Evaluate the limit $\lim_{x\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$ Evaluate the limit:
$$\lim_{n\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$$
Using the fact that ${(1 + x)^{1/2} \approx 1 + x/2}$ for "small" x,
I have that $\sqrt{n+1}\approx\sqrt{n}(\frac{1}{2n}+1)$ then $n\rightarrow \infty$. However, following this procedure I end up with the following limit: $\lim_{n\rightarrow \infty}2n^2=\infty$, but the answer is $\frac{1}{2}$. I would be thankful for any help.
 A: $$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})=$$
$$=\sqrt{n^3}\left(\frac{2}{\sqrt{n+2}+\sqrt{n}}-\frac{2}{\sqrt{n+3}+\sqrt{n+1}}\right)=$$
$$=\frac{2\sqrt{n^3}\left(\sqrt{n+3}-\sqrt{n+2}+\sqrt{n+1}-\sqrt{n}\right)}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n+3})}=$$
$$=\tfrac{2\sqrt{n^3}\left(\frac{1}{\sqrt{n+3}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n}}\right)}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n+3})}=\tfrac{2\left(\frac{1}{\sqrt{1+\frac{3}{n}}+\sqrt{1+\frac{2}{n}}}+\frac{1}{\sqrt{1+\frac{1}{n}}+1}\right)}{(\sqrt{1+\frac{2}{n}}+1)(\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{3}{n}})}\rightarrow\frac{2\left(\frac{1}{2}+\frac{1}{2}\right)}{2\cdot2}=\frac{1}{2}.$$
A: Another way to multiply by conjugates:
$$\begin{align}
\sqrt{n+1}+\sqrt{n+2}-\sqrt n-\sqrt{n+3}&={(n+1+2\sqrt{(n+1)(n+2)}+n+2)-(n+2\sqrt{n(n+3)}+n+3)\over\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}}\\
&={2(\sqrt{n^2+3n+2}-\sqrt{n^2+3n})\over\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}}\\
&={4\over(\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3})(\sqrt{n^2+3n+2}+\sqrt{n^2+3n})}\end{align}$$
so
$$\begin{align}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt n-\sqrt{n+3})
&={4\over\displaystyle\left({\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}\over\sqrt n}\right)\left({\sqrt{n^2+3n+2}+\sqrt{n^2+3n})\over n}\right)}\\
&\to{4\over(1+1+1+1)(1+1)}={1\over2}\end{align}$$
A: `You have to use Taylor's expansion at order $2$:
\begin{align}
&\phantom{=}\sqrt{n^3}\bigl(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3}\bigr)\\&=n^2\Bigl(\sqrt{1+\frac1n}+\sqrt{1+\frac2n}-1-\sqrt{1+\frac3 n}\Bigr) \\
&=n^2\biggl(1+\frac1{2n}-\frac1{8n^2}+1+\frac1n-\frac1{2n^2}-1-1-\frac 3{2n}+\frac9{8n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggr)\\
&=n^2\biggl(\frac4{8n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggr)=\frac12+o(1).
\end{align}
A: Answer :
$\sqrt{n^3} (\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3}) =\sqrt{n^3}\frac{(2\sqrt{n^2 +3n+2}-2\sqrt{n^2 +3n})}{\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3}}=\frac{4 \sqrt{n^3} }{(\sqrt{n^2 +3n+2}+\sqrt{n^2 +3n})(\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3})}=\frac{4 \sqrt{n^3} }{n \sqrt{n + 3} + \sqrt{n^3 + 4 n^2 + 5 n + 2} +\sqrt{n^3 + 5 n^2 + 8 n + 4} + \sqrt{n^3 + 3 n^2 + 2 n} +\sqrt{n^3 + 6 n^2 + 11 n + 6} + \sqrt{n^3 + 4 n^2 + 3 n} + \sqrt{n^3 + 5 n^2 + 6 n} + n^{3/2} + 3 \sqrt{ n}}$ $=\frac{4}{\sqrt{1+\frac{3}{n}}+\sqrt{1+\frac{4}{n}+\frac{5}{n^2 }+\frac{2}{n^3 }} +\sqrt{1+\frac{5}{n}+\frac{8}{n^2} +\frac{4}{n^3 }} +\sqrt{1 +\frac{3}{n}+\frac{2}{n^2 }} +\sqrt{1 +\frac{6}{n}+\frac{11}{n^2 }+\frac{6}{n^3 }} +\sqrt{1+\frac{4}{n^2 }+\frac{3}{n^3 }} +\sqrt{1+\frac{5}{n}+\frac{6}{n^2 }} +1+3\sqrt {\frac{1}{n^2 }}} $
So :
$\lim _{n\to+\infty} \sqrt{n^3} (\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})=\frac{4}{8} =\frac{1}{2}$
A: Let $t={1\over n}.$ Limits are for $t\to 0^+.$ The given limit equals $L=\lim f(t),$ where
$$f(t)={1\over t^2}\big( \sqrt{1+t} + \sqrt{1+2t} - 1 - \sqrt{1+3t}\big).$$
$$\text{Let } g(t)= \sqrt{1+t} + \sqrt{1+2t} + 1 + \sqrt{1+3t}.   $$
$$f(t)g(t)={1\over t^2}\big(1+t+1+2t-(1+1+3t)+2\sqrt{(1+t)(1+2t)}-2\sqrt{1+3t}\big)$$
$$={2\over t^2}\big(\sqrt{(1+t)(1+2t)}-\sqrt{1+3t}\big)$$
$$\text{Let } h(t)= \sqrt{(1+t)(1+2t)}+\sqrt{1+3t}.   $$
$$f(t)g(t)h(t)={2\over t^2}\big( 2t^2 \big)$$
Taking limits here gives $\ L\cdot 4\cdot 2=2\cdot 2.$  Thus, $L={1\over 2}.$
