MAT265 Homework 08 (Solutions)
1. Let θ = sin−1 x, so that sin θ = x = x1 . We then have the following right triangle:
1
θ
√
x
1−x2
From the picture we can see that
x
tan sin−1 (x) = tan(θ) = √
.
1 − x2
2. For simplicity, let g(x) = x2 , h(x) = ln(x), and j(x) = arcsin(x). We then have that
f (x) = g(x) · h(j(x)),
and after applying a product rule and a chain rule, we get
f 0 (x) = g 0 (x) · h(j(x)) + g(x) · h0 (j(x)) · j 0 (x)
1
1
.
= 2x ln(arcsin(x)) + x2 ·
·√
arcsin(x)
1 − x2
3.
a. First,
√
√
3 + 3 x3
3
lim
=
.y
x→∞ 4 + 2x3
2
√
Since arccos(x) is continuous in a neighborhood of 23 , we have
!
!
√
√ !
√
3
3 + 3 x3
π
3 + 3 x3
= arccos lim
= arccos
= .
lim arccos
3
3
x→∞ 4 + 2x
x→∞
4 + 2x
2
6
b. This limit is indeterminate of type 00 , so l’Hospital’s Rule would apply. However, factoring
is an easier method.
x3 + x − 10
(x − 2)(x2 + 2x + 5)
lim
= lim
= lim (x2 + 2x + 5) = 13
x→2
x→2
x→2
x−2
x−2
c. This limit is indeterminate of type 00 , so l’Hospital’s Rule would apply. However, using
trig identities first might make it easier.
limπ
x→ 2
cos(3x)
cos(3x) · sin(5x)
= limπ
cot(5x) x→ 2
cos(5x)
This new limit is still indeterminate of type 00 , so applying l’Hospital’s Rule, we get
limπ
x→ 2
cos(3x) · sin(5x) LH
−3 sin(3x) sin(5x) + 5 cos(3x) cos(5x)
= limπ
x→ 2
cos(5x)
−5 sin(5x)
−3(−1)(1) + 5(0)(0)
=
−5(1)
3
=− .
5
1
MAT265 Homework 08 (Solutions)
d. This limit is indeterminate of type 00 , so applying l’Hospital’s Rule, we get
x3
3x2
LH
= lim
.
x→0 1 − cos(x)
x→0 sin(x)
lim
This new limit is still indeterminate of type 00 , so applying l’Hospital’s Rule again, we
get
6(0)
3x2 LH
6x
=
= 0.
= lim
x→0 sin(x)
x→0 cos(x)
1
lim
e. This limit is indeterminate of type 00 . Before
√ applying l’Hospital’s Rule, we make a
substitution to save us some grief. Let t = x. Then as x → ∞, t → ∞. So
√
ln( x)
ln(t)
lim √
= lim
x→∞
t→∞
t
x
1
LH
= lim
t→∞
t
1
= 0.
f. This limit is super indeterminate, so we first rearrange it to get it into an indeterminate
form that we know.
11 t
t
5
−
1
t
t
5
11 − 5
lim
= lim
.
t→∞
t→∞
t
t
. So, applying
In this form, it is now clear that the limit is indeterminate of type ∞
∞
l’Hospital’s Rule, we get
11 t
11 t
t
t
t 11 t
5
5
ln
5
−
1
−
1
+
5
ln 11
5
5
5
5
LH
lim
. = lim
= ∞.
t→∞
t→∞
t
1
g. This limit is indeterminate of type 0 · ∞, and so l’Hospital’s Rule does not directly apply.
Rearranging the limit, we get
tan x1
1
3
lim x tan
= lim
.
x→∞
x→∞ 1/x3
x
This new limit is now indeterminate of type 00 , and so we can apply l’Hospital’s Rule.
Before applying it, we make a substitution to simplify the derivatives. Let t = x1 , so as
x → ∞, t → 0.
tan x1
tan t
lim
= lim 3
3
x→∞ 1/x
t→0 t
sec2 t
LH
= lim
= ∞.
t→0 3t2
2
MAT265 Homework 08 (Solutions)
h. This limit is indeterminate of type ∞0 , so l’Hospital’s does not apply directly. So, let
1
L = lim (ex + 2x) 2x .
x→∞
By continuity of the natural logarithm, we then have that
1
ln(ex + 2x)
.
ln L = lim ln (ex + 2x) 2x = lim
x→∞
x→∞
2x
This new limit is indeterminate of type
we get
∞
,
∞
so we can apply l’Hospital’s Rule. Doing so,
ln(ex + 2x) LH
lim
= lim
x→∞
x→∞
2x
ex +2
ex +2x
2
ex + 2
= lim x
.
x→∞ 2e + 4x
Again this new limit is indeterminate of type 00 , so we apply l’Hospital’s Rule again to
get
ex
ex + 2 LH
=
lim
x→∞ 2ex + 4
x→∞ 2ex + 4x
ex
= lim x
x→∞ e
2 + e4x
1
1
= lim
= .
x→∞ 2 + 4x
2
e
lim
1
Thus we have ln L = 21 , whence L = e 2 .
4. This limit is indeterminate of type 00 , so applying l’Hospital’s Rule, we get
5x − 4x LH
5x ln 5 − 4x ln 4
=
lim
x→∞ 3x − 2x
x→∞ 3x ln 3 − 2x ln 2
ln 5 − ln 4
=
.
ln 3 − ln 2
lim
3