MATH 668 Midterm: MATH 668 Louisville 668S16E2
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Exam #1 solutions: (8 points) evaluate the following limits; when a limit can not be evaluated, explicitly say so. (a) (2 points) lim t 1. 3t. (b) (2 points) lim x 5 x2 3x 10 x2 10x+25 . Since an exponential function with positive base is continuous throughout, we may use straightforward substitution to see that lim t 1. 0 , and we shall thus try to factor out (x 5) from the numerator and denominator. However, direct substitution in this new limit still yields an incalculable form, which is now 7. 0 ; since this has a nonzero numerator, it will correspond to some sort of in nite discontinuity, at which this limit does not exist. (c) (2 points) lim r 3 r3 1 r2+r . (d) (2 points) lim s . Direct substitution yields a nonzero denominator, so lim r 3 r3 1 r2+r = 33 1.
