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Byju's Answer
Standard XII
Mathematics
Summation by Sigma Method
limn→∞n/n + 1...
Question
lim
n
→
∞
(
(
n
n
+
1
)
α
+
sin
1
n
)
n
when
α
ϵ
Q is equal to
A
e
1
−
α
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B
-
α
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C
e
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D
e
1
+
α
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Solution
The correct option is
A
e
1
−
α
lim
n
→
∞
(
(
n
n
+
1
)
α
+
sin
1
n
)
n
Replace
n
with
1
x
,
Now as
n
→
∞
,
x
→
0
∴
Limit becomes
L
=
lim
x
→
0
(
(
x
+
1
)
−
α
+
sin
x
)
1
x
Taking
log
on both sides, we get
ln
L
=
lim
x
→
0
ln
(
(
x
+
1
)
−
α
+
sin
x
)
x
It os forming the
0
0
form,
So, using L'Hospital's Rule, we get
ln
L
=
lim
x
→
0
−
α
(
x
+
1
)
−
α
−
1
+
cos
x
(
x
+
1
)
−
α
+
sin
x
⇒
ln
L
=
1
−
α
⇒
L
=
e
1
−
α
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0
Similar questions
Q.
lim
n
→
∞
{
(
n
n
+
1
)
α
+
sin
1
n
}
n
(
where
α
∈
N
)
is equal to
Q.
The increasing order of the values of e/m (charge/ mass) is :
Q.
I
f
∫
e
2
[
1
l
o
g
x
−
1
(
l
o
g
x
)
2
]
d
x
=
α
+
β
l
o
g
2
,
t
h
e
n
Q.
The increasing order (lowest first) for the values of
e
/
m
(charge/mass) for electron (
e
), proton (
p
), neutron (
n
) and alpha particle (
α
) is:
Q.
Solve this
(
α
+
1
)
2
(
α
+
1
)
2
−
(
α
+
1
)
(
β
+
1
)
+
(
β
+
1
)
2
(
β
+
1
)
2
−
(
α
+
1
)
(
β
+
1
)
to get
=
α
+
1
α
−
β
+
β
+
1
β
−
α
=
α
−
β
α
−
β
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