lim _(x → π/2) ( tan ²x( (2sin ²x + 3 sin x + 4)^(1/2) - (sin ²x + 6 sin x + 2)^(1/2) ) ) is equal to

Solution

Maths sir · 4 steps

Rationalise the root difference, write tan² x through sin x, and the zero factor (1-sin x) cancels.

1. Substitute sin x. Let s = sin x. As x → π/2, s → 1. The bracket becomes √9 - √9 = 0, while tan² x → ∞, so this is an ∞ × 0 form.
2. Rationalise. Multiply and divide the bracket by the sum of the two roots. The numerator becomes (2s² + 3s + 4) - (s² + 6s + 2) = s² - 3s + 2 = (s-1)(s-2).
3. Rewrite tan squared. tan² x = (sin² x)/(cos² x) = s²/(1-s)(1+s). Since s - 1 = -(1-s), the factor (1-s) cancels and leaves a minus sign.

   lim(s → 1) (- s² (s-2))/(1+s)(√(2s²+3s+4)+√(s²+6s+2))

4. So. Put s = 1. The numerator is -1 × (-1) = 1 and the denominator is 2 × (3 + 3) = 12. The limit is 1/12, option (a).

The idea to keep. When a difference of square roots goes to zero, rationalise it first. Then write the other factor in the same variable so the zero factor cancels cleanly.

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