Solução_Calculo_Stewart_6e

F.
48 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
x 2 + x − 6 (x +3)(x − 2)
11. lim
=lim
=lim(x +3)=2+3=5
x→2 x − 2 x→2 x − 2
x→2
x 2 − x +6
13. lim
does not exist since x − 2 → 0 but x 2 − x +6→ 8 as x → 2.
x→2 x − 2
15. lim
t→−3
t 2 − 9
2t 2 +7t +3 = lim
t→−3
(t +3)(t − 3)
(2t +1)(t +3) = lim
t→−3
t − 3
2t +1 = −3 − 3
2(−3) + 1 = −6
−5 = 6 5
(4 + h) 2 − 16 (16 + 8h + h 2 ) − 16 8h + h 2 h(8 + h)
17. lim
=lim
=lim =lim =lim(8 + h) =8+0=8
h→0 h
h→0 h
h→0 h h→0 h h→0
19. Bytheformulaforthesumofcubes,wehave
lim
x→−2
x +2
x 3 +8 = lim
x→−2
x +2
(x +2)(x 2 − 2x +4) = lim
x→−2
1
x 2 − 2x +4 = 1
4+4+4 = 1
12 .
√ √
9 − t 3+ t 3 − t
21. lim
t→9 3 − √ t =lim
t→9 3 − √ √ √
=lim 3+ t =3+ 9=6
t
t→9
√ √ √ x +2− 3 x +2− 3 x +2+3
(x +2)− 9
23. lim
=lim
· √ = lim
x→7 x − 7 x→7 x − 7 x +2+3 x→7 (x − 7) √ x +2+3
25. lim
x→−4
27. lim
x→16
29. lim
t→0
=lim
x→7
x − 7
(x − 7) √ x +2+3 = lim
x→7
1
4 + 1 x +4
x
4+x = lim 4x
x→−4 4+x = lim
x→−4
4 − √ x
16x − x 2 = lim
x→16
1
t √ 1+t − 1
t
x +4
4x(4 + x) = lim
x→−4
(4 − √ x )(4 + √ x )
(16x − x 2 )(4 + √ x ) = lim
x→16
1
√ x +2+3
=
1
√
9+3
= 1 6
1
4x = 1
4(−4) = − 1 16
16 − x
x(16 − x)(4 + √ x )
1
= lim
x→16 x(4 + √ x ) = 1
16 4+ √ 16 = 1
16(8) = 1
128
1 − √ √ √
1+t 1 − 1+t 1+ 1+t
=lim
t→0 t √ = lim
1+t t→0 t √ t +1 1+ √ 1+t =lim
t→0
=lim
t→0
−1
√ 1+t
1+
√ 1+t
=
−1
√ 1+0
1+
√ 1+0
= − 1 2
−t
t √ 1+t 1+ √ 1+t
31. (a)
lim
x→0
x
√ 1+3x − 1
≈ 2 3
(b)
x
f(x)
−0.001 0.6661663
−0.0001 0.6666167
−0.00001 0.6666617
−0.000001 0.6666662
0.000001 0.6666672
0.00001 0.6666717
0.0001 0.6667167
0.001 0.6671663
The limit appears to be 2 3 .