<s>[INST] Limit of $\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right)$ 
Given the limit:
  $$
\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right) = \alpha
$$
Find the value of $\alpha$


Could not get my head around on how to simplify the nominator.
Anyone feeling like this should be easy?
Thanks in advance.
 [/INST] Let se $x=1+y$ with $y\to0$ then
$$\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right) = \lim\limits_{y\to 0}\left(\frac{\sqrt{y^2+4y+8-8\sqrt{1+y}}}{\log(1+y)}\right)$$
and note that


*

*$\sqrt{1+y}=1+\frac12 y-\frac18y^2 +o(y^2)$

*$\log(1+y)=y+o(y)$


then
$$\frac{\sqrt{y^2+4y+8-8\sqrt{1+y}}}{\log(1+y)}=\frac{\sqrt{y^2+4y+8-8-4y+y^2+o(y^2)}}{y+o(y)}=\frac{\sqrt{2y^2+o(y^2)}}{y+o(y)}=\frac{|y|\sqrt{2+o(1)}}{y+o(y)}=\frac{|y|}{y}\frac{\sqrt{2+o(1)}}{{1+o(1)}}$$
thus the limit doesn't exist.
</s>