Question

lim x=8

2-cubedroot x/ 8-x

How to solve?

Answer 1

\[\lim_{x\to 8}\frac{2-\sqrt[3]{x}}{8-x}\] ??

Answer 2

yes

Answer 3

i tried to solve it but it didn't work out

Answer 4

ok and i assume you have to do this by hand, not l'hopital or recognizing it as a derivative right? then it will take some algebra to do this

Answer 5

yea

Answer 6

if it was the square root you would multiply top and bottom by the conjugate, because \[a^2-b^2=(a+b)(a-b)\]you do something analagous here, using instead
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 7

in your case \(a=2,b=\sqrt[3]{x}\) so to get rid of the radical multiply top and bottom by
\[4+2\sqrt[3]{x}+\sqrt[3]{x}^2\]

Answer 8

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

Answer 9

wait, so i get the 4 because I square the 2, I get the value in the middle because i multiply the values in the first bracket together, and I get the third value by squaring the x cubed root.

Answer 10

the numerator is \(8-x\) so you get
\[\frac{1}{4+\sqrt[3]{x}+\sqrt[3]{x}^2}\]

Answer 11

yes

Answer 12

you are trying to get rid of the radical in the numerator
you use \(a^3-b^3=(a-b)(a^2+ab+b^2)\) to do it, in other words \(a-b=2-\sqrt[3]{x}\)

Answer 13

that way you will get rid of the cubed root up top

Answer 14

ok, so i'm given the first bracket a-b

Answer 15

right \(a-b=2-\sqrt[3]{x}\)

Answer 16

oh, but should it not have the cubed root then? sorry, i know its a stupid question

Answer 17

so in order to get \(8-x\) up top you multiply top and bottom by \((2^2+2\sqrt[3]{x}+\sqrt[3]{x}^2)\)

Answer 18

you get \(2^3-\sqrt[3]{x}^3=8-x\) in the numerator

Answer 19

which by some miracle cancels with the \(8-x\) in the denominator

Answer 20

i am sure you ran across problem like this:
\[\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}\] in that case it is enough to multiply top and bottom by \(\sqrt{x}+2\) because then the radical will be gone up top

Answer 21

oh ok, i get how it works now. So if this was a square root i would multiply the top and bottom by (a+b) since it would be (a-b)

Answer 22

that is using the fact that \((a-b)(a+b)=a^2-b^2\) you are doing the same thing here but you need to use the difference of two cubes, not the difference of two squares

Answer 23

so you would get in my example
\[\frac{\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x+2})}\]
\[=\frac{x-4}{(x-4)(\sqrt{x}+2)}\]
\[=\frac{1}{\sqrt{x}+2}\]

Answer 24

then replace \(x\) by \(4\)
this situation is more complicated, but the idea is identical

Answer 25

oh ok .

Answer 26

second to last step is
\[\frac{1}{4+\sqrt[3]{x}+\sqrt[3]{x}^2}\] last step is to replace \(x\) by \(8\)

Answer 27

i think i get the idea now. So if it is a cubic i would multiply by (a^2 +ab+b^2) , it would either be positive or negative depending on (a-b). same idea if it was a square root, but i take the difference of squares.

Answer 28

yes that is the entire idea

Answer 29

ok. thank you so much

Answer 30

by "cubic" you mean "cubed root"
yw

Answer 31

yea