. Simplifying, we obtain $\left |\dfrac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right |= 
\left |\dfrac{\sqrt{2x-1}-\sqrt{7x-21}}{\sqrt{x-3}} \right |$. Let $\delta = \dfrac{3}{4}$, then $|x-4|<\delta\Rightarrow 3.25<x<4.75\Rightarrow\sqrt{x-3}>\sqrt{3.25-3}=\sqrt{0.25}=0.5\Rightarrow \left |\dfrac{\sqrt{2x-1}-\sqrt{7x-21}}{\sqrt{x-3}} \right |<\left |\dfrac{(\sqrt{2x-1}-\sqrt{7x-21})\cdot(\sqrt{2x-1}+\sqrt{7x-21})}{0.5(\sqrt{2x-1}+\sqrt{7x-21})} \right|=\left |\dfrac{-5(x-4)}{0.5(\sqrt{2x-1}+\sqrt{7x-21})} \right |< \left |\dfrac{-5(x-4)}{0.5(\sqrt{5.5}+\sqrt{1.75})} \right |<\left| \dfrac{-5(x-4)}{0.5(2+1)}\right|=\dfrac{10}{3}|x-4|$. 
Take $\delta = \min\{\dfrac{10}{3}\epsilon,\dfrac{3}{4}\}$ and we have that $\left |\dfrac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right |<\dfrac{10}{3}|x-4|<\dfrac{3}{10}\cdot\dfrac{10}{3}\epsilon=\epsilon$, as required.
