$\lim_{n \to \infty}\left(\sqrt[3]{n^3+3n}-\sqrt{n^2-2n}\right) $

Question

\begin{equation} \lim_{n \to \infty}\left(\sqrt[3]{n^3+3n}-\sqrt{n^2-2n}\right) \end{equation}

He probado esto

\begin{equation} \lim_{n \to \infty}\left(\sqrt[3]{(n+1)^3-3n-1}-\sqrt{(n-1)^2-1}\right) \end{equation}

lo que me lleva de nuevo a la indefinición $0\cdot\infty$

Answer 1

Una pista:

$$\sqrt[3]{n^3+3n}-\sqrt{n^2-2n} =\\= \frac{(\sqrt[3]{n^3+3n})^6-(\sqrt{n^2-2n})^6}{\big(\sqrt[3]{n^3+3n}+\sqrt{n^2-2n}\big)\cdot \big((\sqrt[3]{n^3+3n})^4+(\sqrt[3]{n^3+3n})^2\cdot(\sqrt{n^2-2n})^2+(\sqrt{n^2-2n})^4\big)}$$

Adición. $$\sqrt[3]{n^3+3n}-\sqrt{n^2-2n} =\frac{(\sqrt[3]{n^3+3n})^2-(\sqrt{n^2-2n})^2}{\sqrt[3]{n^3+3n}+\sqrt{n^2-2n}}\ (1)$$

$$(\sqrt[3]{n^3+3n})^2-(\sqrt{n^2-2n})^2 =\\= \frac{(\sqrt[3]{n^3+3n})^6-(\sqrt{n^2-2n})^6}{(\sqrt[3]{n^3+3n})^4+(\sqrt[3]{n^3+3n})^2\cdot(\sqrt{n^2-2n})^2+(\sqrt{n^2-2n})^4}\ (2)$$

denominador de $(1)$ da $$\frac{1}{\sqrt[3]{n^3+3n}+\sqrt{n^2-2n}} \sim \frac{1}{2n}$$ denominador de $(2)$ da $$\frac{1}{(\sqrt[3]{n^3+3n})^4+(\sqrt[3]{n^3+3n})^2\cdot(\sqrt{n^2-2n})^2+(\sqrt{n^2-2n})^4} \sim \frac{1}{3n^4}$$ numerador de $(2)$ tienen la máxima potencia en el miembro $3n^4\cdot 2n$ que conduce a un resultado correcto.

Answer 2

$\sqrt[3]{n^3+3n}-\sqrt{n^2-2n}= \sqrt[6]{(n^3+3n)^2}- \sqrt[6]{(n^2-2n)^3}=a^{1/6}-b^{1/6}$ , donde $a=(n^3+3n)^2, b=(n^2-2n)^3$
Simplifiquemos primero $a^{1/6}-b^{1/6}$ . Tenga en cuenta que $a^{1/6}-b^{1/6}=\frac{a^{1/3}-b^{1/3}}{a^{1/6}+b^{1/6}}=\frac{(a^{1/3})^3-(b^{1/3})^3}{(a^{1/6}+b^{1/6})(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3})}=\frac{a-b}{(a^{1/6}+b^{1/6})(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3})}$

El numerador es $a-b=n^6\left ((1+\frac 3{n^2})^2-(1-\frac 2n)^3\right)=n^6(6/n+ 9/n^4+8/n^3-18/n^2)$

$1$ factor en el denominador = $n\left((1+\frac 3n)^{1/3}+(1-\frac 2n)^{1/2}\right)$
$2$ nd factor en el denominador= $n^4\left ( (1+\frac 3{n^2})^{4/3}+(1+\frac 3{n^2})^{2/3}(1-\frac 2n)+(1-\frac 2n)^{1/2}\right)$

Así que tenemos
$\begin{align}\sqrt[3]{n^3+3n}-\sqrt{n^2-2n}&= \sqrt[6]{(n^3+3n)^2}- \sqrt[6]{(n^2-2n)^3}\\&=a^{1/6}-b^{1/6}\\&=\frac{6+\frac 1{n^3}(9+8n-18n^2)}{\left((1+\frac 3n)^{1/3}+(1-\frac 2n)^{1/2}\right)\left ( (1+\frac 3{n^2})^{4/3}+(1+\frac 3{n^2})^{2/3}(1-\frac 2n)+(1-\frac 2n)^{1/2}\right)}\\& \to\frac {6} {2\times 3}=1\end {align}$

Answer 3

\begin{gather*} \left( n^{3} +3n\right)^{\frac{1}{3}} =n\left( 1+\frac{3}{n^{2}}\right)^{\frac{1}{3}}\\ \left( n^{2} -2n\right)^{\frac{1}{2}} =n\left( 1-\frac{2}{n}\right)^{\frac{1}{2}}\\ \therefore \left( n^{3} +3n\right)^{\frac{1}{3}} -\left( n^{2} -2n\right)^{\frac{1}{2}} =n\left(\left( 1+\frac{3}{n^{2}}\right)^{\frac{1}{3}} -\left( 1-\frac{2}{n}\right)^{\frac{1}{2}}\right) \end{gather*} Recordemos que \begin{gather*} ( 1+x)^{n} =1+nx+\frac{n( n-1)}{2!} x^{2} +..\\ \Longrightarrow \left( 1+\frac{3}{n^{2}}\right)^{\frac{1}{3}} =1+\frac{1}{3} \cdotp \frac{3}{n^{2}} +\frac{1}{3} \cdotp \frac{-2}{3} \cdotp \frac{9}{2n^{4}} +..\\ =1+\frac{1}{n^{2}} -\frac{1}{n^{4}} +( other\ higher\ order\ terms)\\ \Longrightarrow \left( 1-\frac{2}{n}\right)^{\frac{1}{2}} =1+\frac{1}{2} \cdotp \frac{-2}{n} +\frac{1}{2} \cdotp \frac{-1}{2} \cdotp \frac{4}{2n^{2}} +...\\ =1-\frac{1}{n} -\frac{1}{n^{2}} +( other\ higher\ order\ terms) \end{gather*} Por lo tanto, \begin{gather*} \Longrightarrow \left( 1+\frac{3}{n^{2}}\right)^{\frac{1}{3}} -\left( 1-\frac{2}{n}\right)^{\frac{1}{2}} =\frac{1}{n} +\frac{2}{n^{2}} +( other\ terms)\\ \Longrightarrow n\left(\left( 1+\frac{3}{n^{2}}\right)^{\frac{1}{3}} -\left( 1-\frac{2}{n}\right)^{\frac{1}{2}}\right) =1+\frac{2}{n} +( other\ terms)\\ \Longrightarrow \lim _{n\rightarrow \infty } n\left(\left( 1+\frac{3}{n^{2}}\right)^{\frac{1}{3}} -\left( 1-\frac{2}{n}\right)^{\frac{1}{2}}\right) =1 \end{gather*} ¿Puedes demostrar por qué el límite de los "otros términos" sería cero?