Find the limit of function using Taylor series Good evening, 
I'm somehow stuck on solving some easy exercises :
$$\lim_{x\to\infty} x^{3/2}\bigl(\sqrt{x+1}+\sqrt{x-1}-2\,\sqrt{x}\bigr)$$
 A: $$
\sqrt{1+x}=\sqrt{x}\,\Bigl(1+\frac1x\Bigr)^{1/2}=\sqrt{x}\,\sum_{n=0}^\infty\binom{1/2}{n}x^n=\sqrt{x}\,\Bigl(1+\frac12\,\frac1x+\dots\Bigr).
$$
Similarly for $\sqrt{x-1}$.
A: it is the following term $$-1/4-{\frac {5}{64\,{x}^{2}}}-{\frac {21}{512\,{x}^{4}}}+O \left( {x}^
{-5} \right)
$$
A: $$\sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x}+\sqrt{x+1}},\qquad \sqrt{x-1}-\sqrt{x}=-\frac{1}{\sqrt{x}+\sqrt{x-1}},$$
hence:
$$\sqrt{x+1}+\sqrt{x-1}-2\sqrt{x}=\frac{\sqrt{x-1}-\sqrt{x+1}}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x-1})}=\frac{-2}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x-1})(\sqrt{x-1}+\sqrt{x+1})}\approx\frac{-2}{(2\sqrt{x})^3}$$
and the limit is $\color{red}{-\frac{1}{4}}$.
A: $$\begin{align}&x\sqrt x\left(\sqrt{x+1}+\sqrt{x-1}-2\sqrt x\right)=x\left(\sqrt{x^2+x}+\sqrt{x^2-x}-2x\right)=\\{}\\
&=x\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{x^2+x}+\sqrt{x^2-x}+2x}=\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2}=\\{}\\
&=\frac{4(x^4-x^2)-4x^4}{\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\right)\left(2\sqrt{x^4-x^2}+2x^2\right)}\\{}\\
&=\frac{-4x^2}{2x^2\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\right)\left(\sqrt{1-\frac1{x^2}}+1\right)}\xrightarrow[x\to\infty]{}\frac{-2}{(1+1+2)(1+1)}=-\frac14\end{align}$$
