Macroeconomics Exam Review 210

Macroeconomics Exam Review 210 - Solutions for Foundations...

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for every x 𝐵 . Given 𝜖 > 0, there exists 𝑁 such that for every 𝑚, 𝑛 > 𝑁 𝐷𝑓 𝑛 𝐷𝑓 𝑚 < 𝜖/𝑟 and 𝐷𝑓 𝑛 𝑔 < 𝜖 Letting 𝑚 → ∞ 𝑓 𝑛 ( x ) 𝑓 ( x ) ( 𝑓 𝑛 ( x 0 ) 𝑓 ( x 0 ) ) 𝜖 x x 0 (4.45) for 𝑛 𝑁 and x 𝐵 . Applying the mean value inequality to 𝑓 𝑛 , there exists 𝛿 such that 𝑓 𝑛 ( x ) 𝑓 𝑛 ( x 0 ) ∥ ≤ 𝜖 x x 0 (4.46) Using (4.45) and (4.46) and the fact that 𝐷𝑓 𝑛 𝑔 < 𝜖 we deduce that 𝑓 ( x ) 𝑓 ( x 0 ) 𝑔 ( x 0 ) ∥ ≤ 3 𝜖 x x 0 𝑓 is differentiable with derivative 𝑔 . 4.40 Define 𝑓 ( 𝑥 ) = 𝑒 𝑥 + 𝑦 𝑒 𝑦 By the chain rule (Exercise 4.22) 𝑓 ( 𝑥 ) = 𝑒 𝑥 + 𝑦 𝑒 𝑦 = 𝑓 ( 𝑥 ) which implies (Example 4.21) that 𝑓 ( 𝑥 ) = 𝑒 𝑥 + 𝑦 𝑒 𝑦 = 𝐴𝑒 𝑥 for some 𝐴 ∈ ℜ Evaluating at 𝑥 = 0 using 𝑒 0 = 1 gives 𝑓 (0) = 𝑒 𝑦
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