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

The answer is $-\frac{1}{4}$, I don't know how to get it though.

Thanks for your help.

UPDATE

I found solution!

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

Now, $t=\frac{1}{\sqrt{x}}$ and $t \to 0$

$$\lim_{t \to 0} \frac{\sqrt{1+2t}-2\sqrt{1+t}+1}{t^2}=$$ $$\lim_{t \to 0} \frac{(1+2t)^\frac{1}{2}-2(1+t)^\frac{1}{2}+1}{t^2}=[\frac{0}{0}]$$

Using L'Hôpital's rule: $$\lim_{t \to 0} \frac{\frac{1}{2}2(1+2t)^{-\frac{1}{2}}-2\frac{1}{2}(1+t)^{-\frac{1}{2}}}{2t}=\lim_{t \to 0} \frac{(1+2t)^{-\frac{1}{2}}-(1+t)^{-\frac{1}{2}}}{2t}=[\frac{0}{0}]$$

And again: $$\lim_{t \to 0} \frac{-\frac{1}{2}2(1+2t)^{-\frac{3}{2}}+\frac{1}{2}(1+t)^{-\frac{3}{2}}}{2}=\lim_{t \to 0} \frac{-(1+2t)^{-\frac{3}{2}}+\frac{1}{2}(1+t)^{-\frac{3}{2}}}{2}=\frac{-1+\frac{1}{2}}{2}=-\frac{1}{4}$$

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  • $\begingroup$ See this answer for a solution. Variants of this limit have occured here earlier. $\endgroup$ – Jyrki Lahtonen Jul 15 '16 at 11:44
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Since $$\sqrt{x+2}-\sqrt{x+1} = \frac{1}{\sqrt{x+1}+\sqrt{x+2}}, \tag{1}$$ $$\begin{eqnarray*}\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x}&=&\frac{1}{\sqrt{x+1}+\sqrt{x+2}}-\frac{1}{\sqrt{x}+\sqrt{x+1}}\\&=&\frac{\sqrt{x}-\sqrt{x+2}}{(\sqrt{x+1}+\sqrt{x+2})(\sqrt{x}+\sqrt{x+1})}\\&=&-\frac{2}{(\sqrt{x+1}+\sqrt{x+2})(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x+2})}\end{eqnarray*}\tag{2}$$ clearly behaves like $-\frac{1}{4x^{3/2}}$.

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  • $\begingroup$ I found the answer! Have a look at updated question post. $\endgroup$ – Andrei Jul 15 '16 at 11:53
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Set $1/x=h>0$ to find

$$\lim_{h\to0^+}\dfrac{(1+2h)^{1/2}-2(1+h)^{1/2}+1}{h^2}$$

Now use Binomial series , $$(1+x)^n=1+nx+\dfrac{n(n-1)x^2}2+\text{the trems containing higher powers of }x$$

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  • $\begingroup$ I don't think think that it is the easiest way to use binomial series. But by the way. Thank's. $\endgroup$ – Andrei Jul 15 '16 at 11:11
  • $\begingroup$ @Andrei, $$(1+rh)^{1/2}=1+\dfrac{rh}2-\dfrac{(rh)^2}{4}+\cdots$$ Here $r=1,2$. Please don't miss to let me know you find any easier way. $\endgroup$ – lab bhattacharjee Jul 15 '16 at 11:14
  • $\begingroup$ I found the answer! Look at my question again ;) $\endgroup$ – Andrei Jul 15 '16 at 11:53
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Set $1/x=h>0$ to find $$F=\lim_{h\to0^+}\dfrac{\sqrt{1+2h}-\sqrt{1+h}-(\sqrt{1+h}-1)}{h^2}$$

As $1+2h-(1+h)=h=1+h-1,$

$$F=\lim_{h\to0^+}\dfrac{\dfrac{1+2h-(1+h)}{\sqrt{1+2h}+\sqrt{1+h}}-\dfrac {1+h-1}{\sqrt{1+h}+1}}{h^2}$$

$=\lim_{h\to0^+}\dfrac1{(\sqrt{1+2h}+\sqrt{1+h})(\sqrt{1+h}+1)}$$\cdot\lim_{h\to0^+}\dfrac{\sqrt{1+h}+1-(\sqrt{1+2h}+\sqrt{1+h})}h$

$$=-\dfrac1{(\sqrt{1+2\cdot0}+\sqrt{1+0})(\sqrt{1+0}+1)}\cdot\lim_{h\to0^+}\dfrac{\sqrt{1+2h}-1}h$$

Now $$\lim_{h\to0^+}\dfrac{\sqrt{1+2h}-1}h=\lim_{h\to0^+}\dfrac{1+2h-1}{(\sqrt{1+2h}+1)h}=?$$

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  • $\begingroup$ @Downvoter, Would you please pinpoint the mistake? $\endgroup$ – lab bhattacharjee Jul 15 '16 at 11:35

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