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How would you evaluate the following limit: $$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$

I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.

Any hints?

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  • $\begingroup$ Hint: $x^6 + x^5 = (x+\frac{1}{6})^6 - p(x)$ where $p(x)$ is a polynomial of degree 4. Similarly, $x^6 - x^5 = (x-\frac{1}{6})^6 - q(x)$ where $q(x)$ is of degree 4. $\endgroup$ – Thomas Andrews Jun 17 '11 at 20:28
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    $\begingroup$ @Thomas Andrews: How did you come up with $1/6$ in $(x\pm 1/6)^6$? $\endgroup$ – Américo Tavares Jun 18 '11 at 9:08
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    $\begingroup$ @Américo: use the binomial formula. For the second term you'll get $6x^5 ({1\over 6}) = x^5$. So just work in reverse. The idea is similar to completing a square... $\endgroup$ – Marek Jun 18 '11 at 10:10
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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, $$ \lim_{x\to+\infty}\color{red}{\sqrt[n]{x^n+\alpha x^{n-1}+A(x)}-\sqrt[m]{x^m+\beta x^{m-1}+B(x)}}=\color{red}{\frac{\alpha}n-\frac{\beta}m}. $$

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  • $\begingroup$ +1 for "plain analysis". Do you think the "analysis" tag should be added? $\endgroup$ – Américo Tavares Jun 18 '11 at 0:31
  • $\begingroup$ @Américo: Thanks. // No. $\endgroup$ – Did Jun 18 '11 at 5:39
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    $\begingroup$ beautiful answers as always. Thank you. $\endgroup$ – user6163 Jun 18 '11 at 7:44
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    $\begingroup$ Very clever-and proves that you should never forget the baby step algebra you do in either high school or early undergraduate work because you never know when it's going to come in handy. $\endgroup$ – Mathemagician1234 Nov 13 '11 at 1:29
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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,

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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 $b = (x^6 - x^5)^{1/6} = x (1 - 1/x)^{1/6}$.

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Edit: The main idea is to use the expansion

$$\sqrt[6]{1+x^{-1}}=1+\frac{1}{6}x^{-1}+O\left( x^{-2}\right) $$

which can be generated by changing variables $x=1/z$, expanding about $z=0$, $$\sqrt[6]{1+z}=1+\frac{1}{6}z+O\left( z^{2}\right) $$ and restoring $x$ via $% z\rightarrow 1/x$.


Let $f(x)=\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$. If $\lim_{x\rightarrow +\infty }(x)=L$, then $\lim_{x\rightarrow -\infty }f(x)=-L$ because $f(-x)=-f(x)$. We have

$$f(x) =\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}=\left\vert x\right\vert \left( \sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}\right). $$

Since

$$\sqrt[6]{1+x^{-1}} =1+\frac{1}{6}x^{-1}+O\left( x^{-2}\right)\quad\text{ and }\quad\sqrt[6]{1-x^{-1}} =1-\frac{1}{6}x^{-1}+O\left( x^{-2}\right),$$

we get

$$\sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}=\frac{1}{3}x^{-1}+O\left( x^{-2}\right)$$

and for $x>0$

$$x\left( \sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}\right) =\frac{1}{3}+O\left( x^{-1}\right).$$

Therefore

$$L=\lim_{x\rightarrow +\infty }f(x) =\lim_{x\rightarrow +\infty }\left\vert x\right\vert \left( \sqrt[6]{1+x^{-1}}-\sqrt[6]{1-x^{-1}}\right)=\lim_{x\rightarrow +\infty }\frac{1}{3}+O\left( x^{-1}\right) =\frac{1}{3}.$$


Edit 2: graph of $f(x)$

enter image description here

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$$ \begin{aligned} \lim _{x\to \infty }\left(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5}\right) & = \lim _{t\to 0}\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}-\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right) \\& = \lim _{t\to 0}\left(\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}-\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)\cdot \frac{\left(\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}+\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}\right)}{\sqrt[6]{\frac{1}{t^6}+\frac{1}{t^5}}+\sqrt[6]{\frac{1}{t^6}-\frac{1}{t^5}}}\right) \\& = \lim _{t\to 0}\left(\frac{\sqrt[3]{t+1}-\sqrt[3]{-t+1}}{t\left(\sqrt[6]{-t+1}+\sqrt[6]{t+1}\right)}\right) \\& = \lim _{t\to 0}\left(\frac{\left(1+\frac{1}{3}t+o\left(t\right)\right)-\left(1-\frac{1}{3}t+o\left(t\right)\right)}{t\left(2+o\left(1\right)\right)}\right) \\& = \color{red}{\frac{1}{3}} \end{aligned} $$

Solved with substitution $t=\frac{1}{x}$ and Taylor expansion

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