If lim _(x → 0) (aeˣ - b cos x + ce⁻ˣ)/(x sin x) = 2, then a + b + c is equal to ____________.

Solution

Maths sir · 5 steps

Expand everything up to x², because the denominator x sin x behaves like x² near zero.

1. Size of denominator. Near x = 0, sin x ≈ x, so x sin x ≈ x². For a finite limit, the numerator must also start at x². Its constant term and its x term must both vanish.
2. Expand the numerator. Use eˣ ≈ 1 + x + x²/2, e⁻ˣ ≈ 1 - x + x²/2 and cos x ≈ 1 - x²/2. Collect the powers of x.

   aeˣ - b cos x + ce⁻ˣ = (a - b + c) + (a - c) x + (a + b + c)/2 x² + …

3. Match the terms. The constant and x terms must be zero, so a - b + c = 0 and a = c. The limit then becomes ((a+b+c)/2)/1 = 2.
4. Check with L'Hopital. Differentiating top and bottom twice gives (aeˣ + b cos x + ce⁻ˣ)/(2 cos x - x sin x). At x = 0 this is (a+b+c)/2, the same result.
5. So. (a+b+c)/2 = 2, which gives a + b + c = 4.

The common slip: 2. Dropping the 2 that comes from differentiating x sin x twice gives 2, not 4. The second derivative of x sin x is 2 cos x - x sin x, which equals 2 at x = 0.

The idea to keep. Find the power of x in the denominator first. Then expand the numerator up to that power: the lower terms must vanish, and the matching coefficients set the value of the limit.

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