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

$$\sqrt{y+4} =2(1+y/4)^\frac12\le 2(1+y/8)=2+y/4$$

$$3-\sqrt{y+4}\ge 3-2-y/4=1-y/4$$

$$\sqrt{ 3-\sqrt{y+4}}\ge \sqrt{ 1-y/4}$$

$$1-\sqrt{ 3-\sqrt{y+4}}\le 1-\sqrt{ 1-y/4}$$


$$\frac{1-\sqrt{ 3-\sqrt{y+4}}}{y} \le \frac{1-\sqrt{ 1-y/4}}{y}$$

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

$$\frac{1-\sqrt{ 1-y/4+y^2/64}}{y} \le \frac{1-\sqrt{ 3-\sqrt{y+4}}}{y} \le \frac{1-\sqrt{ 1-y/4}}{y}$$

  • $\begingroup$ and let $y=0$ but this doesnt work? $\endgroup$
    – user548054
    Sep 28 '19 at 20:40
  • 1
    $\begingroup$ @Elementary $y\to 0^+$ and certainly it doesn't make sense to $y=0$. $\endgroup$
    – Sebastiano
    Sep 28 '19 at 20:42
  • $\begingroup$ can you show final step? $\endgroup$
    – user548054
    Sep 28 '19 at 20:43
  • $\begingroup$ @Elementary As Sebastiano noted $y=0$ doesn't make sense. It could be useful if you check all the step instead to be sure all is ok. We are using that for $t\to 0^+$ $$1+t/2-t^2/8 \le (1+t)^{\frac12}\le 1+t/2$$ $\endgroup$
    – user
    Sep 28 '19 at 20:47
  • $\begingroup$ Sorry I can not see where $\frac 18$ comes from $\endgroup$
    – user548054
    Sep 28 '19 at 20:57

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