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Mathematics
Mathematics
Problem
Let
a
=
lim
x
→
0
ln
(
cos
2
x
)
3
x
2
a = \lim_{x \to 0} \frac{\ln(\cos 2x)}{3x^2}
a
=
lim
x
→
0
3
x
2
l
n
(
c
o
s
2
x
)
,
b
=
lim
x
→
0
sin
2
2
x
x
(
1
−
e
x
)
b = \lim_{x \to 0} \frac{\sin^2 2x}{x(1-e^x)}
b
=
lim
x
→
0
x
(
1
−
e
x
)
s
i
n
2
2
x
,
c
=
lim
x
→
1
x
−
x
ln
x
c = \lim_{x \to 1} \frac{\sqrt{x-x}}{\ln x}
c
=
lim
x
→
1
l
n
x
x
−
x
. Then
a
,
b
,
c
a, b, c
a
,
b
,
c
satisfy : (a)
a
<
b
<
c
a < b < c
a
<
b
<
c
(b)
b
<
c
<
a
b < c < a
b
<
c
<
a
(c)
a
<
c
<
b
a < c < b
a
<
c
<
b
(d)
b
<
a
<
c
b < a < c
b
<
a
<
c
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