Using only definition: A sequence, t, converges to the real number s provided that for each r > 0 th


Question: Using only definition: A sequence, t, converges to the real number s provided that for each r > 0 there exists a real number N such that for all n an element of N, n>N implies that the absolute value of |t-s| < r. If t converges, then s is called the limit of the sequence t. If a sequence does not converge to a real number it is said to diverge.

Prove the following:

a) For any real number k>0, the limit as n approaches infinity, of 1/n^k = 0

b) \(\underset{n\to \infty }{\mathop{\lim }}\,\frac{3n+1}{n+2}=3\)

c) \(\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sin n}{n}=0\)

d) \(\underset{n\to \infty }{\mathop{\lim }}\,\frac{n+2}{{{n}^{2}}-3}=0\)

Price: $2.99
Solution: The solution file consists of 2 pages
Solution Format: Word Document

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