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Why in Wolfram Mathematica 9.0 I get $\displaystyle \lim_{x \to -1 }\dfrac{1+ \sqrt[5]{x}}{1+ \sqrt[3]{x}}=\frac{1+\sqrt[5]{-1}}{1+\sqrt[3] {-1}}$. But, my solution is: 1. Thanks.

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    $\begingroup$ $\sqrt[5]{}$ and $\sqrt[3]{}$ are ambiguous if you are working over the complex numbers, which Mathematica almost certainly does by default. There should be a way to force it to keep everything real, though. $\endgroup$
    – Ian
    Apr 23 '16 at 15:35
  • $\begingroup$ @Ian To do that, see mathematica.stackexchange.com/questions/3886/… $\endgroup$
    – user332714
    Apr 23 '16 at 15:38
  • $\begingroup$ How looks the command to get the correct result in R? $\endgroup$
    – J. Jam
    Apr 23 '16 at 15:38
  • $\begingroup$ Mathematica has some idiosyncratic defaults with its root functions. $\endgroup$
    – hunter
    Apr 23 '16 at 15:38
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I don't think the correct answer is $\;1\;$. Using $\;x^n+1=(x+1)(x^{n-1}-x^{n-2}+\ldots-x+1)\;$ for odd $\;n\;$

$$\frac{1+\sqrt[5]x}{1+\sqrt[3]x}=\frac{x^{2/3}-x^{1/3}+1}{x^{4/5}-x^{3/5}+x^{2/5}-x^{1/5}+1}\xrightarrow[x\to-1]{}\frac35$$

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You can also use L'Hopital's rule explicitly, here, because both the numerator and denominator approach zero, and their derivatives exist. Hence:

$\displaystyle \lim_{x \to -1 }\dfrac{1+ \sqrt[5]{x}}{1+ \sqrt[3]{x}}=\displaystyle \lim_{x \to -1} \frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{3}x^{-\frac{2}{3}}}=\displaystyle \lim_{x \to -1} \frac{3}{5} x^{-\frac{2}{15}}=\frac{3}{5}$

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With L'Hôpital's rule $$\lim _{ x\to -1 }{ \frac { 1+\sqrt [ 5 ]{ x } }{ 1+\sqrt [ 3 ]{ x } } } =\lim _{ x\to -1 }{ \frac { \frac { 1 }{ 5\sqrt [ 5 ]{ { x }^{ 4 } } } }{ \frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 2 } } } } =\frac { 3 }{ 5 } } $$

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