# Confusion over Mathematica 9.0, when calculating the limit. [closed]

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

• $\sqrt{}$ and $\sqrt{}$ 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.
– Ian
Apr 23 '16 at 15:35
• @Ian To do that, see mathematica.stackexchange.com/questions/3886/…
– user332714
Apr 23 '16 at 15:38
• How looks the command to get the correct result in R? Apr 23 '16 at 15:38
• Mathematica has some idiosyncratic defaults with its root functions. Apr 23 '16 at 15:38

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 $\;n\;$
$$\frac{1+\sqrtx}{1+\sqrtx}=\frac{x^{2/3}-x^{1/3}+1}{x^{4/5}-x^{3/5}+x^{2/5}-x^{1/5}+1}\xrightarrow[x\to-1]{}\frac35$$
$\displaystyle \lim_{x \to -1 }\dfrac{1+ \sqrt{x}}{1+ \sqrt{x}}=\displaystyle \lim_{x \to -1} \frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{3}x^{-\frac{2}{3}}}=\displaystyle \lim_{x \to -1} \frac{3}{5} x^{-\frac{2}{15}}=\frac{3}{5}$
With L'Hôpital's rule $$\lim _{ x\to -1 }{ \frac { 1+\sqrt [ 5 ]{ x } }{ 1+\sqrt [ 3 ]{ x } } } =\lim _{ x\to -1 }{ \frac { \frac { 1 }{ 5\sqrt [ 5 ]{ { x }^{ 4 } } } }{ \frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 2 } } } } =\frac { 3 }{ 5 } }$$