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We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sqrt[3]{x^2})}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=\lim_{x \to -8}\left[\frac{-(x+8)}{x+8}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=-\lim_{x \to -8} \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\\&=-2.\end{align*}

But when I input this

lim\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} as x to -8

into Wolfram|Alpha, it gives the limit $0$.

Why is Wolfram|Alpha making a mistake here?

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    $\begingroup$ You should better find your mistake. $\endgroup$ – gammatester Oct 8 '18 at 9:32
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    $\begingroup$ @gammatester $2+\sqrt[3]{-8}=2+(-2)=0.$ $\endgroup$ – mengdie1982 Oct 8 '18 at 9:46
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    $\begingroup$ No, the principle cube root is $1 +i\sqrt{3} \ne 0$ $\endgroup$ – gammatester Oct 8 '18 at 9:48
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    $\begingroup$ @gammatester We only research the real number domain, not complex analysis. $\endgroup$ – mengdie1982 Oct 8 '18 at 9:50
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    $\begingroup$ But you asked about Wolfram Alpha, and there is lot of information in the answer for your WA query. So you had better asked, why does WA use complex numbers and not what is WA's mistake. $\endgroup$ – gammatester Oct 8 '18 at 9:58
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WolframAlpha understands the expression $\sqrt[3]{x}$ for negative x in a different way than you expect.

Try this: lim\frac{\sqrt{1-x}-3}{2+surd(x,3)} as x to -8

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  • $\begingroup$ you mean that WA thinks \sqrt[3]{x} no longer to be the cubic root for $x$ if $x<0$? $\endgroup$ – mengdie1982 Oct 8 '18 at 9:42
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    $\begingroup$ @mengdie1982: There is no such thing as the cubic root, there are three complex roots. $\endgroup$ – gammatester Oct 8 '18 at 9:46
  • $\begingroup$ I grasp you. Thanks! $\endgroup$ – mengdie1982 Oct 8 '18 at 9:47
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    $\begingroup$ WA gives a cube root in the complex domain. But it will not be the real cube root. The problem is not well stated, really, unless you are told whether to use the real cube root or the principal value cube root in the complex domain, which are different things when the argument is negative. $\endgroup$ – Oscar Lanzi Oct 8 '18 at 9:50
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    $\begingroup$ You should include relevant details in this post, not just a link. Also, it would be beneficial to provide an explanation of the difference here. $\endgroup$ – Em. Oct 9 '18 at 3:42
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If you take the complex roots of $\sqrt[3]{x}$ you get $0$ as the limit, because the denominator is different from zero in this case.

So, Wolfram|Alpha did not make a mistake but just uses a different root of $\sqrt[3]{x}$.

For the real root you get $-2$:

  • $t^3 = -x \Rightarrow \lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = \lim_{t \to 2}\frac{\sqrt{1+t^3}-3}{2-t} = -f'(2) \mbox{ for } f(t) = \sqrt{1+t^3}$

$$f'(t) = \frac{3t^2}{2\sqrt{1+t^3}}\Rightarrow \lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = - f'(2) = -2$$

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  • $\begingroup$ Can you do the same for the other two, imaginary cube roots? $\endgroup$ – Mitch Oct 8 '18 at 20:13
  • $\begingroup$ @Mitch: I am not quite sure what you mean. But, moving along a complex branch would mean: $$\frac{\sqrt{1-x}-3}{2+\sqrt[3]{|x|}\sqrt[3]{-1}} = \frac{\sqrt{1-x}-3}{2+\frac{\sqrt[3]{|x|}}{2}(1\pm i\sqrt{3})} \stackrel{x\in \mathbb{R}, x\to -8}{\longrightarrow} \frac{0}{3 \pm i\sqrt{3}} = 0$$ $\endgroup$ – trancelocation Oct 9 '18 at 6:22
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In Mathematica 11.3 I get

In[1]:= Limit[(Sqrt[1 - x] - 3)/(2 + CubeRoot[x]), x -> -8]
Out[1]= -2

Mathematica Documentation says CubeRoot[x] gives the real-valued cube root of $x$.

Even

In[4]:= -8^(1/3)
Out[4]= -2

Mathematica gives me the correct answers

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I think Wolfram Alpha does the wrong thing here, no matter what everybody else says. The cube root of a negative real number should be real by default. But anyway you can work around it by asking Wolfram Alpha for

$$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2-\sqrt[3]{-x}}$$

Inputting

lim\frac{\sqrt{1-x}-3}{2-\sqrt[3]{-x}} as x to -8

gives $-2$, as expected.

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