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Mathematics
Mathematics
Problem
Let
f
(
x
)
=
lim
n
→
∞
log
(
2
+
x
)
−
x
2
n
sin
x
1
+
x
2
n
f(x)=\lim_{n\rightarrow \infty}\frac{\log(2+x)-x^{2n}\sin x}{1+x^{2n}}
f
(
x
)
=
lim
n
→
∞
1
+
x
2
n
l
o
g
(
2
+
x
)
−
x
2
n
s
i
n
x
. Then (a)
f
f
f
is continuous at
x
=
1
x=1
x
=
1
(b)
lim
x
→
1
+
f
(
x
)
=
log
3
\lim_{x\rightarrow 1^+} f(x)=\log 3
lim
x
→
1
+
f
(
x
)
=
lo
g
3
(c)
lim
x
→
1
+
f
(
x
)
=
−
sin
1
\lim_{x\rightarrow 1^+} f(x)=-\sin 1
lim
x
→
1
+
f
(
x
)
=
−
sin
1
(d)
lim
x
→
1
+
f
(
x
)
\lim_{x\rightarrow 1^+} f(x)
lim
x
→
1
+
f
(
x
)
does not exist
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