Why does Wolfram|Alpha make a mistake here?

Question

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?

I do not now what you are doing. But the numerator is $0$ and the denomiator is $\ne 0$ for $x=-8$. — gammatester, Oct 08 '18 at 09:37
@metamorphy: Typo corrected, you cannot use LH because the denomiator is $\ne 0$ when using the principal cube-root. — gammatester, Oct 08 '18 at 09:41
@gammatester We only research the real number domain, not complex analysis. — WuKong, Oct 08 '18 at 09:50
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. — gammatester, Oct 08 '18 at 09:58
It is easy to make WA get the Wrong Answer for limitless limit questions, even when it interprets the question correctly. — user21820, Oct 08 '18 at 18:25

score 39 · Accepted Answer · edited Oct 09 '18 at 03:39

39

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

edited Oct 09 '18 at 03:39

Em.

16,196

answered Oct 08 '18 at 09:38

Pavel R.

1,398

you mean that WA thinks \sqrt[3]{x} no longer to be the cubic root for $x$ if $x<0$? – WuKong Oct 08 '18 at 09:42
28

@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! – WuKong Oct 08 '18 at 09:47
9

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
4

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
Alternatively, typing the cube root as a rational power, i.e. lim\frac{\sqrt{1-x}-3}{2+x^(1/3)} as x to -8 provides a toggle for assuming either the real-valued or principal root. – user170231 Jun 27 '24 at 21:51

trancelocation · Answer 2 · 2018-10-08T10:35:02.440

14

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 $-2$:

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

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

edited Oct 08 '18 at 10:35

answered Oct 08 '18 at 09:51

trancelocation

33,349
1
21
46

Can you do the same for the other two, imaginary cube roots? – Mitch Oct 08 '18 at 20:13
1

@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

score 4 · Answer 3 · edited Oct 09 '18 at 08:31

4

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

edited Oct 09 '18 at 08:31

grg

1,003

answered Oct 08 '18 at 21:21

user1483

181
3

Why does Wolfram|Alpha make a mistake here?

3 Answers3