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3.1 LIMITS OF FUNCTIONS PURPOSE OF A LIMIT: Differential Calculus is the study of rates of change of a function y = f with respect to an independent variable (usually x or t). We are usually interested in how much y changes whenever x changes by 1 unit. It is easy to quantify average rates of change since there are 2 points involved in calculating the average. The following formula can be used to find the distance covered by an object in motion over time interval t: d = rt where d is the distance traveled in some time period r is the rate that the distance changed during a time period t is the number of time periods If we wanted to know the average rate of change for some time period and we knew the distance covered and the length of the time period, we could use the formula: d . where r is the average rate of change per time period r= t What if we wanted to know how fast an object was moving at a point in time? We know it is in motion since we can clearly see it moving and can easily calculate any average rate of speed that we are interested in. The above formula as it is however, cannot give us an instantaneous rate of speed since there will be no time period and no distance traveled at a point in time. In other words, t = 0 and d=0. To avoid the problem of division by 0, we will purposely refrain from discussing that t might equal 0. Instead, we will only consider values of r when t is very close to 0 but never equal to 0. In other words, we will calculate the rate of speed over very short time periods on the left side of t = 0 and the rate of speed over very short time periods on the right side of t = 0. If r appears to be approaching the same value on the left side as it is on the right side, then it is probably safe to assume that that value is the instantaneous rate of change that we are looking for. This process of finding a value that a function is approaching as its independent variable approaches a particular value is called finding the limit of the function at that value. While calculating the short time periods mentioned above can be helpful in getting an idea of the value of a particular limit, mathematical methods are available to yield the exact value, if it exists. 1 INFORMAL DEFINITION OF A LIMIT: Let f be a function and let a and L be real numbers: If as x takes values closer and closer (but not equal) to a on both sides of a, the corresponding value of f(x) get closer and closer (and perhaps equal) to L, and the value of f(x) can be made as close to L as desired by taking values of x close enough to a; then lim f(x) = L The limit of f(x) as x approaches a is L x→a LEFT AND RIGHT HAND LIMITS: lim f(x) = L means that f(x) approaches real number L x→a as x→a from the LEFT of a. (x < a) lim f(x) = L means that f(x) approaches real number L x→a+ as x→a from the RIGHT of a. (x > a) THINGS TO REMEMBER ABOUT LIMITS: For lim f(x) = L to exist, then the left and right hand limits must exist and be equal. x→a In other words lim f(x) = lim f(x) + x→a x→a The value f(a) does not have to exist for the limit to exist at x→a . (The function may have a hole or be undefined at x = a ) Limits Do Not Exist at Vertical Asymptotes of Rational Functions: Limit DNE Limit DNE Limit DNE Limit DNE lim f(x) = ∞ lim f(x) = ∞ x→a x→a Indeterminate forms: . 0 . ∞ and + . 0 ∞ If a rational function results in an indeterminate form when analyzing a limit at x→a , and there are no common factors between the numerator and denominator that will cancel, then no conclusion may be made about the value or existence of the limit without additional investigation. (see L’Hopitals Rule). This is true of both double and single sided limits. 2 GRAPHIC INTERPRETATION OF LIMITS: ● ● y=2 y=1 ● ● x=0 x=1 x=2 x=3 x=4 x=5 x Function Limit Value Value 0 f(0) = 0 lim f(x) = DNE Since f(x) for x < 0 is undefined. x→0 1 f(1) = 1 lim f(x) = 1 Since left and right hand limits →1. x→1 2 f(2) = lim f(x) = 1 Since left and right hand limits →1. undefined x→2 Doesn’t matter that f(x) is undefined. 3 f(3) = 2 lim f(x) = 1 Since left and right hand limits →1. x→3 Doesn’t matter that f(3) = 2. 4 f(4) = 2 lim f(x) = DNE Since left and right hand limits x→4 are not equal. 5 f(5) = lim f(x) = DNE Vertical Asymptote undefined x→5 3 PROPERTIES OF LIMITS: Let f and g be two functions where lim f(x) = L and lim g(x) = M x→a x→a where L, M, and a are real numbers: 1 lim k f(x) = k lim f(x) = k L for any constant k x→a x→a 2 lim [f(x) + g(x)] = lim f(x) + lim g(x) = L + M x→a x→a x→a lim [f(x) • g(x)] = lim f(x) • lim g(x) = LM 3 x→a x→a x→a lim f(x) 4 lim f(x) = x→a = L . / if M = 0 x→a g(x) lim g(x) M x→a 5 If p(x) is a polynomial, then lim p(x) = p(a) x→a 6 lim [f(x)]k = [lim f(x)] k = L k x→a x→a 7 lim f(x) = lim g(x) if f(x) = g(x) for all x a x→a x→a 8 For any real number b > 0 f(x) L lim b = b lim f(x) = b x→a x→a 9 For any real number b such that 0 < b < 1 or b>1 lim [logb f(x)] = logb [lim f(x)] = logb (L) if L>0 x→a x→a 4 FINDING LIMITS: Finding Limits of Polynomial Functions f(x) as x→a : lim (4x2 2x + 3) = 4(2)2 – 2(2) + 3 = 15 ← Substitute x = 2 x→2 Finding Limits of Radical Functions f(x) as x→a : Caution: Limits of lim ( √ 4x 11 ) = √ 4(9) 11 = 5 ← Substitute x = 9 even radicals do not x→9 exist when the expression is < 0. Finding Limits of Rational Functions R(x) as x→a: If the Denominator is not equal to 0 : x2 4 (1)2 4 lim ( x 2 ) = x 2 =3 ← Substitute x = 1 x→1 If the Denominator = 0 and the Numerator also equals 0 : If there are no Factor and cancel any common factors. common factors, then the function is lim ( x 4 ) = (x+ 2)(x2) = x + 2 = (2) + 2 = 4 2 an indeterminate x→2 x 2 x2 form. If the Denominator = 0 and the Numerator is a binomial containing a radical : Rationalize the numerator. √x 2 √x 2 √x + 2 x - 4 x4 lim ( √x 2) = . 1 .1. 1 ( x-4)(√x 4)(√x + 2) =+2 = (x +2) = = 1/4 . . xx44 x 4 √x + 2 √x √x + 2 √4 + 2 x→4 If the Denominator = 0 and the Numerator does not equal 0 : x2 + 4 lim ( x 2 ) = Does Not Exist ← Vertical Asymptote x→2 Finding Limits of Rational Functions R(x) as x→ ∞ : Divide top and bottom by the highest power of x in the Denominator. 4x2-6x+3 4 – 6 + 3 4x26x+3 . x2 x x2 Lim k p = 0 ( )= 1 . Lim 2x2x+4 . = . = 2 x→o x 2 2x -x+4 2 – 1 + 4 o x→∞ . x2 1 x x2 . 5

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