lim(x → π/4) (8 √2-(cos x+sin x)⁷)/(√2-√2 sin 2 x) is equal to

Solution

Maths sir · 4 steps

Put t = cos x + sin x, because sin 2x is just t² - 1. Then both parts become power differences.

1. Substitute. Let t = cos x + sin x. Squaring gives t² = 1 + sin 2x, so sin 2x = t² - 1. As x → π/4, t → √2, and both numerator and denominator go to 0.
2. Rewrite both parts. Since (√2)⁷ = 8√2, the numerator is (√2)⁷ - t⁷. The denominator is √2 - √2(t² - 1) = √2((√2)² - t²).

   L = 1/√2 lim(t → √2) (t⁷ - (√2)⁷)/(t² - (√2)²)

3. Standard limit. Divide top and bottom by t - √2 and use lim(t → a) (tⁿ - aⁿ)/(t - a) = n aⁿ⁻¹. With a = √2, the top gives 7a⁶ = 56 and the bottom gives 2a = 2√2.
4. So. L = 1/√2 × 56/2√2 = 56/4 = 14, which is option (a).

The idea to keep. When cos x + sin x and sin 2x appear together, set t = cos x + sin x and use t² = 1 + sin 2x. That usually turns the limit into the standard (tⁿ - aⁿ)/(t - a) form.

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