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I need to find: $\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)}{x-8}$

I cannot solve this by substitution because that would cause the denominator to equal 0.

Normally, I would simply use the conjugate trick, however I am uncertain how I would rationalize the numerator.

$$\frac{(\sqrt[3]{x} -2)}{x-8}\times\frac{\sqrt[3]{x}+2}{\sqrt[3]{x}+2}$$

However, clearly this won't help me with anything, as I won't be able to factor anything.

$$\frac{(\sqrt[3]{x^2} -4)}{(x-8)(\sqrt[3]{x^2}+2)}$$

I am unsure about how to continue from here. Perhaps I am on the wrong track entirely. Any form of guidance would be welcome. Thank you.

J. W. Tanner
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Horatio Nelson
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7 Answers7

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The denominator is already $0$ in the limit. That doesn't mean the limit doesn't exist.

Recall the factorization of a difference of two cubes: $$ a^3 - b^3 = (a-b)(a^2+ab+b^2). $$ In particular $$ x-8 = \big(\sqrt[3] x - 2\big)\big( \sqrt[3]x^2 + 2\sqrt[3]x + 4\big). $$ So $$ \frac{\sqrt[3] x - 2}{x-8} = \frac{\sqrt[3]x-2}{\big(\sqrt[3]x-2\big)\big(\sqrt[3]x^2 +2\sqrt[3]x + 4\big)}. $$ Can you take it from there?

Michael Hardy
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  • Thank you! If the limit doesn't exist, then should I even bother with trying to factor it? – Horatio Nelson Jan 21 '20 at 06:05
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    @HoratioNelson : In this case the limit does exist. Whether factoring would be the means of discovering that the limit doesn't exist, in cases where it doesn't, might depend on the nature of the case. – Michael Hardy Jan 21 '20 at 22:40
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Hint:

Using the formula for the difference of cubes,

$x-8=(\sqrt[3]x-2)(x^{2/3}+2\sqrt[3]{x}+4)$.

J. W. Tanner
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Take the steps below$$\lim_{x \to 8} \frac{\sqrt[3]{x} -2}{x-8} =\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}{(x-8)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}$$ $$=\lim_{x \to 8}\frac{x-8}{(x-8)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)} =\lim_{x \to 8}\frac{1}{(\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4} =\frac1{4+4+4}=\frac1{12}$$

Quanto
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You may apply L-Hospital Rule $$L=\lim_{x \rightarrow 8} \frac{x^{1/3}-2}{x-8}= \lim_{x \rightarrow 8}(1/3)(x)^{-2/3}=\frac{1`}{12}.$$

Z Ahmed
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    For some purposes the use of L'Hopital's rule in this context would involve circular reasoning. – Michael Hardy Jan 21 '20 at 05:36
  • @Michael Hardy, you should rather say that you want a non-calculus answer. Saying that L-Hospital' s rule is circular is not appropriate. Cheers! – Z Ahmed Jan 21 '20 at 05:48
  • Since the question involves limits, is any answer a non-calculus answer. L'Hopital's rule is not circular, but all to often students fail to realize that its use in certain contexts is circular. – Michael Hardy Jan 21 '20 at 05:51
  • The other answers are algebraic and Limit is not essentially an issue of calculus. – Z Ahmed Jan 21 '20 at 05:55
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    "Limit is not essentially an issue of calculus"? From one reasonable point of view, finding and understanding limits is the whole of calculus. – Michael Hardy Jan 21 '20 at 05:56
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Let $f(x):=\sqrt[3]{x}.$ Then $ \frac{\sqrt[3]{x} -2}{x-8}= \frac{f(x)-f(8)}{x-8} \to f'(8)= \frac{1}{12}$ as $x \to 8.$

Fred
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Just another way.

Let $x=y+8$ to make $$A=\frac{\sqrt[3]{x} -2}{x-8}=\frac{\sqrt[3]{y+8} -2}{y}$$ Now, use the binomial expansion or Taylor series to get $$A=\frac{1}{12}-\frac{y}{288}+O\left(y^2\right)$$ Back to $x$ $$A=\frac{1}{12}-\frac{x-8}{288}+O\left((x-8)^2\right)$$ WHich shows the limit and how it is approched.

Using your calculator, try for $x=9$, this would give as an exact result $A=0.0772$ while the above truncated formula gives $\frac{11}{144}=0.0764$

Claude Leibovici
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Set $y:=x^{1/3}$, and consider $y \rightarrow 2$.

We have $\dfrac{y-2}{y^3-8}$.

$y-2$ is a factor of $y^3 -8$, since $2^3-8=0$;

Exercise : polynomial long division.

$(y^3-8)÷(y-2)=y^2+2y+4$;

$-(y^3-2y^2)$

$2y^2-8$

$-(2y^2-4y)$

$4y-8$

$ -(4y-8)$

$0$;

Now consider $\lim_{y \rightarrow 2} \dfrac{1}{y^2+2y+4}$.

Peter Szilas
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