27
$\begingroup$

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?

$\endgroup$
10
  • 12
    $\begingroup$ You should better find your mistake. $\endgroup$ Oct 8, 2018 at 9:32
  • 1
    $\begingroup$ can you point out my fault? $\endgroup$ Oct 8, 2018 at 9:34
  • 1
    $\begingroup$ @gammatester $2+\sqrt[3]{-8}=2+(-2)=0.$ $\endgroup$ Oct 8, 2018 at 9:46
  • 1
    $\begingroup$ No, the principle cube root is $1 +i\sqrt{3} \ne 0$ $\endgroup$ Oct 8, 2018 at 9:48
  • 1
    $\begingroup$ @gammatester We only research the real number domain, not complex analysis. $\endgroup$ Oct 8, 2018 at 9:50

3 Answers 3

37
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ you mean that WA thinks \sqrt[3]{x} no longer to be the cubic root for $x$ if $x<0$? $\endgroup$ Oct 8, 2018 at 9:42
  • 28
    $\begingroup$ @mengdie1982: There is no such thing as the cubic root, there are three complex roots. $\endgroup$ Oct 8, 2018 at 9:46
  • $\begingroup$ I grasp you. Thanks! $\endgroup$ Oct 8, 2018 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, 2018 at 9:50
  • 4
    $\begingroup$ 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, 2018 at 3:42
14
$\begingroup$

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 $-2$:

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

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

$\endgroup$
2
  • $\begingroup$ Can you do the same for the other two, imaginary cube roots? $\endgroup$
    – Mitch
    Oct 8, 2018 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, 2018 at 6:22
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.