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Question 45 - Week 7 Practice | Prasnya
Q45
00:00
20 Nov 2022
Q45.
Find the limit of the sequence
{
a
n
}
\{a_n\}
{
a
n
}
such that
a
n
=
4
n
−
1
3
n
2
+
9
a_n=\frac{4n-1}{3n^2+9}
a
n
=
3
n
2
+
9
4
n
−
1
.
@passage
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Q45
00:00
20 Nov 2022
Q45.
Find the limit of the sequence
{
a
n
}
\{a_n\}
{
a
n
}
such that
a
n
=
4
n
−
1
3
n
2
+
9
a_n=\frac{4n-1}{3n^2+9}
a
n
=
3
n
2
+
9
4
n
−
1
.
@passage
Your Answer
Save
Read
Check
Details
