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How to find

$$\lim _{x\rightarrow \infty }\left[ \sqrt [6] {x^{6}+x^{5}}-\sqrt [6] {x^{6}-x^{5}}\right]\;?$$

Elias Costa
  • 14,658

2 Answers2

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$$(x^6+x^5)-(x^6-x^5) = \left [(x^6+x^5)^{1/6} - (x^6-x^5)^{1/6} \right ] \times \\\left [(x^6+x^5)^{5/6} + (x^6+x^5)^{4/6} (x^6-x^5)^{1/6}+ (x^6+x^5)^{3/6} (x^6-x^5)^{2/6}+ (x^6+x^5)^{2/6} (x^6-x^5)^{3/6}+ (x^6+x^5)^{1/6} (x^6-x^5)^{4/6}+ (x^6-x^5)^{5/6} \right ] $$

so as $x \to \infty$, we have

$$(x^6+x^5)^{1/6} - (x^6-x^5)^{1/6} \sim \frac{2 x^5}{6 (x^6)^{5/6}} = \frac13$$

Ron Gordon
  • 138,521
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It may be simpler to manage if we rewrite it as $xA-xB$, where $A=\sqrt[6]{1+1/x}$ and $B=\sqrt[6]{1-1/x}$. Then \begin{align} xA-xB&=x\frac{(A-B)(A^5+A^4B+A^3B^2+A^2B^3+AB^4+B^5)}{A^5+A^4B+A^3B^2+A^2B^3+AB^4+B^5}\\[2ex] &=x\frac{A^6-B^6}{A^5+A^4B+A^3B^2+A^2B^3+AB^4+B^5}\\[2ex] &=x\left(1+\frac{1}{x}-1+\frac{1}{x}\right) \frac{1}{A^5+A^4B+A^3B^2+A^2B^3+AB^4+B^5}\\[2ex] &=\frac{2}{A^5+A^4B+A^3B^2+A^2B^3+AB^4+B^5} \end{align} Computing the limit at the denominator is easy, because $$ \lim_{x\to\infty}A=1=\lim_{x\to\infty}B. $$

Note: it should more properly be $A(x)$ and $B(x)$ rather than $A$ and $B$, but the abbreviated form saves space.

egreg
  • 238,574
  • Not to sound combative, but simpler than what? My solution? Looks pretty much like my solution in spirit, except you take the extra step of factoring out an $x$ from the root, which is really not necessary when you use the difference of sixth powers as we have done. – Ron Gordon Nov 17 '13 at 15:17
  • @RonGordon I'm not saying mine is different from yours; it's an alternative way to look at it and, in my opinion, simpler to manage. I remove the indeterminations at the beginning, basically, rather than at the end. I've changed the starting sentence. – egreg Nov 17 '13 at 15:25