Limits

Quantities of the form 0^0 are indeterminate and must be resolved by taking logarithms to convert the power into a product. Let L = \lim_{x \to 0^+} x^x, so \ln L = \lim_{x \to 0^+} x \ln x. This product is a 0 \cdot (-\infty) form; rewrite it as \frac{\ln x}{1/x} and apply L'Hopital, giving \frac{1/x}{-1/x^2} = -x \to 0. Hence \ln L = 0 and L = e^0 = 1. Option 0 assumes the base drives the result to zero, ignoring that the shrinking exponent counteracts it. Option e confuses this with limits that produce Euler's number through 1^{\infty} forms, which is a different mechanism. The choice Does not exist is wrong because the one-sided limit is perfectly well defined even though x^x is undefined for x \le 0. The governing JEE Advanced technique is the logarithmic transformation of a variable exponent, the same device used for limits of the form f(x)^{g(x)} whenever the base and exponent both approach delicate values. The deeper reason the rewrite works is that the natural logarithm is continuous and strictly increasing, so the limit of the logarithm equals the logarithm of the limit, letting us recover L by exponentiating at the end. The intermediate product x \ln x is decisive because the polynomial factor x decays to zero faster than \ln x diverges to negative \infty, a hierarchy of growth rates that JEE problems repeatedly exploit. Plausibility check: numerically (0.1)^{0.1} \approx 0.794 and (0.01)^{0.01} \approx 0.955, a sequence visibly climbing toward 1, which corroborates the analytic conclusion that the limit equals 1.