Evaluate the limit (1/x)-(1/(x^x-1)) when x tends to 0
Answers
Answered by
0
Answer:
Explanation:
From the binomial expansion
(
1
+
x
)
n
=
1
+
n
x
+
n
(
n
−
1
)
2
!
x
2
+
n
(
n
−
1
)
(
n
−
2
)
3
!
x
3
+
⋯
+
we have
(
1
+
x
)
1
x
=
1
+
1
x
x
+
1
x
(
1
x
−
1
)
2
!
x
2
+
1
x
(
1
x
−
1
)
(
1
x
−
2
)
3
!
x
3
+
⋯
+
=
1
+
1
+
1
(
1
−
x
)
2
!
+
1
(
1
−
x
)
(
1
−
2
x
)
3
!
+
⋯
+
so
lim
x
→
0
(
1
+
x
)
1
x
=
lim
x
→
0
1
+
1
+
1
(
1
−
x
)
2
!
+
1
(
1
−
x
)
(
1
−
2
x
)
3
!
+
⋯
+
=
∞
∑
k
=
0
1
k
!
=
e
Answer link
Answered by
0
Answer:
not defined
Step-by-step explanation:
because
0 can not devide any real number
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