Page 66 - Calculus Workbook For Dummies

50 Part II: Limits and Continuity

Solutions for Problems with Limits

2
a lim x - 9 = 6
x " 3 x - 3
Factor, cancel, plug in.
^ x - 3 ^h x + 3h
= lim
x " 3 ^ x - 3h
x + 3
= lim
x " 3 1
3 + 3
=
1
= 6
b lim x - 1 = 1
2
x
x " 1 x + - 2 3
Factor, cancel, plug in.
^ x - 1h
= lim
x " ^ x - 1 ^h x + 2h
1
1
= lim
x " 1 x + 2
1
=
1 + 2
1
=
3
c lim x + 2 = 1
3
x " - 2 x + 8 12
Factor, cancel, plug in.
^ x + 2h
= lim
2
x " - ^ x + h x - 2 x + 4i
2 _
2
1
= lim 2
x " - 2 x - 2 x + 4
1
= 2
2 +
^ - 2 - ^h 2 - h 4
1
=
12
2
d lim x - 4 = 0
2
x " 2 4 x + 5 x - 6
Did you waste your time factoring the numerator and denominator? Gotcha! Always
0
plug in first! When you plug 2 into the limit expression, you get 20 , or 0 — that’s your
answer.
e lim x - 9 = –6
x " 9 3 - x
1. Multiply numerator and denominator by 3 + x.
^ x - 9h ` 3 + xj
= lim $
x " 9 ` 3 - xj ` 3 + xj