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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!

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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.

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    @simonthumper: I have been around longer.2012-03-30
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$\displaystyle \lim_{x \to \infty}x^{3/2}(\sqrt {x+2}+\sqrt x-2\sqrt{x+1})=\displaystyle \lim_{x \to \infty} \frac{x^2(-2x-2+2\sqrt{x^2+2x})}{\sqrt x(\sqrt {x+2}+\sqrt x+2\sqrt{x+1})}=$

$=\displaystyle \lim_{x \to \infty}\frac{2x^2(\sqrt{x^2+2x}-(x+1))(\sqrt{x^2+2x}+(x+1))}{\sqrt x (\sqrt {x+2}+\sqrt x+2\sqrt{x+1})(\sqrt{x^2+2x}+(x+1))}=$

$=\displaystyle \lim_{x \to \infty} \frac{-2x^2}{\sqrt x (\sqrt {x+2}+\sqrt x+2\sqrt{x+1})(\sqrt{x^2+2x}+(x+1))}=$

$=\displaystyle \lim_{x \to \infty} \frac{-2}{\left(\sqrt{1+\frac{2}{x}}+\sqrt 1+2\sqrt{1+\frac{1}{x}}\right)\left(\sqrt{1+\frac{2}{x}}+1+\frac{1}{x}\right)}=$

$=\frac{-2}{ 4 \cdot 2} =\frac{-1}{4}$