MATH
MA113_S13_Assignment_2_solutions

# MA113_S13_Assignment_2_solutions - MA 113 Calculus I Spring...

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MA 113 – Calculus I – Spring 2013 Written Assignment #2 Solution Due Friday, February 1, 2013, at beginning of lecture Question A: Let function f be defined by f ( x ) = x + 3 for x ≤ - 1 , x 2 for - 1 < x 1 , - x 2 + 2 x + 2 for x > 1 . i) Compute the right- and left-hand limits at x = - 1 and at x = 1; ii) determine whether f is left-continuous, right-continuous or continuous at x = - 1 and at x = 1; iii) sketch the graph of f . (i) Each piece of f ( x ) is a polynomial, thus we can evaluate each limit using direct substitution. lim x →- 1 - f ( x ) = lim x →- 1 - x + 3 = - 1 + 3 = 2 lim x →- 1 + f ( x ) = lim x →- 1 + x 2 = ( - 1) 2 = 1 lim x 1 - f ( x ) = lim x 1 - x 2 = 1 2 = 1 lim x 1 + f ( x ) = lim x 1 + - x 2 + 2 x + 2 = - 1 + 2 + 2 = 3 (ii) For the function values at x = - 1 and 1, we get f ( - 1) = - 1 + 3 = 2 and f (1) = 1 2 = 1. Since f ( - 1) = lim x →- 1 - f ( x ) 6 = lim x →- 1 + f ( x ), f is left-continuous at x = - 1. Similarly, f (1) = lim x 1 - f ( x ) 6 = lim x 1 + f ( x ), thus f is also left-continuous at x = 1. (iii) - 5 - 4 - 3 - 2 - 1 1 2 3 4 - 3 - 2 - 1 1 2 3 4 0 Question B: Let m and n be two positive integers. Evaluate the following limit (algebraically)

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• Fall '09
• Calculus, lim, Continuous function, direct substitution, lim x2, correct inverse function

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