Evaluation of $ \lim_{ x \to \infty} x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$ I am trying to calculate $$ \lim_{ x \to \infty}  x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$$
Whenever I try it all I can seem to reach is the conclusion that it goes to infinity, however when I try it in wolframalpha it gives me the answer -1/4! How is that value reached?! I can't see any way of cancelling x's or using binomial expansion or anything!
 A: Hint: You may find it useful to multiply and divide by $x^{1/2}$, obtaining
$$x^2\left(\left(1+\frac{2}{x}\right)^{1/2} -2\left(1+\frac{1}{x}\right)^{1/2}+1\right).$$
 Then expand the square roots, up to the $1/x^2$ terms, using the (general) Binomial Theorem. 
There are other methods. For example, after the simplification above, use L'Hospital's Rule, perhaps after making the change of variable $u=1/x$. You will probably use the Rule twice, though there are alternatives after the first application of the Rule.
A: $$\displaystyle \lim_{x \to \infty}x^{3/2}(\sqrt {x+2}+\sqrt x-2\sqrt{x+1})=\displaystyle \lim_{x \to \infty} \frac{x^2(-2x-2+2\sqrt{x^2+2x})}{\sqrt x(\sqrt {x+2}+\sqrt x+2\sqrt{x+1})}=$$
$$=\displaystyle \lim_{x \to \infty}\frac{2x^2(\sqrt{x^2+2x}-(x+1))(\sqrt{x^2+2x}+(x+1))}{\sqrt x (\sqrt {x+2}+\sqrt x+2\sqrt{x+1})(\sqrt{x^2+2x}+(x+1))}=$$
$$=\displaystyle \lim_{x \to \infty} \frac{-2x^2}{\sqrt x (\sqrt {x+2}+\sqrt x+2\sqrt{x+1})(\sqrt{x^2+2x}+(x+1))}=$$
$$=\displaystyle \lim_{x \to \infty} \frac{-2}{\left(\sqrt{1+\frac{2}{x}}+\sqrt 1+2\sqrt{1+\frac{1}{x}}\right)\left(\sqrt{1+\frac{2}{x}}+1+\frac{1}{x}\right)}=$$
$$=\frac{-2}{ 4 \cdot 2} =\frac{-1}{4}$$
