©Zarestky
Math 142 Quiz 2 Version B
2/3/2011
NAME:______KEY__________________________________
Use f ( x ) =
1.
( x + 2)( x !5)
3x ( x !5)
for problems 1-5. (2pts each)
( x + 2)( x "5)
( x + 2) 7
= lim
=
x!5
x!5
3x
15
3x ( x "5)
lim f ( x ) = lim
x!5
The ( x !5) terms cancel. Then evaluate the expression for x = 5.
2.
lim" f ( x ) = lim"
x!0
( x + 2)
3x
x!0
= "#
The function has a vertical asymptote at x = 0. Use a graph or table of values to determine the limit.
3.
lim f ( x ) = lim
x!+"
( x + 2)
x!+"
3x
=
1
3
The exponents of the polynomials in the numerator and denominator are equal, thus, f ( x ) has a
1
horizontal asymptote at y = .
3
4.
Where is f ( x ) continuous? Express your answer in interval notation.
(!",0) # (0,5) # (5,+")
5.
(All real numbers except 0 and 5.)
Calculate the average rate of change of f ( x ) from x = 2 to x = 3.
f (3) ! f (2)
3! 2
=
5
9
! 23
1
=!
1
9