Ap - Barrons Chapter 2
Ap - Barrons Chapter 2
Ap - Barrons Chapter 2
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Practice Exercises
-
2
-
x2
42-1 is
2. lim-
X400 X -
x-3
3. lim x.,� - 2x - 3 is
x43
1
(A) 0 (B) 1 (C) - (D) 00
(E) none of these ·
4
4. lim f is
X40
-
x2
6. lim 4
2
X4oo 4X - X - 2
is
x3 +27
7. lim 20x52 + 10x+9 is
x+-oo
(A) -00
(B) -1 (C) 0 (D) 3 (E) 00
3x2 +27
8. lim
X4oo x3 -27
is
(A) 3 (B) 00
(C) 1 (D) -1 (E) 0
9. lim ;��, is
.t4oo
x2 -9
14. The graph of y = -- has
3x-9
sinx
ls• lim 2
x-+0 x +3x
1·s
1
(A) 1 (B) 3 (C) 3 (D) oo (E)
4
.
. I 1s
16. }"1msmx
x-+0
(A) 00
(B) 1 (C) nonexistent (D) -1 (E) none of these
2x2 + 4 ?
17• Wh.1ch statement .1s true about the curve y = + x
2 7 _ 4x2 •
2x2 + I .
18 •
1.tm (2-x)(2+x) IS
X-+oo
20. limxsin:}
x--....
is
(A) 0 (B) 00
(C) nonexistent (D) -1 (E) 1
sin(1t-x)
21. lim 1t-X lS
x--.1t
II. f( 1) exists
III. f is continuous at x = 1
= x::x
23. If {/(x) for x -:I= 0,
/(0) = k,
and iff is continuous at x = 0, then k =
1 1
(A) -1 (B) (C) 0 (D) (E) 1
2 2
3x(x-1)
f(x) = x2-Jx+ forx '# 1, 2,
2
24. Suppose /(1) = -3,
/( 2 ) = 4.
Then f(x) is continuous
(A) except at x = 1 (B) except at x = 2 (C) except at x = 1 or 2
(D) except at x = 0, 1, or 2 (E) at each real number
4
25. The graph of f(x) = x2 _ 1 has
(A) one vertical asymptote, at x = 1
(B) the y-axis as vertical asymptote
(C) the x-axis as horizontal asymptote and x = ±1 as vertical asymptotes
{D) two vertical asymptotes, at x = ±1, but no horizontal asymptote
(E) no asymptote
2x2+2x+3
26. The graph of y = 4X2 - 4X has
I. /(0) exists
II. limf(x)
x--+0
exists
III. f is continuous at x = 0
Part B. Directions: Some of the following questions require the use of a graphing
calculator.
28. If [x] is the greatest integer not greater than x, then x--+1/2
lim[x] is
1
(A) (B) 1 (C) nonexistent
. (D) 0 (E) none of these
30. limsin
.t-+oo
x
0 (x = 0)
(A) is continuous everywhere
(B) is continuous except at x = 0
(C) has a removable discontinuity at x = 0
(D) has an infinite discontinuity at x = 0
(E) has x = 0 as a vertical asymptote
Questions 32-36 are based on the y
function f shown in the graph and
defined below:
1-x {-l�x<O)
2
2x -2 (0� X � 1)
f(x)= -x+2 (1 < X < 2)
1 (x = 2)
2x-4 (2 < X � 3)
32. limf(x)
x-+2
39. If y = - -
1 , then limy is
1
- x---.0
I CHALLENGE I
2 + lQx
1 l
(A) 0 (B) (C) (D) (E) nonexistent
12 2 3
2 3
-1
40. For what value(s) of a is it true that lj.!1JJ(x) exists andj(a) exists, but
lj.!1JJ(x) :t=f (a)? It is possible that a=
41. limf(x)
X-+a
does not exist for a=
I. limj(x)
X-+1-
exists.
II. limf(x)
X-+I'
exists.
III. limf(x)
X-+I
exists.
13. (B) Because the graph of y = tan x has vertical asymptotes at x = ± 'i , the graph
of the inverse function y = arctan x has horizontal asymptotes at = ± i .
y
19. (B) Since lxl = x if x > 0 but equals -x if x < 0, lim l::1
+
= lim £=1 while
+
x-+0 X x-+0 X
1.1m-=
IXI 1·1m-=-. -x l
x-+0- X x-+0- X
x(x-I) x-I
23. (B) f(x)=--
2 X =--,
2 for all x -:t: 0. For/to be continuous at x=0, limf(x) �O
24. (B) Only x=1 and x=2 need be checked. Since f(x)= }:_2 for x -:t: 1, 2, and
f is not continuous at x = 2.
25. (C) As x � ±oo, y=f(x) � 0, so the x-axis is a horizontal asymptote. Also, as
x � ±1, y � oo, so x=±1 are vertical asymptotes.
26. (C) As x � oo, y � I ; the denominator (but not the numerator) of y equals 0
at x=0 and at x= 1.
x2 +x
27. (D) The function is defined at O to be 1, which is also lim
x-+0
=lim(x + 1).
X x-+0
30. (E) As x � oo, the function sin x oscillates between -1 and 1; hence the limit
does not exist.
/
33. (E) Verify thatfis defined atx = 0, 1,2,and 3 ( as well as at all other points in
[-1,3]).
34. (C) Note· that lill!f(x) = limf (x) = 0. However,f(2) = 1. Redefiningf(2) as 0
+
x--.2 x--.2
removes the discontinuity.
35. (B) The function is not continuous at x = 0, 1,or 2.
36. (B) limf (x) = 1.
limf(x) = 0 -:I: x--.1•
x�i-
40. (A) limf(x) = l, butf(-1) =2. The limit does not exist at a= 1 andf(2) does
X--+-1
not exist.
same, lim
x--+I
f(x) does not exist.