If lim x → 0 (a x² eˣ-b log ₑ(1+x)+c x e⁻ˣ)/(x² sin x)=1, then 16(a²+b²+c²) is equal to ________.

Solution

Maths sir · 4 steps

The denominator behaves like x³, so expand the top to x³ and match coefficients.

1. Size of denominator. Near x = 0, sin x ≈ x, so x² sin x behaves like x³. For the limit to be the finite value 1, the numerator must have no x or x² term, and its x³ coefficient must be 1.
2. Expand each term. Up to x³: ax²eˣ = ax² + ax³, -b logₑ(1+x) = -bx + b/2x² - b/3x³, and cxe⁻ˣ = cx - cx² + c/2x³. Collecting powers gives the numerator below.

   (c-b) x + (a + b/2 - c)x² + (a - b/3 + c/2)x³ + …

3. Match coefficients. c - b = 0 gives c = b. Then a + b/2 - b = 0 gives a = b/2. Finally b/2 - b/3 + b/2 = 2b/3 = 1, so b = 3/2, c = 3/2, a = 3/4.
4. So. a² + b² + c² = 9/16 + 9/4 + 9/4 = 81/16, and 16(a²+b²+c²) = 81.

The idea to keep. When a limit with unknown constants must be finite, find the power of x in the denominator. Every lower power in the numerator must vanish, and the matching power gives the limit's value.

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