Help finding the limit of the following function as x tends to 0 $$
\lim_{x\to0}\left(\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}\,\right)
$$
From Demidovich 5000 problems in mathematical analysis
 A: You can try with $x=t^2$ (the domain is $x>0$, we assume also $t>0$), so your function becomes
$$
\sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2}}}} - \sqrt{\frac{1}{t^2} - \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2}}}} =
\frac{\sqrt{1+\sqrt{t^2+t^3}}-\sqrt{1-\sqrt{t^2+t^3}}}{t}
$$
and your limit is reduced to computing a derivative.
You can simplify the task observing that the function is also writable as
$$
\frac{2\sqrt{t^2+t^3}}{t}\frac{1}{\sqrt{1+\sqrt{t^2+t^3}}+\sqrt{1-\sqrt{t^2+t^3}}}
$$
and the limit is essentially immediate.
A: $\lim \limits_{x\rightarrow0} \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}=?$
Let $u=\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}$ then the function is
$\sqrt{\frac{1}{x}+ u} -\sqrt{\frac{1}{x}-u}=\frac{1}{\sqrt{x}}(1+ux)^{\frac{1}{2}}-\frac{1}{\sqrt{x}}(1-ux)^{\frac{1}{2}}$ 
Using the generalized Binominal theorem we get: 
$\frac{1}{\sqrt{x}}\sum \limits_{k=0}^\infty\dbinom{\frac{1}{2}}{k}(xu)^k-\frac{1}{\sqrt{x}}\sum \limits_{k=0}^\infty\dbinom{\frac{1}{2}}{k}(-1)^k(xu)^k$
Dependig on that $k$ odd or even the expression becomes for zero or twice one.
When $k=2i+1$ (k, i are integers) we get:
$\frac{2}{\sqrt{x}}\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(xu)^{2i}(xu)=2\sqrt{x}u\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(xu)^{2i}$
and substitue back the expression of $u$:
$2(1+\sqrt{x})^{\frac{1}{2}}\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i}=2\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i+\frac{1}{2}}$ 
Separate $i=0$ case:
$2\dbinom{\frac{1}{2}}{1}(x)^0(1+\sqrt{x})^{\frac{1}{2}}+2\sum \limits_{i=1}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i+\frac{1}{2}}$
As 
$\dbinom{\frac{1}{2}}{1}=\frac{\Gamma(\frac{3}{2})}{\Gamma(\frac{1}{2})}=\frac{\frac{1}{2}\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2})}=\frac{1}{2}$ the function is the following:
$(1+\sqrt{x})^{\frac{1}{2}}+2\sum \limits_{i=1}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i+\frac{1}{2}}$ and take its limit:
$\lim \limits_{x\rightarrow0}(1+\sqrt{x})^{\frac{1}{2}}\rightarrow1$ and for all other $i$ the sum keeps to $0$
Finally
$\lim \limits_{x\rightarrow0} \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}=1$
