Limits

The denominator x \sin x behaves like x^2 near 0, so for a finite limit the numerator must vanish through order one and leave a clean x^2 coefficient. Expanding, a e^x = a(1 + x + x^2/2), c e^{-x} = c(1 - x + x^2/2), and b \cos x = b(1 - x^2/2). The constant term a - b + c must be 0 and the linear coefficient a - c must be 0, giving a = c and a - b + c = 0, so b = 2a. The x^2 coefficient is a/2 + c/2 + b/2 = a/2 + a/2 + a = 2a, and dividing by the denominator coefficient 1 forces 2a = 2, so a = 1, c = 1, b = 2. Therefore a + b + c = 1 + 2 + 1 = 4. Option 0 reports the forced relation a - b + c instead of the requested sum a + b + c. Option 2 uses only the value of b. Option 6 over-counts by doubling a term. The JEE pattern is equating coefficients order by order to fix the unknown constants. Plausibility check: with a = c = 1, b = 2 the numerator's leading term is 2x^2 and x \sin x \sim x^2, so the quotient tends to 2 exactly as required, confirming the coefficients and hence a + b + c = 4.