Calculating limit of $\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$ As the title says we want to calculate:
$$\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$$
By multiplying nominator and denominator in their conjugates 
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(x+1+x+2\sqrt{x(x+1)}-4(x+2))}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(x+2+x-4+2\sqrt{(x+2)(x-4)})-4x)}$
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$
I think now we can take $$2x\approx2\sqrt{x^2+x}\approx2\sqrt{x^2-2x-8}\\[2ex]
\sqrt{x}\approx\sqrt{x+1}\approx\sqrt{x+2}\approx\sqrt{x-4}$$ as $x$ goes to infinity. Hence the limit of above fraction would be $\dfrac{7}{2}$, but wolframalpha gives me $\dfrac{3}{2}$ as the limit of the above fraction.
What am I doing wrong?
 A: Divide top and bottom by $\sqrt{x}$
$$ \lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}} = \lim_{x\to\infty}\dfrac{\sqrt{1+\frac1{x}}-2\sqrt{1+\frac{2}{x}}+\sqrt{1}}{\sqrt{1+\frac{2}{x}}-2\sqrt{1}+\sqrt{1-\frac{4}{x}}}$$
$$ \to \frac{0}{0}$$
so use L'Hopitals rule differentiate top and bottom
$$\lim_{x\to\infty}\dfrac{-\frac1{x^2} \frac1{2} \frac1{\sqrt{1 + \frac1{x} }}+ \frac{2}{x^2}\frac1{\sqrt{1 + \frac{2}{x}}}}{ - \frac1{x^2} \frac1{\sqrt{1 + \frac{2}{x}}} + \frac{2}{x^2} \frac1{\sqrt{1 - \frac{4}{x}}}} $$
Multiply top and bottom by $x^2$ and take the limit
$$ \dfrac{-\frac1{2} + 2}{-1 + 2} = \frac{3}{2}$$
A: You need one more conjugate-multiplication.
You already have
$$\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$$
First, note that
$$\dfrac{\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4}}{\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x}}=\frac{\sqrt{1+\frac 2x}+2+\sqrt{1-\frac 4x}}{\sqrt{1+\frac 1x}+2\sqrt{1+\frac 2x}+1}\to 1\ (x\to \infty)$$
Now multiplying 
$$\dfrac{-2x-7+2\sqrt{x^2+x}}{-2x-2+2\sqrt{x^2-2x-8}}$$
by
$$\frac{-2x-7-2\sqrt{x^2+x}}{-2x-7-2\sqrt{x^2+x}}\cdot\frac{-2x-2-2\sqrt{x^2-2x-8}}{-2x-2-2\sqrt{x^2-2x-8}}\ (=1)$$
gives
$$\frac{(-2x-2-2\sqrt{x^2-2x-8})((-2x-7)^2-4(x^2+x)}{(-2x-7-2\sqrt{x^2+x})((-2x-2)^2-4(x^2-2x-8))}=\frac{(-2x-2-2\sqrt{x^2-2x-8})(24x+49)}{(-2x-7-2\sqrt{x^2+x})(16x+36)}=\frac{(-2-\frac 2x-2\sqrt{1-\frac 2x-\frac{8}{x^2}})(24+\frac{49}{x})}{(-2-\frac 7x-2\sqrt{1+\frac 1x})(16+\frac{36}{x})}\to \frac 32\ (x\to\infty)$$
A: You can write the fraction as $$\frac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}=\frac{(\sqrt{x+1}-\sqrt{x+2})+(\sqrt{x}-\sqrt{x+2})}{(\sqrt{x+2}-\sqrt{x})+(\sqrt{x-4}-\sqrt{x})}=\frac{-\frac{1}{\sqrt{x+2}+\sqrt{x+1}}-\frac{2}{\sqrt{x}+\sqrt{x+2}}}{\frac{2}{\sqrt{x+2}+\sqrt{x}}-\frac{4}{\sqrt{x-2}+\sqrt{x}}}=\frac{-\frac{\sqrt{x}}{\sqrt{x+2}+\sqrt{x+1}}-\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{x+2}}}{\frac{2\sqrt{x}}{\sqrt{x+2}+\sqrt{x}}-\frac{4\sqrt{x}}{\sqrt{x-2}+\sqrt{x}}},$$ and as $x\to\infty$, this converges to $$\frac{-\frac{1}{2}-\frac{2}{2}}{\frac{2}{2}-\frac{4}{2}}=\frac{3}{2}.$$
