Question

Find lim x->2 of

(x^3 - 8)/(x^4 -16)

Answer 1

You did this one before I believe where the top and bottom both came out to zero....Do you remember what you ended up doing to solve it?

Answer 2

nope....no idea....

Answer 3

Something called....L'Hopital rule perhaps?

Answer 4

i believe you can factor the numerator and denominator

Answer 5

find the limit algebraically, not using L'hopitals rule

Answer 6

Yep, you can do that too

Answer 7

http://www.wolframalpha.com/input/?i=+lim+x-%3E2+of+++%28x^3+-+8%29%2F%28x^4+-16%29 Try that

Answer 8

maybe get rid of the (x - 2) term

Answer 9

factorize numerator and denominator and cancel the terms. you leave be left be some polynomials. substitute x=2 and you will get the answer.

Answer 10

the numberator is a difference of 2 cubes, numerator is a difference of 2 squares (twice)

Answer 11

*you will be left with some polynomials.

Answer 12

numberator* lol numerator

Answer 13

or if you want to do it algebraically, the top factors to (x-2)(x^2+2x+2)

Answer 14

the bottom factors to (x^2+4)(x-2)(x+2)

Answer 15

cancelling you have (x^2+2x+2)/(x^2+4)(x+2)

Answer 16

the numerator does not factor to (x-2)(x^2+2x+2)

Answer 17

multiplying those together gives you x^3 - 4x -4

Answer 18

now since, the zero in the denominator has been cancelled, you can just plug in the values and get whatever

Answer 19

except that you dont because you didnt factor correctly

Answer 20

it's (x^2+2x+4)/(x^2+4)(x+2)

Answer 21

oops, the numerator factors to (x-2)(x^2+2x+4)

Answer 22

that works...

Answer 23

you still have a zero-free denominator

Answer 24

plugging in, you get 12/32, reduced to 3/8

Answer 25

Yeah, I forgot to multiply

Answer 26

Yes, correct, it's 3/8

Answer 27

:)

Answer 28

tell me this....how do you figure out that x^4 - 16 is equal to (x^2 + 4)(x-2)(x+2) ? What are the steps involved...?

Answer 29

difference of squares

Answer 30

(x^2+4)(x^2-4)

Answer 31

(x^2+4)(x^2-4) = (x^2+4)(x+2)(x-2)

Answer 32

and then x^2-4 factors to (x+2)(x-2)

Answer 33

the formula for the difference of squares is:
\[a^2-b^2 = (a+b)(a-b)\]

Answer 34

alright.....been awhile since i had to do anything like that.....

Answer 35

how about the numerator? is that difference of cubes?

Answer 36

yup

Answer 37

alright, i looked those up....just forgot those equations existed...

Answer 38

Formula for the difference of cubes is:
\[a^3-b^3 = (a-b)(a^2+ab+b^2)\]

Answer 39

\[a ^{3} - b ^{3} = (a-b)(a ^{2}+ab-b ^{2})\]