lim x->2 [(x^5 - 32) / ( x - 2)]
Limit of x approaching to 2
i can't factorize the x^5 - 32..would anyone be so kind to help me please? thanks!
2006-07-02 16:07:45 · 5 answers · asked by Mathy Troublemaker 1 in Education & Reference ➔ Homework Help
Do it like long division. The spacing is messed up...
x^4 +2x^3+4x^2+8x+16
. _________________
(x-2) /x^5 - 32
x^5 - 2x^4
2x^4 -32
2x^4 -4x^3
4x^3- 32
4x^3 -8x^2
8x^2-32
8x^2 -16x
16x-32
16x - 32
Unless I messed up in writing this out (what a pain), the answer is just plugging 2 into the quotient for x. Looks like 80.
2006-07-02 16:11:00 · answer #1 · answered by Bill S 6 · 1⤊ 0⤋
Quickest way is L'hopital's rule:
Take an expression that is a quotient and differentiate the top and bottom separately until the limit can be evaluated by plugging in the number:
[(x^5 - 32) / ( x - 2)] ----> Cannot evaluate at x = 2
Apply the rule:
[(5x^4 - 0) / ( 1 - 0)] = 5x^4 ----> CAN evaulate at x = 2
lim x->2 [(x^5 - 32) / ( x - 2)] = 5(2^4) = 80
2006-07-03 23:09:58 · answer #2 · answered by Anonymous · 0⤊ 0⤋
This should get you started
use long division to divide (x^5-32) by (x-2) . If it is divisible (and it is) then you will have "factorized" as you put it, the (x^5-32) into two factors, one of which is (x-2)
I bet you can go from there.
2006-07-02 23:15:43 · answer #3 · answered by enginerd 6 · 0⤊ 0⤋
can you believe I got 100 at my exam about limits a month ago and now in holidays I cant even remember what those things are for? :P
2006-07-02 23:13:14 · answer #4 · answered by aoc10010001100 2 · 0⤊ 0⤋
cant u use long division to simplify it?
2006-07-02 23:11:50 · answer #5 · answered by Ali 1 · 0⤊ 0⤋