F.3 Maths

a) Simplify (x^4-1)/(x^3-1).
(x^4-1)/(x^3-1)
= (x^2-1)(x^2+1)/(x-1)(x^2+x+1)
= (x-1)(x+1)(x^2+1)/(x-1)(x^2+x+1)
= (x+1)(x^2+1)/(x^2+x+1)

b) Estimate the value of (x^4-1)/(x^3-1) if x is very close to 1.
lim(x->1)(x^4-1)/(x^3-1)
= lim(x->1) (x+1)(x^2+1)/(x^2+x+1)
= (2)(2)/(3)
= 4/3


2012-01-29 17:11:28 補充：
其實 佢Question b 果到就係希望你用番 a part 果到去計....
所以 用番a part就有我以上方法

2012-01-29 17:12:38 補充：
另一個方法係有的 , 別L'Hospital Rule
lim(x->1)(x^4-1)/(x^3-1)
= lim(x->1) (4x^3)/(3x^2) [ By L'Hospital Rule, since it's in 0/0 type]
= 4(1)^3 / 3(1)^2
= 4/3

u r lkkc f3, right?

You can sub. x=1 directly actually.