Question

Use the limit laws to calculate limn→[infinity] an, for the following sequence (an):

(a) an = n n 2 ;
(b) an = 3n 2 n−5 ;
(c) an = n−2 n2 n−5 ;
(d) an = 2n3 n 2 n3−5 ;
(e) an = 2−n n2 1 ;

Answer 1

For the sequences provided, the limits can be calculated as follows: (a) 0, (b) infinity, (c) 1, (d) 2, and (e) 0.

Step-by-step explanation:

To calculate the limit of a sequence as n approaches infinity, we can apply the limit laws. Let's calculate the limits for the given sequences:

(a) lim(n→∞) an = lim(n→∞) n/(n^2) = lim(n→∞) 1/n = 0

(b) lim(n→∞) an = lim(n→∞) (3n^2)/(n-5) = lim(n→∞) (3n)/(1-5/n) = ∞

(c) lim(n→∞) an = lim(n→∞) (n-2)/(n^2/(n-5)) = lim(n→∞) (n-2)/(n-5) = 1

(d) lim(n→∞) an = lim(n→∞) (2n^3)/(n^2(n^3-5)) = lim(n→∞) 2/(1-5/n^3) = 2

(e) lim(n→∞) an = lim(n→∞) (2^-n)/(n^2/1) = lim(n→∞) (2^-n) = 0

Answer 2

(a) limₙ→∞ aₙ = ∞ (The sequence an = n^n diverges to infinity).

(b) limₙ→∞ aₙ = ∞ (The sequence an = 3n² / (n - 5) diverges to infinity).

(c) limₙ→∞ aₙ = 0 (The sequence an = (n - 2) / (n² * (n - 5)) tends to zero).

(d) limₙ→∞ aₙ = ∞ (The sequence an = 2n³ / (n² * n³⁻⁵) diverges to infinity).

(e) limₙ→∞ aₙ = 0 (The sequence an = 2^(-n) / (n²) approaches zero).

Step-by-step explanation:

(a) For an = n^n, as n approaches infinity, the sequence grows without bound, leading to the divergence of the sequence to infinity.

(b) In the sequence an = 3n² / (n - 5), when n tends to infinity, the numerator grows faster than the denominator, resulting in the sequence also diverging to infinity.

(c) For the sequence an = (n - 2) / (n² * (n - 5)), applying limit laws, the highest power term in the denominator dominates, leading to the sequence approaching zero.

(d) In an = 2n³ / (n² * n³⁻⁵), as n approaches infinity, the numerator grows faster than the denominator, causing the sequence to diverge to infinity.

(e) For the sequence an = 2^(-n) / (n²), the term 2^(-n) decreases rapidly as n grows, and the denominator grows much faster than the numerator, resulting in the sequence converging to zero as n tends to infinity.