Why the two limits are equivalent?

Question

I was reading @user164524's answer here. I wonder why are the two following limits equivalent,

$$\lim_{x\to\infty}\dfrac{\sqrt[3]{x^5+x^4}-\sqrt[3]{x^5-x^4}}{\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}}=\lim_{x\to0}\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x\sqrt[6]{1+x}+x\sqrt[6]{1-x}}$$

It seems we used the substitution $u=\frac1x$ and as $x\to\infty$ , $u\to0$. So after dividing the numerator and denominator by $\sqrt[3]{x^5}$ I get, $$\lim_{u\to0}\frac{\sqrt[3]{1+u}-\sqrt[3]{1-u}}{\sqrt[6]{u^5+u^6}+\sqrt[6]{u^5-u^6}}=\lim_{u\to0}\frac{\sqrt[3]{1+u}-\sqrt[3]{1-u}}{|u|\sqrt[6]{\frac1u+1}+|u|\sqrt[6]{\frac1u-1}}$$ But it is different from the limit I mentioned.

Answer 1

The limits are not equivalent. There are some mistakes in the referred question. Note, the title in the referred question states \begin{align*} \lim_{x\to\infty}\left(\left(x^\color{blue}{5}+x^\color{blue}{4}\right)^{\frac{1}{6}}-\left(x^\color{blue}{5}-x^\color{blue}{4}\right)^{\frac{1}{6}}\right) \end{align*} whereas the body of the referred question states \begin{align*} \lim_{x\to\infty}\left(\left(x^\color{blue}{6}+x^\color{blue}{5}\right)^{\frac{1}{6}}-\left(x^\color{blue}{6}-x^\color{blue}{5}\right)^{\frac{1}{6}}\right) \end{align*}

Taking the formula from the title we obtain \begin{align*} \lim_{x\to\infty}&\left(\left(x^5+x^4\right)^{\frac{1}{6}}-\left(x^5-x^4\right)^{\frac{1}{6}}\right)\\ &=\lim_{x\to\infty}\frac{\left(\sqrt[6]{x^5+x^4}-\sqrt[6]{x^5-x^4}\right)\left(\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}\right)} {\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}}\\ &=\lim_{x\to\infty}\frac{\sqrt[3]{x^5+x^4}-\sqrt[3]{x^5-x^4}} {x^{\frac{5}{6}}\left(\sqrt[6]{1+\frac{1}{x}}+\sqrt[6]{1-\frac{1}{x}}\right)}\\ &=\lim_{x\to\infty}\frac{x^{\frac{5}{6}}\left(\sqrt[3]{1+\frac{1}{x}}-\sqrt[3]{1-\frac{1}{x}}\right)} {\sqrt[6]{1+\frac{1}{x}}+\sqrt[6]{1-\frac{1}{x}}}\\ \end{align*}

Hint: One answer in the referred question mixes up expressions $x^5-x^4$ with $x^6-x^5$ and similar things.

Here we show \begin{align*} \color{blue}{\lim_{x\to\infty}\left(\left(x^5+x^4\right)^{\frac{1}{6}}-\left(x^5-x^4\right)^{\frac{1}{6}}\right)=0}\tag{1} \end{align*}

We recall \begin{align*} a^6-b^6&=\left(a-b\right)\left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)\\ &=(a-b)\sum_{j=0}^5a^{5-j}b^{j}\tag{2} \end{align*} Letting \begin{align*} a&=\left(x^5+x^4\right)^{\frac{1}{6}}=x^{\frac{5}{6}}\left(1+\frac{1}{x}\right)^{\frac{1}{6}}\\ b&=\left(x^5-x^4\right)^{\frac{1}{6}}=x^{\frac{5}{6}}\left(1-\frac{1}{x}\right)^{\frac{1}{6}}\\ \end{align*}

we obtain expanding according to (2) \begin{align*} \color{blue}{\lim_{x\to\infty}}&\color{blue}{\left(\left(x^5+x^4\right)^{\frac{1}{6}}-\left(x^5-x^4\right)^{\frac{1}{6}}\right)}\\ &=\lim_{x\to\infty}\frac{\left(\left(x^5+x^4\right)^{\frac{1}{6}}-\left(x^5-x^4\right)^{\frac{1}{6}}\right) \sum_{j=0}^5\left(x^5+x^4\right)^{\frac{5-j}{6}}\left(x^5-x^4\right)^{\frac{j}{6}}} {\sum_{j=0}^5\left(x^5+x^4\right)^{\frac{5-j}{6}}\left(x^5-x^4\right)^{\frac{j}{6}}}\\ &=\lim_{x\to\infty}\frac{\left(x^5+x^4\right)-\left(x^5-x^4\right)} {\sum_{j=0}^5\left(x^5+x^4\right)^{\frac{5-j}{6}}\left(x^5-x^4\right)^{\frac{j}{6}}}\\ &=\lim_{x\to\infty}\frac{2x^4} {x^{\frac{25}{6}}\sum_{j=0}^5\left(1+\frac{1}{x}\right)^{\frac{5-j}{6}}\left(1-\frac{1}{x}\right)^{\frac{j}{6}}}\\ &=\lim_{x\to\infty}\frac{2} {x^{\frac{1}{6}}\sum_{j=0}^5\left(1+\frac{1}{x}\right)^{\frac{5-j}{6}} \left(1-\frac{1}{x}\right)^{\frac{j}{6}}}\\ &\,\,\color{blue}{=0} \end{align*} and the claim (1) follows.

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Hint: The following is valid \begin{align*} \color{blue}{\lim_{x\to\infty}\left(\left(x^6+x^5\right)^{\frac{1}{6}}-\left(x^6-x^5\right)^{\frac{1}{6}}\right)=\frac{1}{3}} \end{align*} Some nice answers can be found here.