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Let g: R -> R be a differentiable function. It is given that lim_(alpha -> pi/2) g(2009 + cos(alpha)) = 3 and lim_(n -> infinity) n * [g(2009 + 1/n) - g(2009)] = 27. Find the value of lim_(t -> 0) [g(2009) / g(2009 + t)]^(1/t).

1. e³
2. e²
3. e⁻²
4. e⁻⁹

Correct answer: e⁻⁹

Solution

From the first limit (cos(alpha)->0 as alpha->pi/2): g(2009+0) = g(2009) = 3. From the second limit: n*[g(2009+1/n)-g(2009)] = g'(2009) = 27 (definition of derivative). The third limit is lim_(t->0) (3/g(2009+t))^(1/t) = e^(lim_(t->0) ln(3/g(2009+t))/t). Since g(2009)=3, this is a 1^inf form. Using L'Hopital or expansion: (3 - g(2009+t))/(t * g(2009+t)) -> -g'(2009)/g(2009) = -27/3 = -9. So the limit = e^(-9).

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