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?

  • 11
    $\begingroup$ You should better find your mistake. $\endgroup$ Oct 8 '18 at 9:32
  • 1
    $\begingroup$ can you point out my fault? $\endgroup$ Oct 8 '18 at 9:34
  • 1
    $\begingroup$ @gammatester $2+\sqrt[3]{-8}=2+(-2)=0.$ $\endgroup$ Oct 8 '18 at 9:46
  • 1
    $\begingroup$ No, the principle cube root is $1 +i\sqrt{3} \ne 0$ $\endgroup$ Oct 8 '18 at 9:48
  • 1
    $\begingroup$ @gammatester We only research the real number domain, not complex analysis. $\endgroup$ Oct 8 '18 at 9:50

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

  • $\begingroup$ you mean that WA thinks \sqrt[3]{x} no longer to be the cubic root for $x$ if $x<0$? $\endgroup$ 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. $\endgroup$ Oct 8 '18 at 9:46
  • $\begingroup$ I grasp you. Thanks! $\endgroup$ Oct 8 '18 at 9:47
  • 9
    $\begingroup$ 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. $\endgroup$ 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

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 $-2$:

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

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

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

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 $x$.


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

Mathematica gives me the correct answers


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

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


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

gives $-2$, as expected.


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