MATH 151 Engineering Math I, Spring 2014 JD Kim Week3 Section 2.2, 2.3, 2.5-2.6, (quantitative) definition of limit, calculation of limits, limits at infinity, continuity. Section 2.2 The limit of a function Let’s investigate the behavior of the funtion f defined by f (x) = x2 − x + 2 for values of x near 2. The following table gives values of f (x) for values of x close to 2 but not equal to 2. x f (x) x f (x) 1.0 2.000000 3.0 8.000000 1.5 2.750000 2.5 5.750000 1.8 3.440000 2.2 4.640000 1.9 3.710000 2.1 4.310000 1.95 3.852500 2.05 4.152500 1.99 3.970100 2.01 4.030100 1.995 3.985025 2.005 4.015025 1.999 3.997001 2.001 4.003010 From the table above, we see that when x is close to 2 (on either side of 2), f (x) is close to 4. In fact, it appears that we can make the value of f (x) as close as we like to 4 by taking x sufficiently close to 2. We express this by saying ”the limit of the function f (x) = x2 −x+2 as x approaches 2 is equal to 4”. The notation for this is lim (x2 − x + 2) = 4 x→2 1 Definition. We write lim f (x) = L x→a and say ”the limit of f (x), as x approaches a, equals L” if we can make the values of f (x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a but not equal to a. This says that the value of f (x) get closer and closer to the number L as x gets closer and closer to the number a (from either side of a), but x 6= a. Ex1) limx→1 x−1 =? x2 − 1 undefined at x = 1, but there is limit, because the definition of limx→a f (x) says that we consider values of x that are close to a but not equal a. x<1 f (x) x>1 f (x) 0.5 0.666667 1.5 0.400000 0.9 0.526316 1.1 0.476190 0.99 0.502513 1.01 0.497512 0.999 0.500250 1.001 0.499750 0.9999 0.500025 1.0001 0.499975 From the table and figure, we make the guess that the equation approaches 0.5 as x approaches 1. 2 Ex2) Let f (x) = 2 x x=1 , find limx→1 f (x). otherwise Note, this function is not continuous. 3 Definition. A function f is continuous at a number a if lim f (x) = f (a) x→a or lim f (x) = lim f (x) = f (a) x→a+ Ex3) Let f (x) = 2−x x 4 4−x x→a− if x < −1 if − 1 ≤ x < 1 , if x = 1 if x > 1 sketch the graph of f (x) and find all values of x for which the limit does not exist. In sum, lim f (x) = L if and only if x→a lim f (x) = L and x→a− Infinite Limits and Vertical asymtotes. Let’s start from f (x) = 1 . x 4 lim f (x) = L x→a+ Ex4) f (x) = 1 , find limx→0 f (x). x2 Ex5) Find the limits; 5-1) limx→5− 6 x−5 5-2) limx→5+ 6 x−5 5-3) limx→5 6 x−5 5-4) limx→0− 5-5) limx→0+ 5-6) limx→0 x−1 + 2) x2 (x x−1 + 2) x2 (x x−1 + 2) x2 (x 5 x+1 . For each vertical − 2x − 3 asymtote, describe the behavior of f (x) near the asymtote. Ex6) Find the vertical asymtotes for f (x) = 6 x2 Section 2.3. laws Calculating limits using the limit Limit Laws Suppose that c is a constant and the limits lim f (x) and lim g(x) x→a x→a exist. Then 1. lim [f (x) + g(x)] = lim f (x) + lim g(x) x→a x→a x→a 2. lim [f (x) − g(x)] = lim f (x) − lim g(x) x→a x→a x→a 3. lim [cf (x)] = c lim f (x) x→a x→a 4. lim [f (x) · g(x)] = lim f (x) · lim g(x) x→a 5. x→a x→a limx→a f (x) f (x) = if lim g(x) 6= 0 x→a x→a g(x) limx→a g(x) lim 6. lim [f (x)]n = [lim f (x)]n x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim xn = an x→a 10. lim x→a √ n 7 x= √ n a 11. lim x→a p n f (x) = p n f (a) where n is a positive integer. (If n is even, we assume that limx→a f (x) > 0). Ex7) Find limx→−2 x3 + 2x2 − 1 . 5 − 3x Ex8) Find limx→−2(x2 + x + 1)5 Ex9) Find limx→4− √ Ex10) Find limx→4+ 16 − x2 √ 16 − x2 8 Ex11) Find limx→1 x4 + x2 − 6 x4 + 2x + 3 Ex12) Find limx→−3 Ex13) Find limx→9 x2 − x − 12 x+3 x2 − 81 √ x−3 Ex14) Find limx→3 √ x−3 x2 + 7 − 4 9 Ex15) Find limx→2 x−4 (x − 2)2 Ex16) If it ispknown that limx→2 f (x) = 3, find f (x) + 6 16-1) limx→2 f (x) + 2x + 11 16-2) limx→2 (f (x))2 Ex17) limt→1 1−t , 2t t2 − 1 10 Ex18) Show that limx→0 |x| = 0 Ex19) Prove that limx→0 |x| does not exist. x 11 Ex20) Let f (x) = x2 + 3x , find limx→−3 f (x). |x + 3| The Squeeze Theorem If f (x) ≤ g(x) ≤ h(x) for all x in open interval that contains a (except possibly at a) and lim f (x) = lim h(x) = L x→a x→a then lim g(x) = L x→a 12 Ex21) Show that limx→0 x sin 1 = 0. x 13 Section 2.5 Continuity Definition A function f is continuous at a number a if lim f (x) = f (a) x→a Note Definition implicitly requires three things if f is continuous at a; 1. f (a) is defined (that is, a is in the domain of f and has a function value at a) 2. limx→a f (x) exists, i.e., limx→a+ f (x) = limx→a− f (x). 3. limx→a f (x) = f (a) Polynomial is always continuous Ex22) Show that f (x) = x2 − 3x + 2 is continuous at any point. 14 In each case the graph cannot be drawn without lifting the pen from the paper, because a hole or break or jump occurs in the graph. Definition The kind of discontinuity illustrated first three is called removable becuase we could remove the discontinuity by redefining f at 2. The discontinuity in second figure is called an infinite discontinuity. The discontinuityies in the last figure are called jump discontinuities because the function ”jumps” from one value to another. Definition We say f (x) is continuous from the right at x = a if limx→a+ f (x) = f (a). similarly, we say f (x) is continuous from the left at x = a if limx→a− f (x) = f (a). 15 Ex23) From the accompaning figure, state the numbers at which f is discontinuous. For each of the numbers stated, state whether f is continuous from the right, or from the left, or neither. Ex24) Explain why the following functions are not continuous at the indicated values of x. 24-1) f (x) = −1 , x=1 (1 − x)2 16 24-2) f (x) = 2x + 1 if x ≤ 0 3x , x=0 if x > 0 x2 − 2x − 8 if x 6= 4 x−4 , x=4 24-3) f (x) = 3 if x = 4 17 Ex25) Find all points of discontinuity for 2x + 1 if 3x if f (x) = x2 + 2 if 4 if 18 x < −1 −1<x<1 x>1 x=1 Ex26) If g(x) = continuous? x2 − c2 if x < 4 cx + 20 if x ≥ 4 19 . For what value(s) of c make(s) g(x) x2 + c if x > 1 Ex27) If f (x) = 4 if x = 1 . For what value(s) of c is f (x) continuous, 4cx + 3 if x < 1 if any? 20 Ex28) Which of the following functions has removable discontinuity at x = a? If the discontinuity is removable, find a function g that agrees with f for x 6= a and is continuous at x = a. 28-1) f (x) = x4 − 16 , x = 2. x−2 28-2) f (x) = 1 , x = 1. x−1 28-3) f (x) = x2 if x < 1 , x=1 2x + 4 if x ≥ 1 21 Intermediate Value Theorem Suppose that f is continuous on the closed interval [a, b] and let N be any number strictly between f (a) and f (b). Then there exists a number c in (a, b) such that f (c) = N. 2. Ex29) Show there is a root of the equation 4x3 − 6x2 + 3x − 2 = 0 between 1 and Ex30) If g(x) = x5 − 2x3 + x2 + 2, show there a number c so that g(c) = −1. 22