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ExamsJEE AdvancedMaths

Let g: R -> R be a differentiable function such that lim(x->pi/2) g(2009 + cos x) = 3 and lim(n->inf) n * [g(2009 + 1/n) - g(2009)] = 27. Find the value of lim(t->0) [g(2009) / g(2009+t)]^(1/t).

  1. e⁻²
  2. e⁻⁹

Correct answer: e⁻⁹

Solution

g(2009) = 3 (from first limit) and g'(2009) = 27 (from second limit, which is derivative definition). The expression becomes lim [g(2009)/g(2009+t)]^(1/t) = e^(lim ln[3/g(2009+t)]/t) = e^(-g'(2009)/g(2009)) = e^(-27/3) = e^(-9).

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