lim _(t → 0) ( 1^(1/(sin ²t)) + 2^(1/(sin ²t)) + ... + n^(1/(sin ²t)) )^(sin ²t) is equal to

Solution

Maths sir · 4 steps

Pull the biggest term, n, out of the bracket. Every other term fades away and only n survives.

1. Change variable. Put x = 1/(sin² t). As t → 0, sin² t → 0⁺, so x → +∞. The limit becomes (1ˣ + 2ˣ + … + nˣ)^(1/x).
2. Factor out the largest. The term nˣ is the biggest, so take it out of the bracket. Since (nˣ)^(1/x) = n, an n comes outside.

   (1ˣ + 2ˣ + … + nˣ)^(1/x) = n[(1/n)ˣ + (2/n)ˣ + … + ((n-1)/n)ˣ + 1]^(1/x)

3. Squeeze the bracket. Call the bracket S. It has n terms, each at most 1, and the last term is exactly 1, so 1 ≤ S ≤ n. Then 1 ≤ S^(1/x) ≤ n^(1/x), and n^(1/x) → 1 because 1/x → 0. So S^(1/x) → 1.
4. So. The limit is n × 1 = n, which is option (b).

The idea to keep. For (a₁ˣ + a₂ˣ + … + aₖˣ)^(1/x) with x → ∞ and positive bases, the largest base wins. Factor it out and squeeze the leftover bracket to 1.

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