Cubed root denominator limit question

[ochocinco1992]

Homework Statement

Evaluate lim x->8 (x-8)/(cubed root of x) -2)

Homework Equations

The Attempt at a Solution

I multiplied both numerator and denominator by (cubed root of x) +2. In the denominator, I wrote it as ((x^1/3)-2)*((x^1/3)+2) which simplifies to (x^2/3)-4. I have ((x-8)*((cubed root of x) +2))/((x^2/3)-4). The answer should be 12 but I cannot reach this answer as I get an answer of 0 in the denominator

 

[Anti-Meson]

Use L'Hôpital's rule...

 

[ochocinco1992]

which is?

 

[Anti-Meson]

http://mathworld.wolfram.com/LHospitalsRule.html

 

[ochocinco1992]

I don't understand how to use it

 

[ochocinco1992]

I found this question on another website and they factored x-8 which canceled the ((x^1/3)-8) in the denominator and then you plug 8 into the rest of the numerator to get 12. Is it not possible to just multiply numerator and denominator by (x^1/3)+2?

 

[Anti-Meson]

Okay, as both the as both the numerator and denominator in your function equal 0, when x = 8, you end up with 0/0 which is undefined, so we must use L'Hôpital's rule. Differentiating the numerator with respect to (wrt) x and the numerator wrt x we get.

[tex]\frac{1}{\frac{1x^{\frac{-2}{3}}}{3}}[/tex] = [tex] \frac{3}{x^{\frac{-2}{3}}} [/tex]

when x = 8 the above expression equals 12.

I hope this example helps you understand L'Hôpital's rule, which is a very powerful mathematical tool for functions. Take sinc x as another example which is (sin x)/x. As x approaches 0 sinc x approaches 0/0, but using L'Hôpital's rule, we can say that when x = 0 sinc 0 = 1.

 

[Bohrok]

I found this question on another website and they factored x-8 which canceled the ((x^1/3)-8) in the denominator and then you plug 8 into the rest of the numerator to get 12. Is it not possible to just multiply numerator and denominator by (x^1/3)+2?

If you had [itex]\sqrt{x} - 2[/itex] with a square root, you could multiply by its conjugate [itex]\sqrt{x} + 2[/itex]. However, you have [itex]\sqrt[3]{x} - 2[/itex] with a cube root which don't really have a conjugate.

So all you can do without using l'Hôpital's rule from Calc II is use the fact that x³ - y³ = (x - y)(x² + xy + y²) and factor x - 8, or multiply [itex]\sqrt[3]{x} - 2[/itex] by something that will turn it into x - 8.

 

[rsa58]

do as bohrok suggests note that x-8 = (x^(1/3))^3 - 2^3 and use the formula he wrote

 

[Anti-Meson]

do as bohrok suggests note that x-8 = (x^(1/3))^3 - 2^3 and use the formula he wrote

That is an acceptable way for this problem, but some expressions will not factor out as easily as this. That is why I have emphasised learning L'Hôpital's rule as it is an easy way to find the limit for all most all expressions.