Question

u(sub n) = (1 + c(n^2))/(2n + 3 + 2sin(n))^2

The limit of this sequence as n approaches infinity is 5. What is the value of c?

I don't even know where to begin with this. Thanks in advance for your help, Open Studiers!

Answer 1

It is this sequence, I think:\[u_n=\frac{ 1+cn^2 }{ (2n+3+2\sin n)^2 }\]
The problem is sin(n), because as n goes off to infinity, sin n always keeps taking on all kinds of values between -1 and 1. It doesn't go to a particular value.
Because 2n is also in the denominator, you can imagine that for very large n, +3+2sin(n) is not that important anymore, in fact they are completely irrelevant, compared to 2n!

Answer 2

The 1 in the numerator doesn't look impressive either...
So, you could just as well replace the whole thing with:\[\frac{ cn^2 }{ (2n)^2 }=\frac{ cn^2 }{ 4n^2 }\]This means: if you make n large enough, only the n² is important, ignore the rest.
But: n² is in the numerator and the denominator! so they cancel...and now you can easily see what the limit is. If it has to be 5, you can come up with a suitable c, I guess...