lim n tends to infinite 1/n sigma j =1 to n 2j-1+8n/2j-1+4n

To find the limit as n tends to infinity of the given expression, we can simplify it step by step.

First, let's simplify the numerator: 1/n * sigma j=1 to n (2j-1 + 8n)

Expanding the sigma notation, we get: 1/n * (2(1)-1 + 2(2)-1 + ... + 2(n)-1 + 8n)

Simplifying further, we have: 1/n * (1 + 3 + 5 + ... + (2n-1) + 8n)

The sum of the arithmetic series from 1 to (2n-1) can be calculated using the formula: Sum = (n/2) * (first term + last term)

Applying this formula, we get: 1/n * (n/2) * (1 + (2n-1))

Simplifying, we have: 1/2 * (1 + 2n - 1)

The 1 and -1 terms cancel out, leaving us with: 1/2 * 2n

Finally, simplifying further, we get: n

Therefore, the limit as n tends to infinity of the given expression is n.