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Problems
Problem
Problem
If
f
(
x
)
=
x
⋅
2
x
−
x
1
−
cos
x
f(x) = \frac{x \cdot 2^x - x}{1 - \cos x}
f
(
x
)
=
1
−
c
o
s
x
x
⋅
2
x
−
x
and
g
(
x
)
=
2
x
⋅
sin
(
log
2
2
x
)
g(x) = 2^x \cdot \sin\left(\frac{\log 2}{2^x}\right)
g
(
x
)
=
2
x
⋅
sin
(
2
x
l
o
g
2
)
, then a.
lim
x
→
0
f
(
x
)
=
log
2
\lim_{x \to 0} f(x) = \log 2
lim
x
→
0
f
(
x
)
=
lo
g
2
b.
lim
x
→
∞
g
(
x
)
=
log
4
\lim_{x \to \infty} g(x) = \log 4
lim
x
→
∞
g
(
x
)
=
lo
g
4
c.
lim
x
→
0
f
(
x
)
=
log
4
\lim_{x \to 0} f(x) = \log 4
lim
x
→
0
f
(
x
)
=
lo
g
4
d.
lim
x
→
∞
g
(
x
)
=
log
2
\lim_{x \to \infty} g(x) = \log 2
lim
x
→
∞
g
(
x
)
=
lo
g
2
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