Infinity Limitation

Let the limit be $\mathfrak H$ . Substitute $\star~2^n=x$ followed by using binomial approximation $\star~(1+\color{#20A900}{\alpha})^n=1+n\color{#20A900}{\alpha}~~(\text{as }\alpha\rightarrow 0)$ .

$$ $\begin{aligned}\mathfrak{H}=& \lim_{x\to\infty}\left(\sqrt[3]{x^3+x^2+x}-x\right)\\=&\lim_{x\to\infty}\left(x\left(\sqrt[3]{1+\color{#20A900}{\dfrac{1}{x}+\dfrac{1}{x^2}}}-1\right)\right)\\=&\lim_{x\to\infty}\left(x\left(\cancel{1}+\dfrac{1}{3}\left(\color{#20A900}{\dfrac{1}{x}+\dfrac{1}{x^2}}\right)-\cancel{1}\right)\right)\\=&\lim_{x\to\infty}\left(\dfrac{1}{3}+\underbrace{\dfrac{1}{3x}}_0\right)\approx\large\color{#0C6AC7}{0.333}\end{aligned}$

Let $x=\left(8^n+4^n+2^n\right)^{\frac{1}{3}}$ and $y=2^n=\left(8^n\right)^{\frac{1}{3}}$ . Since $x^3-y^3=(x+y)\left(x^2+xy+y^2\right)$ , we have $x-y=\frac{x^3-y^3}{x^2+xy+y^2}$ , which means that $\left(8^n+4^n+2^n\right)^{\frac{1}{3}} - 2^n = \frac{4^n+2^n}{ \left(8^n+4^n+2^n\right)^{\frac{2}{3}} +\left(8^n+4^n+2^n\right)^{\frac{1}{3}}\times 2^n +4^n } = \frac{1+\left(\frac{1}{2}\right)^n}{ \left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n \right)^{\frac{2}{3}} +1 + 1\left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n\right)^{\frac{1}{3}}}$

This means that $\lim_{n \to \infty}\left(\left(8^n+4^n+2^n\right)^{\frac{1}{3}} - 2^n\right) = \lim_{n \to \infty} \frac{1+\left(\frac{1}{2}\right)^n}{ \left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n \right)^{\frac{2}{3}} +1 + 1\left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n\right)^{\frac{1}{3}}} =\frac{1}{3}$