MAT137Y1 Study Guide - Midterm Guide: Classification Of Discontinuities

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20 Nov 2015
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MAT137Y1 Full Course Notes
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[4 points] calculate the following two limits: (a) lim x 2. 4 + x2 x3 x2 + 2x. Solution: the numerator and denominator are both polynomials and hence are con- tinuous. Furthermore, the denominator is non-zero at x = 2 since 23 22 + 2(2) = 8. is continuous at 2 and we may determine the. 4 + x2 limit by simple evaluation: x3 x2 + 2x. 4 + x2 x3 x2 + 2x lim x 2. Solution: unlike part (a), the denominator in this question does evaluate to zero, mean- ing that the function is not continuous and we are required to be a bit more clever. We instead proceed by factoring the numerator and denominator into their linear constituents before evaluating the limit: 4 x2 x3 x2 2x lim x 2. (x 2)(x + 2) x(x 2)(x + 1)

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