Math 113 – Homework 7 – Solutions

Math 113 – Homework 7 – Solutions
Due: 20 December 2005 Tuesday.
15 tan x − 15x − 5x3 − 2x5
Q-1) lim
x→0 24x cos x − 24x + 12x3 − x5
=
17x7
lim 21 7
x→0 − x
30
+ ···
+ ···
=−
x→0
= lim
x→0
=
lim
x→0
+ · · · ) − 15x − 5x3 − 2x5
24x(1
+ · · · ) − 24x + 12x3 − x5
T aylor
=
lim
x→0
2x3 sec x
2x − x2 − 2(x − x2 /2 + x3 /3 − · · · )
2x3 sec x
= −3. (Note here that sec 0 = 1.)
−2x3 /3 + · · ·
(esin x − 1) sin x3
(esin x − 1)
sin x3
=
lim
lim
x→0
x3 tanh x
tanh x x→0 x3
Q-3) lim
x→0
5
7
x3
+ 2x
+ 17x
3
15
315
2
4
x6
− x2 + x24 − 720
15(x +
170
.
7
2x3 sec x
2x − x2 − 2 ln(1 + x)
Q-2) lim
T aylor
arctan x
arctan x
1
Q-4) lim
= lim
lim
x→0 sinh x cos x
x→0
x→0
sinh x
cos x
L0 H ôpital
=
L0 H ôpital
=
lim
x→0
lim
x→0
1
1+x2
cosh x
2
e−1/x
Q-5) Find lim
, where n is any integer.
x→0
xn
If n ≤ 0, then this limit is clearly 0. So we now assume that n > 0.
2
e−1/x
lim
x→0
xn
= lim
t→±∞
(t= x1 )
=
tn
e t2
lim
t→±∞
L0 H ôpital
=
lim
t→±∞
ntn−1
2tet2
tn−1
n!
1
n
lim
= · · · = lim n n lim t2 = 0.
2
t
t→±∞ 2 t t→±∞ e
2t t→±∞ e
2
Conclusion: lim
x→0
e−1/x
= 0, where n is any integer.
xn
Comments and questions to [email protected]
esin x cos x
· 1 = 1.
sech2 x
· 1 = 1.