Find the limit of the sequences?

Question

(a) {n+2/n-1}

(b) {2n^2-1/n^2}

could you pls explain the process to find the limit of these sequences.

Answer 1

(a) Limit as n → ∞ of (n + 2) / ( n - 1)

What you do is divide the numerator
and the denominator, each by n.

Numerator : (n + 2) / n = 1 + 2/n

Denominator : (n - 1) / n = 1 - 1/n

So you have : Limit as n → ∞ of (1 + 2/n) / (1 - 1/n)

Now, as n → ∞, 2/n and 1/n both tend to zero.

Therefore, Limit as n → ∞ is (1 + 0) / (1 - 0) = 1 / 1 = 1
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(b) Limit as n → ∞ of (2n^2 - 1) / n^2

This time, divide top and bottom by n^2, or,
what is the same process, dividing the two
top terms separately by n^2.

This gives : 2 - 1/n^2

Now as n → ∞, n^2 → ∞, and 1/n^2 → 0.

So, Limit = 2 - 0 = 2.

Answer 2

(a) {n+2/n-1}
assuming that the sequence is:
(n+2)/(n-1), the limit n-> infinity is 1

(b) {2n^2-1/n^2}
again if you mean:
(2n^2-1)/n^2 = 2 - 1/n^2 and the limit n->infinity is simply 2 .