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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?

Em.
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mengdie1982
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3 Answers3

37

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

Em.
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Pavel R.
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  • you mean that WA thinks \sqrt[3]{x} no longer to be the cubic root for $x$ if $x<0$? – mengdie1982 Oct 08 '18 at 09:42
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    @mengdie1982: There is no such thing as the cubic root, there are three complex roots. – gammatester Oct 08 '18 at 09:46
  • I grasp you. Thanks! – mengdie1982 Oct 08 '18 at 09:47
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    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. – Oscar Lanzi Oct 08 '18 at 09:50
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    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. – Em. Oct 09 '18 at 03:42
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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 $-2$:

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

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

trancelocation
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  • Can you do the same for the other two, imaginary cube roots? – Mitch Oct 08 '18 at 20:13
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    @Mitch: I am not quite sure what you mean. But, moving along a complex branch would mean: $$\frac{\sqrt{1-x}-3}{2+\sqrt[3]{|x|}\sqrt[3]{-1}} = \frac{\sqrt{1-x}-3}{2+\frac{\sqrt[3]{|x|}}{2}(1\pm i\sqrt{3})} \stackrel{x\in \mathbb{R}, x\to -8}{\longrightarrow} \frac{0}{3 \pm i\sqrt{3}} = 0$$ – trancelocation Oct 09 '18 at 06:22
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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

grg
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user1483
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