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Math 1100: Quantitative Analysis Limits of Rational Functions Limits of Polynomial Functions Let f (x) be a polynomial (that is, a function of the type f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 ). Then lim f (x) = f (c) x→c Example 1. lim x3 − 2x + 1 = (−3)3 − 2(−3) + 1 x→−3 = −27 + 6 + 1 = −20 Example 2. lim 2x2 − x = 2(1)2 − 1 x→1 =2−1 =1 Limits of Polynomial Functions Suppose that we instead want to find the limit of a (x) , where f (x) and g(x) are polynomials. There are a few possibilities in rational function fg(x) this case. • If g(x) 6= 0, then f (x) f (c) = x→c g(x) g(c) This case corresponds to a point where the function is continuous. lim f (x) does not exist. This corresponds to a vertical x→c g(x) • If g(x) = 0 and f (x) 6= 0, then lim asymptote of the graph. • If g(x) = 0 and f (x) = 0, then we can factor (x − c) out of the numerator and denominator and proceed with a new function. This case may correspond to a hole in the graph (in which case the limit exists) or to a vertical asymptote (in which case it does not). 1 The next few examples will use the function h(x) = shown below. x2 −5x+6 . x2 −x−2 The graph of this function is 5 2.5 -7.5 -5 -2.5 0 2.5 5 7.5 10 -2.5 -5 We define f (x) = x2 − 5x + 6 and g(x) = x2 − x − 2. x2 − 5x + 6 x→4 x2 − x − 2 Example 3. Compute lim f (x) f (c) = . Thus x→c g(x) g(c) Since g(4) = 42 − 4 − 2 = 10, we have lim x2 − 5x + 6 42 − 5(4) + 6 = x→4 x2 − x − 2 42 − 4 − 2 2 = 10 1 = 5 lim x2 − 5x + 6 x→−1 x2 − x − 2 Since g(−1) = (−1)2 − (−1) − 2 = 0 and f (x) = (−1)2 − 5(−1) + 6 = 12, the limit does not exist. Graphically, we can see that there is an asymptote at x = −1. Example 4. Compute lim 2 x2 − 5x + 6 x→2 x2 − x − 2 2 Since g(2) = 2 − 2 − 2 = 0 and f (x) = 22 − 5(2) + 6 = 0, we are in the third indeterminate case. Example 5. Compute lim x2 − 5x + 6 (x − 3) = lim 2 x→2 x − x − 2 x→2 (x + 1) x−3 = lim x→2 x + 1 −1 = 3 lim Graphically, we can see that there is a hole in the graph at the point (2, − 13 ) 3