Page 311 - Calculus Demystified

298

2 Solutions to Exercises
+ lim 1 dx
(→0 + −1+( (x + 1) 1/3
(x + 1) 2/3 −1−( (x + 1) 2/3 2
= lim + lim
(→0 + 2/3 −2 (→0 + 2/3 −1+(
(−() 2/3 (−1) 2/3 3 2/3 (() 2/3
= lim + 2/3 − 2/3 + lim + 2/3 − 2/3
(→0
(→0
3
= 2 · 3 2/3 − 1 .
6 −2−(
(d) x dx = lim x dx
−4 (x − 1)(x + 2) (→0 + −4 (x − 1)(x + 2)
0 x 1−( x
+ lim + −2+( (x−1)(x+2) dx+ lim + 0 (x−1)(x+2) dx
(→0
(→0
6
+ lim x dx. Now
(→0 + 1+( (x − 1)(x + 2)
x 1/3 2/3
TEAMFLY = x − 1 + x + 2 .
(x − 1)(x + 2)

Therefore
6 x dx
−4 (x−1)(x+2)
−2−( 1/3 2/3 0 1/3 2/3
= lim + x−1 + x+2 dx+ lim + x−1 + x+2 dx
(→0
(→0
−4 1−( −2+(
6

+ lim 1/3 + 2/3 dx+ lim 1/3 + 2/3 dx
(→0 + 0 x−1 x+2 (→0 + 1+( x−1 x+2
1 2 −2−(
= lim + 3 ln|x−1|+ ln|x+2| −4
3
(→0
0
2
+ lim 1 ln|x−1|+ ln|x+2|
(→0 + 3 3 −2+(
1 2 1−(
+ lim + 3 ln|x−1|+ ln|x+2|
3
(→0
0
6
2
+ lim 1 ln|x−1|+ ln|x+2| .
(→0 + 3 3 1+(

Team-Fly
®