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Correct answer: 0 <= a < 1, L = 0
Interpreting the exponent as x*(x²+3)/(x²+1) ~ x as x->inf: base -> a. If 0<=a<1: a^x -> 0. If a=1: 1^x = 1 (finite but exponent not needed). If a>1: a^x -> inf (not finite). So for L to be finite with a>=0, we need 0<=a<1, giving L=0. Answer: option D.