Test1s

# Test1s

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MATH1821 Mathematical Methods for Actuarial Science I Test 1 1. (6 points) Using limit laws (and any known limits discussed in lectures, if necessary), evaluate the following limits. (a) lim x 1 sin - 1 ( x - 1) x 2 - 1 (It is given that sin - 1 x is continuous at x = 0.) (b) lim x →-∞ x 2 + 4 - 2 x Solution. (a) As sin - 1 x is continuous at x = 0, lim x 1 sin - 1 ( x - 1) = 0. Hence lim x 1 sin - 1 ( x - 1) x 2 - 1 = lim x 1 sin - 1 ( x - 1) ( x - 1)( x + 1) = lim x 1 sin - 1 ( x - 1) sin(sin - 1 ( x - 1)) · 1 x + 1 = 1 2 . (b) lim x →-∞ x 2 + 4 - 2 x = lim x →-∞ x 2 x ( x 2 + 4 + 2) = lim x →-∞ 1 - 1 + 4 /x 2 + 2 /x = - 1 2. (5 points) Use the definition to show that lim x 4 x = 2. Solution. Note that | x - 2 | = | x - 4 | x + 2 | x - 4 | 2 . Hence for any ϵ > 0, let δ = 2 ϵ . Then δ > 0 and 0 < | x - 4 | < δ | x - 2 | = | x - 4 | x + 2 | x - 4 | 2 < δ 2 = ϵ. This shows lim x 4 x = 2. 3. (4 points) Is it possible to define f (3) in a way that extends f ( x ) = x + 1 - 2 x - 3 to be continuous at x = 3? Give the value of f (3) if it is possible. Solution. Since lim x 3 x + 1 - 2 x - 3 = lim x 3 ( x + 1 - 2)( x + 1 + 2) ( x - 3)( x + 1 + 2) = lim x 3 1 x + 1 + 2 = 1 4 , f is continuous at 3 if we define

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