Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$

So many clever solutions, so many convenient algebraic manipulations to remember... For those of us with small-sized memories, let us come back to plain analysis. The method is automatic, it requires zero idea and it wins.

To wit, the main term in $x^6\pm x^5$ is $x^6$ hence $\sqrt[6]{x^6\pm x^5}=x+o(x)$. Oops... this only gives that the difference is $(x+o(x))-(x+o(x))=o(x)$, but we are after the limit so $o(x)$ is not enough, not even to show that the limit exists. We must go (at least) one step further in the expansions, so let us do that.

Note first that, if $u=o(1)$, then $\sqrt[6]{1+u}=1+\frac16u+o(u)$ because $(1+u)^a=1+au+o(u)$ for every $a$ and because $\sqrt[6]{1+u}=(1+u)^{1/6}$.

Furthermore, $\sqrt[6]{x^6\pm x^5}=x\sqrt[6]{1\pm u}$ with $u=\frac1x$, hence $\sqrt[6]{x^6\pm x^5}=x\left(1\pm\frac16u+o(u)\right)$, that is, $\sqrt[6]{x^6\pm x^5}=x\pm\frac16+o(1)$.

This proves that the difference is $(x+\frac16+o(1))-(x-\frac16+o(1))=\frac13+o(1)$, in other words the limit exists and is $\frac13$. End of the proof.

Corollary: Consider some positive real numbers $n$ and $m$, real numbers $\alpha$ and $\beta$, and functions $A$ and $B$ such that $A(x)$ is $o(x^{n-1})$ and $B(x)$ is $o(x^{m-1})$. Then, $$ \lim_{x\to+\infty}\color{red}{\sqrt[n]{x^n+\alpha x^{n-1}+A(x)}-\sqrt[m]{x^m+\beta x^{m-1}+B(x)}}=\color{red}{\frac{\alpha}n-\frac{\beta}m}. $$

Hint: See this FAQ page. One way to apply what was done there directly is rewrite your limit as $$\lim_{x\to \infty}\left(\sqrt[6]{x^{6}+x^{5}}-x\right)- \lim_{x\rightarrow\infty}\left(\sqrt[6]{x^{6}-x^{5}}-x\right).$$ In any case, take a look at what was done, because the ideas used on that thread will solve your problem.

Hope that helps,

Hint: use $a^6 - b^6 = (a - b)(a^5 + a^4 b + a^3 b^2 + a^2 b^3 + a b^4 + b^5)$ with $a =(x^6 + x^5)^{1/6} = x (1 + 1/x)^{1/6}$ and $b = (x^6 - x^5)^{1/6} = x (1 - 1/x)^{1/6}$.