MATH 1A - SOLUTION TO 4.4.14, 4.4.15, 4.4.49, AND 4.4.71
PEYAM RYAN TABRIZIAN
1. 4.4.14
lim
θ→ π
2
1 − sin(θ)
− cos(θ)
= limπ − cos(θ) (l’Hopital’s rule)
θ→ 2
csc(θ)
2
sin (θ)
2
= limπ sin (θ)(cancelling out)
θ→ 2
=1
2. 4.4.15
ln(x)
√ = lim
x→∞
x→∞
x
lim
1
x
1
√
2 x
by L’Hopital’s rule
√
2 x
= lim
invert and multiply rule for fractions
x→∞ x
2
= lim √
x→∞
x
=0
Date: Thursday, October 28th, 2010.
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PEYAM RYAN TABRIZIAN
3. 4.4.49
lim
x→∞
p
x2
√x 2 + x + x
+ x − x = lim
+x−x · √
(use conjugate form)
x→∞
x2 + x + x
x2 + x − x2
= lim √
x→∞
x2 + x + x
x
= lim √
2
x→∞
x +x+x
x
q
= lim √
(factor out x2 out of the square root)
x→∞
1
2
x · 1+ x +x
√
x
= lim q
( x2 = |x| = x, since x > 0)
x→∞
x 1 + x1 + x
x
= lim q
x→∞
x
1 + x1 + 1
p
= lim q
x→∞
=√
=
x2
1
1+
1
x
+1
1
1+1
1
2
4. 4.4.71
4.1. a. Here, you want to calculate limt→∞ , so treat t as our x, and leave everything
else as a constant!!!!. In particular, we get:
ct
lim e− m = 0
t→∞
because c > 0 and m > 0
So, we get:
lim v =
t→∞
mg
mg
(1 − 0) =
c
c
MATH 1A - SOLUTION TO 4.4.14, 4.4.15, 4.4.49, AND 4.4.71
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4.2. b. Here, we let c → 0+ , so we treat c as our x, and leave everything else as a
constant!.
Then, we have:
!
ct
1 − e− m
lim v = lim mg
c
c→0+
c→0+
− ct !
t
− −m
e m
= lim+ mg
1
c→0
!
t
0
m e
=mg
1
!
t
m 1
=mg
1
mgt
m
=gt
=
(by l’Hopital’s rule)