Calculus Volume 1

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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