How can I calculate the limit without using the L'Hopital's rule $$ \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} $$
I have tried to multiply with conjugates or use auxiliary variables but it did not arrive at all simple

$$\lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}$$

Let's let $z = x-2$ (or $x = z + 2$). Then we get
$$\lim_{z \to 0} \frac{1-\sqrt{1-2\sqrt z}}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}}$$
\begin{align}
   1-\sqrt{1-2\sqrt z}
   &= \dfrac{1-(1-2\sqrt z)}{1+\sqrt{1-2\sqrt z}} \\
   &= \dfrac{2\sqrt z}{1+\sqrt{1-2\sqrt z}}
\end{align}

\begin{align}
   \frac{1}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}}
   &=\frac{1+\sqrt{1-\sqrt{\frac{z}{z+4}}}}
          {1-\left(1-\sqrt{\frac{z}{z+4}}\right)} \\
   &=\frac{1+\sqrt{1-\sqrt{\frac{z}{z+4}}}}
          {\sqrt{\frac{z}{z+4}}} \\
   &=\frac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
          {\sqrt{z}} \\
\end{align}

\begin{align}
   \dfrac{1-\sqrt{1-\sqrt{4x-8}}}
         {1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}
   &= \dfrac{2\sqrt z}
            {1+\sqrt{1-2\sqrt z}}\cdot
      \dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
            {\sqrt{z}}\\
   &= 2\dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
             {1+\sqrt{1-2\sqrt z}}
\end{align}

If we now let $z=0$, we get
\begin{align}
   \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}
                       {1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}
   &= 2 \lim_{z \to 0} \dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
             {1+\sqrt{1-2\sqrt z}} \\
   &= 2 \lim_{z \to 0} \dfrac{2+2}{1+1} \\
   &= 4 \\
\end{align}
