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Byju's Answer
Standard XII
Mathematics
Existence of Limit
Evaluate x ...
Question
Evaluate
l
i
m
x
→
1
[
100
∑
K
=
1
x
K
]
−
100
x
−
1
Open in App
Solution
lim
x
→
1
[
100
∑
K
=
1
x
K
−
100
]
x
−
1
=
lim
x
→
1
[
x
+
x
2
+
x
3
+
.
.
.
.
.
.
+
x
100
−
100
x
−
1
]
=
lim
x
→
1
[
1
+
2
x
+
3
x
2
+
.
.
.
.
.
.
+
100
x
99
1
]
(Using L' Hospital's)
=
5050
.
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0
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