<s>[INST] Help with limit of function 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?
 [/INST] 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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