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
 A: First, let $x=u+2$ to get
$$\lim_{u\to0}\frac{1-\sqrt{1-\sqrt{4u}}}{1-\sqrt{1-\sqrt{\frac u{u+4}}}}$$
The numerator may be handled with binomial expansion:
$$\sqrt{1-2\sqrt u}=1-\sqrt u+\mathcal O(u)$$
And likewise the denominator:
$$\sqrt{1-\sqrt{\frac u{u+4}}}=1-\frac12\sqrt{\frac u{u+4}}+\mathcal O\left(\frac u{u+4}\right)$$
which leads us to conclude that
$$\lim_{u\to0}\frac{1-\sqrt{1-\sqrt{4u}}}{1-\sqrt{1-\sqrt{\frac u{u+4}}}}=\lim_{u\to0}\frac{\sqrt u+\mathcal O(u)}{\frac12\sqrt{\frac u{u+4}}+\mathcal O\left(\frac u{u+4}\right)}=\lim_{u\to0}\frac{1+\mathcal O(\sqrt u)}{\frac1{2\sqrt{u+4}}+\mathcal O\left(\frac{\sqrt u}{u+4}\right)}=\frac1{\frac1{2\sqrt4}}=4$$
A: $$\lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}$$

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}
A: HINT:
We need $x\to2^+$
$$\lim_{x \to 2}\frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\dfrac{x-2}{x+2}}}}=\lim_{x \to 2}\dfrac{2\sqrt{x-2}}{\sqrt{\dfrac{x-2}{x+2}}}\cdot\dfrac{1+\sqrt{1-\sqrt{\dfrac{x-2}{x+2}}}}{1+\sqrt{1-\sqrt{4x-8}}} $$
Cancel out $\sqrt{x-2}$ as $\sqrt{x-2}\ne0\iff x\ne2$ as $x\to2$  
Can you take it from here?
A: HINT: multiply numerator and denominator by $$1+\sqrt{1-\sqrt{\frac{x-2}{x+2}}}$$
and by $$1+\sqrt{1-\sqrt{4x-8}}$$ and simplify the term
A: Let $u = (4x-8)^{1/2}, v = [(x-2)/(x+2)]^{1/2}.$ The expression equals
$$\frac{u}{v}\left (\frac{1-(1-u)^{1/2}}{u}\right )\big /\left(\frac{1-(1-v)^{1/2}}{v}\right ).$$
Now $u/v = 2(x+2)^{1/2} \to 4$ as $x\to 2^+.$ Because both $u,v \to 0$ as $x\to 2^+,$ both fractions in parentheses approach the same nonzero limit, as you can verify by using the conjugate trick, or simply by recognizing the definition of the derivative of $(1-u)^{1/2}$ at $0.$ It follows that the limit is $4\cdot 1=4.$
A: First use $t=x-2$ substitution, to send it to a limit at point $0$:
$$ \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} \ =\ \lim_{t\to 0^+}\frac{1-\sqrt{1-2\sqrt{t}}}{1-\sqrt{1-\sqrt{\frac{t}{t+4}}}}=\dots$$
Now multiply the numerator and denominator by $1+\sqrt{1-\sqrt{\frac{t}{t+4}}}$ and we get
$$\dots=\lim_{t\to 0^+}\frac{\left(1-\sqrt{1-2\sqrt{t}}\,\right)\left(1+\sqrt{1-\sqrt{\frac{t}{t+4}}}\right)}{\sqrt{\frac{t}{t+4}}}$$
The next step: observe that the crucial part is
$$\lim_{t\to 0^+}\frac{\left(1-\sqrt{1-2\sqrt{t}}\,\right)}{\sqrt{t}}$$
because the rest tends to $4$ as $t\to 0$ (since in the rest you can just put $t=0$ to obtain the limit).
It is a much easier limit now, can you finish from here? 
