HW 2 – 5B:
Math 400
Name____________________
The limit of f (x) as x approaches a
where there is a point at ( a , f (a)) and
f (x) is continuous in the “neighborhood of x = a
Theorem One:
The limit of the function f (x) as x approaches a
from the left OR right side where there is a point at ( a , f (a))
and f (x) is continuous in the “neighborhood of x = a
1A)
lim
f (x) = f (a) if f (a)∈REALS AND f (x) is continuous in the “neighborhood of x = a
x→ a
1B)
lim
f (x) = f (a) if f (a) ∈REALS AND f (x) is continuous in the “neighborhood of x ≥ a
x → a+
1C)
lim
f (x) = f (a) if f (a) ∈REALSAND f (x) is continuous in the “neighborhood of x ≤ a
x → a−
Check to see if f (x) is continuous in the “neighborhood given in the limit. If it is then find f (a).
Determine each limit if it exists.
1.
lim
3x − 7 !
x→ 3
2.
lim
− 3x + 2 !
x → − 4+
3.
lim
x 2 + 2x − 2
x → 2−
4.
lim
x 3 − 2x 2 + 3x − 4 !
x→−1
5.
lim
4
!
x → 3 x −1
6.
lim
x → 2+
7.
lim
3x + 4
!
−
x → 2 8x −1
8.
lim
−3
!
x → 3 x2
9.
lim
−6
x → 4 x2 − 4
Math 400 HW 2 – 5B !
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3x + 4
1− x
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10.
lim
x → 4+
x−4!
11.
lim
x → 4−
x−4!
12.
lim
x→ 4
x−4
13.
lim
x → 3+
3− x !
14.
lim
x → 3−
3− x !
15.
lim
x→ 3
3− x
16.
lim
x→4
−x
!
x−3
17.
lim
x → 12
x
!
x−3
18.
lim
x→1
x
x+3
19.
lim 3
3x + 6 !
x→ 7
20.
lim
3
+ 7x + 6 !
x →−2
21.
lim
3 5 + 2x
x→−3
5−x
22.
lim
cos( x ) !
x → 5π / 2
23.
lim
sec ( x ) !
x → 7π / 6
24.
lim
sin ( x )
x → 5π / 3
Math 400 HW 2 – 5B !
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25.
lim
csc ( x ) !
x → 5π / 6
26.
lim
tan ( x ) !
x → −π /6
27.
lim
cot ( x )
x → − 3π / 2
28.
lim
sin ( 3x ) !
x → −π /2
29.
lim
cot (2x ) !
x → −π /6
30.
lim
cos (5x )
x → −π
31.
lim
tan( x π ) !
x→ 3
32.
lim
⎛ xπ ⎞
sec
!
⎝ 3⎠
x → 11
33.
lim
⎛ xπ ⎞
tan
⎝ 6 ⎠
x→−7
34.
lim
ln( x ) !
x→ e
35.
lim
⎛ 1⎞
ln
!
x → e ⎝ x⎠
36.
lim
e x cos ( 3x )
x→ 0
Note: loge (x) = ln(x)
Math 400 HW 2 – 5B !
2. logb x a = a • logb x
3. ln(e) = 1
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Vertical Asymptotes
Given
lim
f (x)
f (a)
non zero number
where
is a
x → a g(x)
g(a)
zero
then there is a “vertical asymptote” in the graph at x = a
Theorem 2A
The limit of the function as x approaches a
from the left OR right side of a vertical asymptote at x= a
lim
f (x)
= +∞ or − ∞
−
x→a
g(x)
based on the sign of f (a − Δx)
and
lim
f (x)
= +∞ or − ∞
+
x→a
g(x)
based on the sign of f (a + Δx)
The limit of the function as x approaches a from the left or right side of a vertical asymptote at x = a
is either +∞ or − ∞ . To determine the sign of ∞ , plug in a number “close to a“ on the side
indicated and find the sign of that number.
Theorem 2B
The limit of the function as x approaches a
from the left AND right sides of a vertical asymptote at x= a
lim
f (x)
= +∞ or − ∞ or DNE
x → a g(x)
based on the signs of f (a − Δx) and f (a + Δx)
lim
f (x)
= +∞ or − ∞ or DNE
x → a g(x)
If f (a − Δx) and f (a + Δx) are both positive then the limit as x → a is + ∞
If f (a − Δx) and f (a + Δx) are both negitive then the limit as x → a is – ∞
If f (a − Δx) and f (a + Δx) have different signs then the limit as x → a DNE
The limit of the function as x approaches a from both the left and left sided of a vertical asymptote at
x = a is either +∞ or − ∞ . Determine the sign of ∞ , for the left side x → a − and the right side x → a +
of the vertical asymptote as shown above. If the left and right sided limits are the same that is the
limit as x → a . If they are not the same the limit as x → a (from both sides) is DNE.
Math 400 HW 2 – 5B !
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Determine each limit if it exists.
37.
lim
x → 1+
x
!
x −1
38.
40.
lim
x +1
!
+
x → − 2 (x + 2) 2
43.
lim
x → 2+
46.
lim
x+2
!
+
x → 1 1−x
49.
lim
x → 0+
−3x
!
x−2
x+3
x
2
!
Math 400 HW 2 – 5B !
lim
x → 1−
x
!
x −1
39.
lim
x
x → 1 x −1
42.
lim
x +1
x → − 2 (x + 2) 2
41.
lim
x +1
!
−
x → − 2 (x + 2) 2
44.
lim
x → 2−
−3x
!
x−2
45.
47.
lim
x+2
!
−
x → 1 1−x
48.
50.
lim
x → 0−
x+3
x
2
!
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51.
lim
−3x
x→ 2 x −2
lim x + 2
x → 1 1− x
lim
x→ 0
x+3
x2
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The Indeterminate Cases for Theorem 1
Many times the use of Theorem 1 to find the limit as x approaches a produces a real number value.
We saw this with the functions in problems 1 to 36. This result allows us to find the limit as x
approaches a immediately. Sometimes the use of Theorem 1 to find the limit as x approaches a
f (a) zero
produces a value of
=
. We call this the Indeterminate case for Theorem 1
g(a) zero
There are 2 outcomes for the indeterminate case.
Either the graph of the function at x = a has a “hole in the graph at x = a and the limit as x
approaches a is a real number value. or
the graph of the function at x = a has a vertical asymptote at x = a and the limit of the function as x
approaches a from the left or right side of a vertical asymptote at x = a is either +∞ or − ∞ .
How do you determine if the
zero
cases describes a “holeʼ or a vertical asymptote in the graph?
zero
If the use of Theorem1 for the
lim
f (x)
x → a g(x)
then
produces
f (a) zero
=
g(a) zero
f (x)
h(x)
to
and RETEST x = a,
g(x)
k(x)
one of two things MUST happen
"reduce"
Case 1(hole)
h(a)
If
is a REAL number then there is a "hole" in the graph at x = a and y =
k(a)
and
lim
f (x)
= the limit of the reduced function
x → a g(x)
lim
lim
lim
h(x) h(a)
h(x) h(a)
=
and
=
and
+
−
x→a
x → a k(x) k(a)
x→a
k(x) k(a)
h(a)
k(a)
h(x) h(a)
=
k(x) k(a)
OR Case 2 (vertical asymptote)
h(a)
non zero number
If
is a
then there is a "vertical asymptote" in the graph at x = a
k(a)
zero
and
lim
lim
f (x)
f (x)
= +∞ or − ∞ and
= +∞ or − ∞
−
+
x→a
x→a
g(x)
g(x)
lim
f (x)
= +∞ or − ∞ or DNE
x → a g(x)
Math 400 HW 2 – 5B !
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Determine each limit if it exists.
52.
lim
x +2
!
+
x → 2 x2 − 4
53.
lim
x2 − 9
55.
!
x → 3+ x 2 − 6x + 9
58.
61.
lim
x → 0+
lim
x → 0+
−2x
4
x − 3x
3
−4 x
3
x − 3x
Math 400 HW 2 – 5B !
lim
x → 2−
x +2
x2 − 4
!
54.
lim
x2 − 9
56.
!
x → 3− x 2 − 6x + 9
!
59.
lim
x → 0−
!
2
62.
lim
x → 0−
−2x
4
x − 3x
−4 x
3
x − 3x
lim
x2 − 9
57.
x → 3 x 2 − 6x + 9
!
60.
!
2
63.
3
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lim
x +2
!
x → 2 x2 − 4
lim
x →0
lim
x→0
−2 x
− 3x 3
x4
−4 x
3
x − 3x 2
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lim
4x2 − 9
64.
!
x → − 3 2 2x + 3
3x − 1
!
9x 2 − 1
66.
lim
x→1 3
68.
lim
x3 + 8
!
x→− 2 x+2
Math 400 HW 2 – 5B !
lim
x2 − 9
65.
!
x → − 3 2x 2 + 7x + 3
67.
lim 3x 2 − 8x − 16
x → 4 2x 2 − 9x + 4
69.
64x − 1
x → 1 4 4x − 1
lim
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3
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70.
lim
x→9
x −3
!
x−9
lim
5− x
72.
!
x → 25 25 − x
74.
lim
x→ 0
x+2− 2
!
x
Math 400 HW 2 – 5B !
71.
lim
x → 16
x −4
!
x − 16
lim 3 x − 1
73.
x → 1 x −1
75.
lim 2 − 4 − x
x→ 0
x
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