Limits

Both numerator and denominator vanish like x^3 near zero, so the limit is governed by comparing their leading cubic coefficients via Taylor expansion. Using \sin x = x - x^3/6 + \cdots, the numerator x - \sin x = x^3/6 + \cdots. Using \tan x = x + x^3/3 + \cdots, the denominator x - \tan x = -x^3/3 + \cdots. The ratio of leading coefficients is (1/6)/(-1/3) = (1/6) \cdot (-3) = -1/2. Hence the limit is -1/2. Option 1/2 drops the negative sign that arises because x - \tan x is negative for small positive x. Option -2 inverts the coefficient ratio. Option 2 both inverts and loses the sign. The governing JEE pattern is dividing two cubic-order infinitesimals by matching their dominant terms, a technique that hinges on recognising that both expressions are odd functions whose first surviving Taylor terms are cubic. Conceptually, x - \sin x measures how the sine curve lags behind the line y = x, while x - \tan x measures how the tangent curve races ahead of it, so the two corrections point in opposite directions and their ratio inherits a minus sign. Higher-order terms beyond x^3 contribute nothing to the limit because they vanish faster once divided by x^3, which is why retaining only the cubic coefficients is rigorous rather than approximate. Plausibility check: for small positive x, x - \sin x > 0 while x - \tan x < 0, so the quotient must be negative, immediately eliminating the two positive options and supporting the value -1/2 in both sign and magnitude.