Question

Does the sequence converge or diverge? `lim sin(4n)/(7+sqrtn)`?

What is `lim_(n->oo) sin(4n)/(7+sqrtn)`?

I got all the way up to the limit of `sin(4n)/sqrtn = oo/oo`, but I saw the answer and it says its 0. I don't understand how it could be 0.

Answer 1

`sin(4n)/(7+sqrtn)` converge and
`lim_(n to oo) sin(4n)/(7+sqrtn)=0`

Explanation:

We will use what we know about `sin(4n)`.
`-1<=sin(4n)<=1`
`=>(-1)/(7+sqrtn)<=sin(4n)/(7+sqrtn)<=1/(7+sqrtn)`
And because `lim_(n to oo) (-1)/(7+sqrtn)=lim_(n to oo) 1/(7+sqrtn)=0`, using squeeze theorem, `limsin(4n)/(7+sqrtn)=0`
So `sin(4n)/(7+sqrtn)` converge.

Answer 2

As:

`abs (sin(4n)) <= 1`

we have that for `n>=1`:

`0 < abs ( sin(4n)/(7+sqrtn)) < 1/sqrtn`

Clearly:

`lim_(n->oo) 0 = 0`

and also:

`lim_(n->oo) 1/sqrtn = 0`

By the squeeze theorem we can conclude that:

`lim_(n->oo) abs ( sin(4n)/(7+sqrtn)) = 0`

but then:

`lim_(n->oo) sin(4n)/(7+sqrtn) = 0`