Math 142
Quiz 2
No Work-No Credit
Due: 9/12/2011-Not Accepted Late!
Fall 2011
Name
Last 4 Digits
Solutions
Work one (1) stating all details. You may use:
lim cos x = 1
x→0
lim sin x = 0
x→0
sin x
, plus the usual trig. identities, if needed!
=1
x
1 − cos x
lim
=0
x→0
x
lim
x→0
sin 3x sin2x
x→0
x sin5x
1) lim
€
x 2 − 2x
x → 0 sin 3x
2) lim
1 − sec 2 x
x→0
x2
3) lim
€
⎛ 1 ⎞
⎛ sin 3x sin2x ⎞
sin 3x sin2x
sin 3x
sin2x ⎜ x ⎟
€⎜
€⋅ lim
= lim
⋅
1) lim
⎟ = 3lim
⎜ ⎟ =
x→0
x → 0⎝
x→0
x → 0 sin5x 1
x sin5x
x
sin5x ⎠
x
⎜ ⎟
⎝ x ⎠
€
sin2x
sin2x
lim
sin 3x 2
2
2x
→
0
2x = 6 .
3 lim
⋅ lim 2x = 3⋅
3x → 0 3x
5 x → 0 sin5x
5 lim sin5x 5
5x → 0 5x
5x
x 2 − 2x
x ( x − 2)
x −2
−2
−2
2
2) lim
= lim
= lim
=
=
=− .
⎛
⎞
sin
3x
x → 0 sin 3x
x → 0 sin 3x
x → 0 sin 3x
sin 3x 3
3
lim
⎜ ⎟ 3 3xlim
→
0
x→0
x
3x
x ⎝ 3 ⎠
€
€
⎛ ⎛ sin x ⎞ 2
⎛ sin 2 x
1 − sec 2 x
tan 2 x
1 ⎞
1 ⎞
3) lim
⎜
=
−lim
=
−lim
⋅
=
−
lim
⋅
lim
⎜
⎟
⎜
⎟
2
2 ⎟ =
x→0
x→0
x → 0⎝ x
x2
x2
cos 2 x ⎠
⎝ x → 0⎝ x ⎠ x → 0 cos x ⎠
€
2
⎛
sin x ⎞
1
= −⎜ lim
=
⎟ = −1, since lim
2
x → 0 cos x
⎝ x → 0 x ⎠
1
(lim cos x)
x→0
€
€
2
= 1.