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Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$? It's in indeterminant form, 0/0 when $x$ approaches $-8$. So I used LHopital's rule and got $$-\frac{3x^{\frac{2}{3}}}{2\sqrt{1-x}}$$ plug in $-8$ it is $-2(-1)^{2/3}$ which is imaginary. I used wolframalpha, the answer is $0$. So, which is correct?

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    You may consider $\lim_{x\to 8}\frac{\sqrt{1+x}-3}{2-\sqrt[3]{x}}$ instead. – 2017-02-18
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    @ Eclipse Sun then the LHopital answer is $\frac{6i}{7}$. Is it 0? – 2017-02-18
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    ${(-8})^{(2/3)}=((-8)^2)^{(1/3)}=(64)^{(1/3)}=4$ – 2017-02-18
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    It is not imaginary, and if the answer is zero, your L'Hopital result is not correct either. – 2017-02-18

3 Answers 3

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Wolfram Alpha is not correct.

You're almost correct, except that we have

$$\left(-8\right)^{2/3}=(64)^{1/3}=4$$ or alternatively $$\left(-8\right)^{2/3}=(-2)^{2}=4$$

Hence, applying L'Hospital's Rule

$$\begin{align} \lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(-\frac{3x^{2/3}}{2\sqrt{1-x}}\right)\\\\ &=-\frac{12}{6}\\\\ &=-2 \end{align}$$

An alternative approach would be to rationalize the numerator and denominator such that

$$\begin{align} \lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(\left(\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}\right)\left(\frac{\sqrt{1-x}+3}{\sqrt{1-x}+3}\right)\left(\frac{x^{2/3}-2x^{1/3}+4}{x^{2/3}-2x^{1/3}+4}\right)\right)\\\\ &=\lim_{x\to -8}\left(-\frac{x^{2/3}-2x^{1/3}+4}{\sqrt{1-x}+3}\right)\\\\ &=-2 \end{align}$$

as expected!

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    Thanks for the first line about WA. People in general and students in particular should not think that WA is the omniscient God of mathematics. +1 – 2017-02-18
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    @ParamanandSingh First of all, thank you. Second, I agree. But I am a bit shocked that WA returned a wrong result for such a simple limit. – 2017-02-18
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    WA does not make mistakes like these ones. But who knows there might be a bug somewhere. As a software engineer I can attest to the fact that products of software engineering are far less reliable compared to products of other conventional branches of engineering. Luckily I have not yet encountered any serious bug on this website. – 2017-02-18
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Use Taylor's polynomial at order $1$:

Set $x=-8+h$ $\;(h\to 0)$. Then

  • $\sqrt{1-x}-3=\sqrt{9-h}-3=3\biggl(\sqrt{1-\dfrac h9}-1\biggr)=3\biggl(1 -\dfrac h{18}+o(h)-1\biggr)=-\dfrac h{6}+o(h)$,
  • $2+\sqrt[3] x=2+\sqrt[3]{-8+h}=2\biggl(1-\sqrt[3]{1-\dfrac h8}\biggr)=2\biggl(1-\Bigl(1-\dfrac h{24}+o(h)\Bigr)\biggr)=\dfrac h{12}+o(h)$

So we obtain $$\frac{\sqrt{1-x}-3}{2+\sqrt[3] x}=\frac{-\dfrac h{6}+o(h)}{\dfrac h{12}+o(h)}=\frac{-\dfrac 1{6}+o(1)}{\dfrac 1{12}+o(1)}=-2+o(1).$$

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With substiuation $x+8=t$ we have $$\lim_{x\to-8}\dfrac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}=\lim_{t\to0}\frac{\sqrt{9-t}-3}{\sqrt[3]{t-8}+2}=\frac{\lim_{t\to0}\frac{\sqrt{9-t}-3}{t}}{\lim_{t\to0}\frac{\sqrt[3]{t-8}+2}{t}}=\frac{(\sqrt{9-t})'\Big|_{t=0}}{(\sqrt[3]{t-8})'\Big|_{t=0}}=\frac{-\frac16}{\frac{1}{12}}=\color{blue}{-2}$$