The value of lim _(n → ∞) 1/nΣⱼ₌₁ⁿ ((2j - 1) + 8n)/((2j - 1) + 4n) is equal to :

Solution

Maths sir · 4 steps

Divide top and bottom by n, spot (2j-1)/2n as the point x, and the sum becomes a clean integral.

1. Divide by n. Divide the numerator and the denominator of each term by n. Each term becomes ((2j-1)/n + 8)/((2j-1)/n + 4).
2. Find the x. Write (2j-1)/n = 2xⱼ with xⱼ = (j - 1/2)/n. As j runs from 1 to n, xⱼ is the midpoint of each strip of width 1/n in [0, 1]. So 1/nΣ becomes ∫₀¹ … dx, and the term becomes (2x + 8)/(2x + 4) = (x + 4)/(x + 2).

   ∫₀¹ (x+4)/(x+2) dx = ∫₀¹ (1 + 2/(x+2))dx

3. Integrate. This gives [x + 2 ln(x+2)]₀¹ = 1 + 2 ln 3 - 2 ln 2 = 1 + 2 logₑ(3/2).
4. So. The limit is 1 + 2 logₑ(3/2), which is option (d).

The idea to keep. In a limit of a sum, pull out 1/n as dx and write each term in j/n as x. Shifts like j - 1/2 do not change the limits; x still runs from 0 to 1.

Answer from the NTA final answer key. Solution by PrepWiser.