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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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2 Answers 2

up vote 7 down vote accepted

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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It's interesting to note that this algebraic alteration turns the problem into $\lim_{h\to0^+} ( \sqrt{1+2h} - 2\sqrt{1+h} + \sqrt{1} )/h^2$, which suggests that the answer should indeed be the second derivative of $\sqrt{1+x}$ at $x=0$. –  Greg Martin Mar 29 '12 at 17:19
    
Ahhh dammit why can't I spot these things for myself! :P –  simonthumper Mar 30 '12 at 9:02
    
@simonthumper: I have been around longer. –  André Nicolas Mar 30 '12 at 13:26

$$\displaystyle \lim_{x \to \infty}x^{3/2}(\sqrt {x+2}+\sqrt x-2\sqrt{x+1})=\displaystyle \lim_{x \to \infty} \frac{x^2(-2x-2+2\sqrt{x^2+2x})}{\sqrt x(\sqrt {x+2}+\sqrt x+2\sqrt{x+1})}=$$

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

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

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

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

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