Find the limit of function using Taylor series

Question

Good evening,

I'm somehow stuck on solving some easy exercises :

$$\lim_{x\to\infty} x^{3/2}\bigl(\sqrt{x+1}+\sqrt{x-1}-2\,\sqrt{x}\bigr)$$

Answer 1

$$ \sqrt{1+x}=\sqrt{x}\,\Bigl(1+\frac1x\Bigr)^{1/2}=\sqrt{x}\,\sum_{n=0}^\infty\binom{1/2}{n}x^n=\sqrt{x}\,\Bigl(1+\frac12\,\frac1x+\dots\Bigr). $$ Similarly for $\sqrt{x-1}$.

Answer 2

it is the following term $$-1/4-{\frac {5}{64\,{x}^{2}}}-{\frac {21}{512\,{x}^{4}}}+O \left( {x}^ {-5} \right) $$

Answer 3

$$\sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x}+\sqrt{x+1}},\qquad \sqrt{x-1}-\sqrt{x}=-\frac{1}{\sqrt{x}+\sqrt{x-1}},$$ 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 $\color{red}{-\frac{1}{4}}$.

Answer 4

$$\begin{align}&x\sqrt x\left(\sqrt{x+1}+\sqrt{x-1}-2\sqrt x\right)=x\left(\sqrt{x^2+x}+\sqrt{x^2-x}-2x\right)=\\{}\\ &=x\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{x^2+x}+\sqrt{x^2-x}+2x}=\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2}=\\{}\\ &=\frac{4(x^4-x^2)-4x^4}{\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\right)\left(2\sqrt{x^4-x^2}+2x^2\right)}\\{}\\ &=\frac{-4x^2}{2x^2\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\right)\left(\sqrt{1-\frac1{x^2}}+1\right)}\xrightarrow[x\to\infty]{}\frac{-2}{(1+1+2)(1+1)}=-\frac14\end{align}$$