Answer Key
793
3( x −6) 2( x −6)
3 x −18 2 x −12 =
18−18 12−12 =
0 0 ; then, lim x →6
3 x −18 2 x −12 = lim x →6
95 . lim x →6 97 . lim x →9 99 . lim θ → π
= 3 2
t −9
0 0 ; then, lim t →9
t +3 t +3
t −9
t −9
= 9−9 3−3 =
( t +3) =6
= lim t →9
= lim t →9
t −3
t −3
t −3
sin θ tan θ = lim θ → π
sin θ sin θ cos θ
π tan π = 0 0 ; then,
sin θ tan θ = sin
lim θ → π
= lim
cos θ =−1
θ → π
3 2 −2 1−1 =
1 2 +
2 x 2 +3 x −2
(2 x −1)( x +2)
2 x 2 +3 x −2
5 2
0 0 ;
then, lim
101 . lim 103 . −∞ 105 . −∞ 107 . lim x →6 109 . lim x →6 ⎛ x →1/2
2 x −1 = lim x →1/2
2 x −1 =
2 x −1 =
x →1/2
2 f ( x ) g ( x ) =2 lim x →6
f ( x ) lim x →6
g ( x ) =72
⎞ ⎠ = lim x →6
⎝ f ( x )+ 1 3
g ( x )
f ( x )+ 1
g ( x ) =7
3 lim x →6
g ( x )− f ( x ) = lim x →6
g ( x )− lim x →6
f ( x ) = 5
111 . lim x →6 113 . lim x →6 ⎡
⎛ ⎝ lim
⎞ ⎠
⎛ ⎝ lim
⎞ ⎠ =28
⎣ ( x +1) f ( x ) ⎤
( x +1)
f ( x )
⎦ =
x →6
x →6
115 .
a. 9; b. 7 117 .
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