Limits

Both numerator and denominator vanish at x = 0, an indeterminate 0/0 form that yields cleanly to either L'Hopital's rule or the Maclaurin series of e^x. Expanding e^x = 1 + x + x^2/2 + x^3/6 + \cdots, the numerator e^x - 1 - x = x^2/2 + x^3/6 + \cdots, so dividing by x^2 leaves 1/2 + x/6 + \cdots, which tends to 1/2. Alternatively, applying L'Hopital once gives \frac{e^x - 1}{2x}, still 0/0, and a second application gives \frac{e^x}{2} \to 1/2. Option 0 results from differentiating only once and stopping prematurely at a still-indeterminate form. Option 1 mistakenly keeps the first nonzero coefficient without halving. Option 2 inverts the factor 1/2. The governing JEE pattern is resolving a quadratic-order 0/0 limit through repeated L'Hopital or second-order expansion, and a key lesson is that L'Hopital must be reapplied as long as the form stays indeterminate, stopping only once a determinate value appears. The series viewpoint explains why the answer is exactly the second Maclaurin coefficient of e^x: subtracting the constant and linear terms strips away the part of e^x already matched by its tangent line, leaving the curvature term that the x^2 denominator is precisely scaled to measure. This interpretation connects the limit to the idea that the second derivative controls how fast a function pulls away from its own tangent. Plausibility check: at x = 0.01 the numerator \approx 0.00005 and dividing by 0.0001 gives 0.5008, snugly approaching 1/2, which corroborates the analytic value.