Let f(x)=lim _(n → ∞) Σᵣ₌₀ⁿ((tan (x / 2ʳ⁺¹)+tan ³(x / 2ʳ⁺¹))/(1-tan ²(x / 2ʳ⁺¹))) Then lim(x → 0) (eˣ-eᶠ⁽ˣ⁾)/(x-f(x)) is equal to ___________.

Solution

Maths sir · 4 steps

Spot that each term is tan 2θ - tan θ, and the whole sum telescopes down to tan x.

1. Rewrite one term. Put t = tan θ with θ = x/2ʳ⁺¹. Since tan 2θ = 2t/(1-t²), we get tan 2θ - tan θ = (2t - t + t³)/(1-t²) = (t + t³)/(1-t²), which is exactly the term in the sum.

   Σᵣ₌₀ⁿ(tan x/2ʳ - tan x/2ʳ⁺¹) = tan x - tan x/2ⁿ⁺¹

2. Let n grow. As n → ∞, x/2ⁿ⁺¹ → 0, so tan x/2ⁿ⁺¹ → 0. Hence f(x) = tan x.
3. Factor the limit. Take out eᵗᵃⁿˣ: the expression becomes eᵗᵃⁿˣ × (eˣ⁻ᵗᵃⁿˣ - 1)/(x - tan x). As x → 0, u = x - tan x → 0, so the fraction tends to lim(u → 0)(eᵘ - 1)/u = 1, and eᵗᵃⁿˣ → e⁰ = 1.
4. So. The limit is 1 × 1 = 1.

The idea to keep. When a sum has halving angles like x/2ʳ, try writing each term as a difference g(2θ) - g(θ). Then it telescopes, and only the first and last terms survive.

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