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How can I calculate the limit $$\lim_{x \to \infty} x^{3/2}( \sqrt{x+1}+ \sqrt{x-1}-2 \sqrt{x})$$

I had ideas like using Laurent series, but I dont think I am allowed since its an elementary course, I tried to play around with the terms but I didnt manage. Help anyone?

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    $\begingroup$ Why the downvotes? It looks like a good question to me. $\endgroup$ – Jack D'Aurizio May 28 '15 at 11:15
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    $\begingroup$ $$x^{3/2}( \sqrt{x+1}+ \sqrt{x-1}-2 \sqrt{x})=-2\frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x-1}}\cdot\frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}+\sqrt{x-1}}$$ $\endgroup$ – Did May 28 '15 at 11:16
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HINT:

Set $1/x=h\implies h\to0^+$ to get

$$\lim_{h\to0^+}\dfrac{\sqrt{1+h}+\sqrt{1-h}-2}{h^2}$$

Now use Binomial series for $\sqrt{1+h}=(1+h)^{1/2}=1+\dfrac h2+\dfrac{1/2(1/2-1)h^2}{2!}+O(h^3)$

and for $\sqrt{1-h}=(1-h)^{1/2}$

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Classic limits involving $\sqrt{1+h}$ and $\sqrt{1-h}$ are \begin{eqnarray*} \lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1+h}-1-\frac{1}{2}h\right) }{% h^{2}} &=&-\frac{1}{8}, \\ \lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1-h}-1+\frac{1}{2}% h\right) }{h^{2}} &=&-\frac{1}{8}. \end{eqnarray*} Therefore it suffices to write the expression as the sum of those classic limits as follows \begin{eqnarray*} \frac{\sqrt{1+h}+\sqrt{1-h}-2}{h^{2}} &=&\frac{\left( \sqrt{1+h}-1-\frac{1}{2% }h\right) +\left( \sqrt{1-h}-1+\frac{1}{2}h\right) }{h^{2}} \\ &=&\frac{\left( \sqrt{1+h}-1-\frac{1}{2}h\right) }{h^{2}}+\frac{\left( \sqrt{% 1-h}- 1+\frac{1}{2}h\right) }{h^{2}} \end{eqnarray*} Therefore \begin{eqnarray*} \lim_{h\rightarrow 0^{+}}\frac{\sqrt{1+h}+\sqrt{1-h}-2}{h^{2}} &=&\lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1+h}-1-\frac{1}{2}h\right) }{% h^{2}}+\lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1-h}- 1+\frac{1% }{2}h\right) }{h^{2}} \\ &=&\left( -\frac{1}{8}\right) +\left( -\frac{1}{8}\right) =-\frac{1}{4}. \end{eqnarray*}

(Each classic limit above can be computed by l'Hospital's rule (twice) for example.

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we will use the binomial theorem $$(big + small)^{1/2} = (big)^{1/2} + \frac12(big)^{-1/2}small -\frac18 (big)^{-3/2}\small^2+\cdots $$ so that $$\begin{align} (x+1)^{1/2} &= x^{1/2} + \frac12 x^{-1/2}-\frac18 x^{-3/2}+\cdots \\ (x-1)^{1/2} &= x^{1/2} - \frac12 x^{-1/2}-\frac18 x^{-3/2}+\cdots \\ \end{align}$$

therefore $$ \sqrt{x+1}+ \sqrt{x-1}-2 \sqrt{x} = -\frac14x^{-3/2}+\cdots$$ and $$\lim_{x \to \infty} x^{3/2}( \sqrt{x+1}+ \sqrt{x-1}-2 \sqrt{x}) = -\frac14.$$

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