©Zarestky
Math 151 Quiz 3 Version B
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!3
lim
x!"5
( x "3)( x " 2)
x 2 "5x + 6
= lim
= lim ( x " 2) = 1
x!3
x!3
x "3
x "3
"1
= "#
( x + 5)
2
Note that
!1
( x + 5)
2
is always negative so the left- and right-hand limits must both equal negative infinity
and thus the limit exists. It is not necessary to show work.
3.
lim (sec x ) =
x!
2
!
6
3
=
2 3
3
sec x is defined at x =
to show work.
4.
!
so the limit is just the function evaluated at that point. It is not necessary
6
2 (7 " x )
#
&
7 " x ((
%%
( = 2 ("1) = "2
lim
= lim+
= 2 % lim+
%% x!7 7 " x (((
x!7+ 7 " x
x!7
7"x
$
'
14 " 2x
Factor out the 2. The value of
7!x
7!x
7!x
7!x
is either 1 or −1. As x approaches 7 from the right, the value of
is −1 because (7 – x) is negative and |7 – x| is positive.
©Zarestky
5.
Math 151 Quiz 3 Version B
$1! x
&
&
& 2
(2 pts) Identify any point(s) of discontinuity for f ( x ) = &
%x
&
&
&
&
'4x
Possible points of discontinuity are 0 and 4.
2/8/2010
for x " 0
for 0 < x < 4
for x # 4
At x = 0, 1! x " x 2 so the function is discontinuous.
At x = 4, x 2 = 4x so the function is continuous.
Partial credit:
• 0.5 pts each for recognizing that x = 0 and x = 4 are the “trouble spots”
• 0.5 pts for continuous at x = 4
• 0.5 pts for discontinuous at x = 0