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.

2 Answers
May 13, 2018

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.

May 13, 2018

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