MATH 251 (PHILLIPS): SOLUTIONS TO WRITTEN HOMEWORK 9.
This sheet is part of the homework for Week 9, and is due in class on Friday 27 May 2016.
All the requirements in the sheet on general instructions for homework apply. In particular,
show your work (unlike WeBWorK), give exact answers (not decimal approximations; again,
unlike WeBWorK), and use correct notation. Some of the grade will be based on correctness
of notation in the work shown.
Write your answers on a separate sheet of paper.
Find the exact values of the following limits, or explain why they do not exist. (Use ±∞ as
appropriate.) Give full reasons in all cases. (Your reasons may not rely on calculator graphs.)
cos(πx)
x→1/2 1/2 − x
1. lim
Solution 1: Set f (x) = cos(πx) for all real x. Recognizing the definition of f 0 (1/2) at the
second step and using the chain rule at the third step, we have
cos(πx)
cos(πx) − cos(π/2)
= − lim
= −f 0 (π/2) = cos0 (π/2) · π = π sin(π/2) = π.
x→1/2 1/2 − x
x→1/2
x − 1/2
lim
Solution 2: The limit has the indeterminate form “ 00 ”. Therefore we may use L’Hospital’s
Rule, getting
− sin(πx) · π
lim
.
x→π/2
−1
Since this limit exists and is equal to π sin(π/2) = π, L’Hospital’s Rule tells that
cos(x)
= π.
x→π/2 π/2 − x
lim
x−2
x→0 sin(x) − 17
2. lim
Solution: Both the numerator and denominator are defined and continuous at x = 0, and
the denominator is not zero there. Therefore
x−2
0−2
−2
2
lim
=
=
= .
x→0 sin(x) − 17
sin(0) − 17
−17
17
Note that L’Hospital’s rule does not apply, because the limit does not have an indeterminate
form. If you try to apply it anyway, you get
1
lim
= 1,
x→0 cos(x)
which is not the correct answer.
Date: 27 May 2016.
1
arctan(ax − a)
, where a is a constant.
x→1
tan(x − 1)
3. lim
0
Solution: Trying to substitute x = 0 gives the undefined expression , so more work is
0
0
needed. Since is an indeterminate form, we may use L’Hospital’s Rule. Using the chain rule
0
on the numerator and denominator, we get
d
dx
1
·a
(arctan(ax − a))
a
1+(ax−a)2
=
.
=
d
2
2 ) sec2 (x − 1)
sec
(x
−
1)
(1
+
(ax
−
a)
(tan(x
−
1))
dx
Now
a
a
a
=
=
= a.
x→1 (1 + (ax − a)2 ) sec2 (x − 1)
(1 + 02 ) sec2 (0)
(1 + 0) · 1
Since this limit exists, L’Hospital’s Rule tells that
lim
arctan(ax − a)
a
= lim
= a.
x→1
x→1 (1 + (ax − a)2 ) sec2 (x − 1)
tan(x − 1)
lim
3x2 − 2
x→−∞ 7x2 − 2 sin(x)
4. lim
Solution: The limit has the indeterminate form “ ∞
”. Therefore more work is needed. We
∞
2
factor out x from both the numerator and denominator, and then use the limit laws:
3 − limx→−∞ x22
3 − x22
3x2 − 2
=
.
=
lim
2 sin(x)
x→−∞ 7x2 − 2 sin(x)
x→−∞ 7 + 2 sin(x)
7
+
lim
x→−∞
2
2
x
x
lim
Certainly limx→−∞ x22 = 0. Since −2 ≤ 2 sin(x) ≤ 2 for all x, and x2 → ∞ as x → −∞, we get
= 0. Therefore
limx→−∞ 2 sin(x)
x2
3x2 − 2
3
lim
= .
x→−∞ 7x2 − 2 sin(x)
7
Here is a different way to arrange essentially the same calculation:
1
(3x2 − 2)
3 − x22
3x2 − 2
2
x
lim
= lim
= lim
1
x→−∞ 7x2 − 2 sin(x)
x→−∞
(7x2 − 2 sin(x)) x→−∞ 7 + 2 sin(x)
x2
x2
=
2
x2
limx→−∞ 2 sin(x)
x2
3 − limx→−∞
7+
=
3+0
3
= .
7+0
7
The hypotheses of L’Hospital’s Rule are satisfied, but it doesn’t help. Applying it once gives
6x
lim
.
x→−∞ 14x − 2 cos(x)
Applying it a second time gives
6
.
x→−∞ 14 + 2 sin(x)
This limit does not exist: the function oscillates between
lim
6
16
=
3
8
and
6
12
= 12 .
cos(3x) − 1
x→0
7x2
5. lim
0
Solution: The limit has the indeterminate form . Therefore we may use L’Hospital’s Rule.
0
d
The chain rule gives
cos(3x) − 1 = −3 sin(3x), so
dx
cos(3x) − 1
−3 sin(3x)
,
lim
= lim
2
x→0
x→0
7x
14x
0
if the second limit exists. The second limit also has the indeterminate form . Therefore we
0
may use L’Hospital’s Rule again. A similar calculation shows that
−3 sin(3x)
−9 cos(3x)
= lim
,
x→0
x→0
14x
14
if the limit on the right exists. But the limit on the right is equal to −9 cos(3 · 0)/14 = −9/14.
Therefore
cos(3x) − 1
9
lim
=
−
.
x→0
7x2
14
lim
x
x→2 e3x
6. lim
Solution: Both the numerator and denominator are defined and continuous at x = 2, and
the denominator is not zero there. Therefore
x
2
2
lim
= 3·2 = 6 .
x→2 e3x
e
e
Note that L’Hospital’s rule does not apply, because the limit does not have an indeterminate
form. If you try to apply it anyway, you get
1
1
= 6,
3x
x→2 3e
3e
lim
which is not the correct answer.
x3
7. lim
x→0 sin(7x3 )
0
Solution 1: The limit has the indeterminate form “ ”. Therefore we may use L’Hospital’s
0
d
3
sin(7x ) = 21x2 cos(7x3 ), so
Rule. The chain rule gives
dx
x3
3x2
=
lim
,
x→0 sin(7x3 )
x→0 21x2 cos(7x3 )
lim
if the second limit exists. To evaluate the second limit, cancel the factors of x2 and 3, getting:
3x2
3
3
3
1
= lim
=
=
= .
x→0 21x2 cos(7x3 )
x→0 21 cos(7x3 )
21 cos(7 · 0)
21
7
lim
(You must simplify
3
21
to 71 , either here or later.) Therefore
1
x3
= .
3
x→0 sin(7x )
7
lim
Solution 2: We reduce this problem to
t
1
1
= = 1.
lim
= lim t→0 sin(t)
t→0 sin(t)
1
t
Since 7x3 → 0 as x → 0, we can let t = 7x3 and get
t
7x3
= lim
= 1.
t→0 sin(t)
x→0 sin(7x3 )
lim
Therefore
x3
1
7x3
1
=
lim
=
.
x→0 sin(7x3 )
7 x→0 sin(7x3 )
7
lim
Solution 3: As in Solution 1, L’Hospital’s Rule applies to show that
x3
3x2
=
lim
,
x→0 sin(7x3 )
x→0 21x2 cos(7x3 )
lim
if the second limit exists. If you don’t cancel x2 , you can still solve the problem by two more
applications of L’Hospital’s Rule. (This solution is provided for reference, in case anyone did
this; it is not recommended because it takes much more time and give too many chances for
algebraic errors.)
0
The second limit has the indeterminate form , so L’Hospital’s Rule applies. Thus,
0
2
6x
3x
= lim
lim
2
3
3
x→0 21 [2x cos(7x ) − x2 sin(7x3 ) · 21x2 ]
x→0 21x cos(7x )
6x
,
= lim
3
x→0 21 [2x cos(7x ) − 21x4 sin(7x3 )]
if the limit on the right hand side exists. At this point, it is again possible to cancel x from the
numerator and denominator, giving
6
lim
.
x→0 21 [2 cos(7x3 ) − 21x3 sin(7x3 )]
Then substitute x = 0, giving the same answer as before. However, since the limit on the right
0
hand side above still has the indeterminate form , L’Hospital’s Rule applies again, and gives
0
6x
lim
3
x→0 21 [2x cos(7x ) − 21x4 sin(7x3 )]
6
= lim
3
3
2
x→0 21 [2(cos(7x ) − x sin(7x ) · 21x ) − 21(4x3 sin(7x3 ) + x4 cos(7x3 ) · 21x2 )]
6
= lim
x→0 (42 − 213 x6 ) cos(7x3 ) − 6 · 212 x3 sin(7x3 )
if the limit on the new right hand side exists. This last limit can be computed by substituting
x = 0, and is
6
6
1
6
=
=
=
.
lim
x→0 (42 − 213 x6 ) cos(7x3 ) − 6 · 212 x3 sin(7x3 )
(42 − 0) cos(0) − 0 · sin(0)
42
7
(The final simplification is necessary.)
8. lim x2 e−3x
x→∞
x2
∞
Solution: We can rewrite this as lim 3x . In this form, it has the indeterminate form
.
x→∞ e
∞
Therefore we may use L’Hospital’s Rule. Thus
x2
2x
lim 3x = lim 3x ,
x→∞ e
x→∞ 3e
∞
, so we apply
if the second limit exists. The second limit also has the indeterminate form
∞
L’Hospital’s Rule again:
2x
2
lim 3x = lim 3x ,
x→∞ 3e
x→∞ 9e
if the limit on the right hand side exists. But this limit is clearly 0. So
x2
2 −3x
lim x e
= lim 3x = 0.
x→∞
x→∞ e
x2 − x + 1
x→−7
x−7
9. lim
Solution: Both the numerator and denominator are continuous at −7, and the denominator
is not zero there. Therefore the limit can be evaluated by simply substituting x = −7. That is,
x2 − x + 1
(−7)2 − (−7) + 1
49 + 7 + 1
57
lim
=
=
=− .
x→−3
x−7
(−7) − 7
−14
14
Note that L’Hospital’s Rule doesn’t apply, since the limit does not have an indeterminate
form. If you try to use it anyway, you get
2x − 1
lim
= −15,
x→−7
1
which is the wrong answer.