數學m1難題

1
lim x→1 (x^3+x^2+2x-2)/(x^2+x+2)
= (1+1+2-2)/(1+1+2)
= 2/4
=1/2

**concept!! 記住當你除去個LIM的時候，第一個=其實係「大概=」!!
下面題目一樣

2.
lim h→0 ((1+h)既立方-1)/h
= lim h→0 [(1+h)^3-1]/h
= lim h→0 (h^3+3h^2+3h+1-1)/h
= lim h→0 h(h^2+3h+3)/h
= lim h→0 (h^2+3h+3)
=0+0+3
=3

** (a+b)^3 = (1)a^3 + 3(a^2+b^1) +3(a+b^2) +1(b^3)

3.
lim r→3 (1/r-1/3)/(r-3)
= lim r→3 [(3-r)/3r]/(r-3)
=lim r→3 [-(r-3)/3r]/(r-3)
=lim r→3 [-(r-3)/3r(r-3)]
=lim r→3 [-1/3r]
=-1/3(3)
=-1/9

** 負號係可以抽出黎放在LIM前面，USUALLY係要LIM的數好大舊就有用，SIMPLY D ga~

2010-10-23 12:14:38 補充：
至於何解可以抽出是因為「-」其實係「-1」乘果舊數，而-1係一個常數，CONSTANT，lim 只會vary d variable。 ps eg
=lim r→3 [-1/3r]
你可以唔只抽-1 --> = -lim r-->3 [1/3r]
你可以抽成 --> =-1/3lim r-->3 [1/r]
'.'-1/3 is a constant!

1. lim x→1 (x^3+x^2+2x-2)/(x^2+x+2)

= (1+1+2-2)/(1+1+2)
= 2/4
=1/2

2. lim h→0 ((1+h)既立方-1)/h

= lim h→0 [(1+h)^3-1]/h
= lim h→0 (h^3+3h^2+3h)/h
= lim h→0 h(h^2+3h+3)/h
= lim h→0 (h^2+3h+3)
=0+0+3
=3

3. lim r→3 (1/r-1/3)/(r-3)

= lim r→3 [(3-r)/3r]/(r-3)
=lim r→3 [-(r-3)/3r]/(r-3)
=lim r→3 -(r-3)/3r(r-3)
=lim r→3 -1/3r
=-1/3(3)
=-1/9