How can I find this limit applying squeeze theorem? How can I evaluate this limit applying squeeze theorem?

$$\lim_{x\to6}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}$$

I found this limit with standart way:

$\lim_{x\to6}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}=\lim_{x\to6}\frac{\left(1+\sqrt{ 3-\sqrt{x-2}} \right)\times \left( 1-\sqrt{ 3-\sqrt{x-2}}\right)}{(x-6)\times \left( 1+\sqrt{ 3-\sqrt{x-2}}\right)}=\frac 18$

I want to write this limit using squeeze theorem.

First step let wlog $y=x-6\to 0^+$ thus
$$\lim_{x\to6^+}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}=\lim_{y\to0^+}\frac{1-\sqrt{ 3-\sqrt{y+4}}}{y}$$
then by Bernoulli inequality
$$\sqrt{y+4} =2(1+y/4)^\frac12\le 2(1+y/8)=2+y/4$$
$$3-\sqrt{y+4}\ge 3-2-y/4=1-y/4$$
$$\sqrt{ 3-\sqrt{y+4}}\ge \sqrt{ 1-y/4}$$
$$1-\sqrt{ 3-\sqrt{y+4}}\le 1-\sqrt{ 1-y/4}$$
then
$$\frac{1-\sqrt{ 3-\sqrt{y+4}}}{y} \le \frac{1-\sqrt{ 1-y/4}}{y}$$
For the other inequality use that
$$\sqrt{y+4} =2(1+y/4)^\frac12\ge 2(1+y/8-y^2/128)=2+y/4-y^2/64$$
$$3-\sqrt{y+4}\le 3-2-y/4+y^2/64$$
$$\sqrt{ 3-\sqrt{y+4}}\le \sqrt{ 1-y/4+y^2/64}$$
$$1-\sqrt{ 3-\sqrt{y+4}}\ge 1-\sqrt{ 1-y/4+y^2/64}$$
Then finally
$$\frac{1-\sqrt{ 1-y/4+y^2/64}}{y} \le \frac{1-\sqrt{ 3-\sqrt{y+4}}}{y} \le  \frac{1-\sqrt{ 1-y/4}}{y}$$
