Evaluate $\lim\limits_{x\to \infty}x\left[2x-\left(x^3+x^2+x\right)^{\frac{1}{3}}-\left(x^3-x^2+x\right)^{\frac{1}{3}}\right]$
My Approach:
Formula I used $(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^2+…..\infty$ where $x\in(-1,1)$
$\lim\limits_{x\to \infty}x\left[2x-x\left(1+\left(\frac{1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}-x\left(1+\left(\frac{-1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}\right]$
$\implies\lim\limits_{x\to \infty}x\left[2x-x\left(1+\left(\frac{1}{3x}+\frac{1}{3x^2}\right)\right)-x\left(1+\left(\frac{-1}{3x}+\frac{1}{3x^2}\right)\right)\right]$
$\implies\lim\limits_{x\to\infty}x\left(-\frac{2}{3x}\right)=-\frac{2}{3}$
But given answer is $\frac{2}{9}$. I am attaching given solution below. Also I am attaching my two more solutions in Image formart.
Best Answer
The formula you used is correct, however you made a couple of mistakes when computing the expansion at infinity, in fact you have: $$\left(1+\frac 1x+\frac 1{x^2}\right)^{1/3}\sim 1+\frac 1{3x}+\frac 2{9x^2} \\ \left(1-\frac 1x+\frac 1{x^2}\right)^{1/3}\sim 1-\frac 1{3x}+\frac 2{9x^2} $$ Hence \begin{align} &\lim_{x\to \infty} x\left[2x-x\left(1+\left(\frac{1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}-x\left(1+\left(\frac{-1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}\right]= \\ &\lim_{x\to \infty} x\left[2x-x\left(1+\frac 1{3x}+\frac 2{9x^2}\right)-x\left(1-\frac 1{3x}+\frac 2{9x^2}\right)^{\frac{1}{3}}\right]= \\ &\lim_{x\to \infty} x\left(2x-x-\frac13-\frac 2{9x} -x+\frac13-\frac 2{9x} \right)= -\frac49 \end{align}