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
Guillaume L.
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.

Andrea S.
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#

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