Find $\lim_{x\to\infty} \sqrt{x^3} \left(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x}\right)$ I would like to calculate
$$\lim_{x\to\infty} \sqrt{x^3} \left(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x}\right).$$
I know this is an indeterminate $\infty\cdot 0$, but when I rewrite it as an indeterminate $\frac{\infty}{\infty}$ and use l'Hôpital, I get again an indeterminate $\infty \cdot 0$. Using again l'Hôpital doesn't help.
 A: Hint: $$\lim_{x\to\infty} \sqrt{x^3} \left[(\sqrt{x+2}-\sqrt{x+1})-(\sqrt{x+1}-\sqrt{x})\right]=\lim_{x\to\infty} \sqrt{x^3} \left(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}-\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)$$
A: I like to think about this as a central difference:
$$\begin{align}f(x-1)&=f(x)-f^{\prime}(x)+\frac12f^{\prime\prime}(x)-\frac16f^{\prime\prime\prime}(x)+\frac1{24}f^{(4)}(\xi_1)\\
f(x+1)&=f(x)+f^{\prime}(x)+\frac12f^{\prime\prime}(x)+\frac16f^{\prime\prime\prime}(x)+\frac1{24}f^{(4)}(\xi_2)\end{align}$$
So
$$f(x-1)-2f(x)+f(x+1)=f^{\prime\prime}(x)+\frac1{12}f^{(4)}(\xi_3)$$
Where $x-1<\xi_1<\xi_3<\xi_2<x+1$ so with $f(x)=\sqrt{x+1}$
$$\lim_{x\rightarrow\infty}x^{3/2}\left(\sqrt{x+2}-2\sqrt{x+1}+\sqrt x\right)=\lim_{x\rightarrow\infty}x^{3/2}\left(-\frac1{4(x+1)^{3/2}}-\frac5{64}(\xi_3+1)^{-7/2}\right)=-\frac14$$
Because
$$0<\frac{x^{3/2}}{(\xi_3+1)^{7/2}}<\frac{x^{3/2}}{x^{7/2}}$$
A: $$L=\lim_{x \rightarrow \infty} x^{3/2}[x^{1/2}(1+2/x)^{1/2}-2(1+1/x)^{1/2}+x^{1/2}]$$
$$L=\lim_{x \rightarrow \infty}x^2([1+x^{-1}-(1/2)x^{-2}+()x^{-3}]-2(1+(1/2)x^{-1}-(1/8)x^{-2}+()x^{-3}]+1)$$
$$L=\lim_{x \rightarrow \infty}x^2[(-1/4)x^{-2}+()x^{-3}]=-1/4$$
A: Hint
Let $1/x=h,h\to0^+$ to find
$$\lim_{h\to0}\dfrac{\sqrt{1+2h}-2\sqrt{1+h}+1}{h^2}$$
$$=\lim_h\dfrac{(\sqrt{1+h}-1)^2}{h^2}+\lim_h\dfrac{\sqrt{1+2h}-(1+h)}{h^2}$$
Now rationalize the numerator of each limit
