# Thread: real sequences analysis help

1. ## real sequences analysis help

1. assume that an->2 and bn-->3 evaluate lim x->inf (2an -3bn)
a) -5 (b)0 (c) the limit doesnot exist (d) the limit cannot be determined

2. Assume that an-->2 and bn-->3 Evaluate lim x->inf (an-2)/bn
(a)0 (b) the limit doesnot exist (c) the limit canot be determined

3.We know from trignometrythat |sin(n)<=1 for all n.This shows that 0<= |sin(n)/n|<= 1/n and hence that -1/n <=|sin(n)/n <= 1/n. We can also evaluate the lim x->inf (sin(n)/n by using
(a) product rule for limit of sequences.
(b) quotient rule for limits of sequences.
(c) the squeeze theorem for sequence

(4) find lim x-->inf sin(n)/n
(a)1 (b) 0 (c) limit doesnot exist

2. Originally Posted by bobby77
1. assume that an->2 and bn-->3 evaluate lim x->inf (2an -3bn)
Use the theorem about sequences, i.e. direct substitution.
2(2)-3(3)=4-9=-5

2. Assume that an-->2 and bn-->3 Evaluate lim x->inf (an-2)/bn
Since lim bn does not equal to zero. We can use division rule.
(2-2)/3=0

3.We know from trignometrythat |sin(n)|<=1 for all n.This shows that 0<= |sin(n)/n|<= 1/n and hence that -1/n <=|sin(n)/n| <= 1/n. We can also evaluate the lim x->inf (sin(n)/n by using
My favorite, the "squeeze theorem".
(Other favorite is composite function rule).
(4) find lim x-->inf sin(n)/n
We have that,
|sin(n)/n|<1/n
Thus,
-1/n<sin(n)/n<1/n
Squeeze theorem.
Both -1/n and 1/n ---> 0
Thus, limit is zero.