©Zarestky
Math 151 Quiz 3 Version A
2/8/2010
NAME:____KEY____________________________________
Section (circle one):
507
508
509
510
511
512
For 1-4, evaluate the limit. 2 pts each.
Partial credit for each of 1-4:
• 1 pt for showing work (if applicable).
• 1 pt for correct final answer.
1.
2.
lim
x!2
lim
x!"7
( x " 2)( x " 4)
x 2 "6x + 8
= lim
= lim ( x " 4) = "2
x!2
x!2
x"2
x"2
1
( x + 7)
2
Note that
= +#
1
( x + 7)
2
is always positive so the left- and right-hand limits must both equal positive infinity
and thus the limit exists. It is not necessary to show work.
3.
lim ( tan x ) = 3
x!
!
3
tan x is defined at x =
to show work.
4.
lim+
x!4
12 "3x
4" x
= lim+
x!4
!
so the limit is just the function evaluated at that point. It is not necessary
3
3(4 " x )
#
&
4 " x ((
%
( = 3("1) = "3
= 3%% lim+
%% x!4 4 " x (((
4" x
$
'
Factor out the 3. The value of
4! x
4! x
4! x
4! x
is either 1 or −1. As x approaches 4 from the right, the value of
is −1 because (4 – x) is negative and |4 – x| is positive.
©Zarestky
5.
Math 151 Quiz 3 Version A
#x 2
%
%
%
(2 pts) Identify any point(s) of discontinuity for f ( x ) = %
$x
%
%
%
%
&3x
Possible points of discontinuity are 0 and 3.
2/8/2010
for x ! 0
for 0 < x < 3
for x " 3
At x = 0, x 2 = x and the function is continuous.
At x = 3, x ! 3x so the function is discontinuous.
Partial credit:
• 0.5 pts each for recognizing that x = 0 and x = 3 are the “trouble spots”
• 0.5 pts for continuous at x = 0
• 0.5 pts for discontinuous at x = 3