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?
 A: 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 $\sqrt{1-h}=(1-h)^{1/2}$
A: 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.
A: 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 
$$\lim_{x \to \infty} x^{3/2}( \sqrt{x+1}+ \sqrt{x-1}-2 \sqrt{x}) = -\frac14.$$
