ntroduction o Syllabus 9.1, 9.2, 9.3. Vera Babe!,ku Math 11-2. Lecture 1
9.1 Limits. Application Preview Although everyone recognizes the value of eliminating any and all particulate pollution from smokestack emissions of factories, company owners are concerned about the cost of removing this pollution. Suppose that USA Steel has shown that the cost C of removing p percent of the particulate pollution from the emissions at one of its plants is C = C(p) 73 p = 1 p To investigate the cost of removing as much of the pollution as possible, we can evaluate the limit as p (the percent) approaches 1 from values less than 1. Using a limit is important in this case, because this function is undefined at p = 1 (it is impossible to remove 1% of the pollution). Verd Habenka Math 11-2. Lecture
o n various applications we have seen the importance of the slope of a line as a rate of change. n particular, the slope of a linear total cost, total revenue, or profit function for a product tells us the marginals or rates of change of these functions. When these functions are not linear, how do we define marginals [and slope]? o We can get an idea about how to extend the concept of slope (and rate of change) to functions that are not linear. Observe that for many curves, if we take a very close (or zoom-in) view near a point, the curve appears straight. We can think of the slope of the straight line as the slope of the curve. The mathematical process used to obtain this zoom-in view is the process of taking limits. Vera Babenko Math 11-2. Lecture 1
x. Concept of a Limit We have used the notation 1(c) to indicate the value of a function 1(x) at x c. f we need to discuss a value that 1(x) approaches as x approaches c, we use the idea of a limit. For example, if 1(x) = 6 x+2 then we know that x = 2 is not in the domain of 1(x), so f( 2) does not exist even though f(x) exists for every value of x 2. r Ve,a Babenko Math 1GU-2. Lecture
f we draw the graph of this function, the figure should show the graph of y = f(x) with an open circle where x = 2. The open circle indicates that f( 2) does not exist but shows that points near x 2 have functional values that lie on the line on either side of the open circle. Even though f( 2) is not defined, the figure shows that x approaches 2 from either side of 2, the graph approaches the open circle at ( 2, 5) and the values of f(x) approach 5. Thus 5 is the limit of f(x) as x approaches 2, write lim f(x) 5, or f(x) * 5 asx * 2 Vera Bbenko Math 11-2. Lecture 1 Lt2 - :1 t - t vq
LMT. Definition Let f(x) be a function defined on an open interval containing c, except perhaps at c. Then urn f(x) = L x +c is read the limit of f(x) as x approaches c equals L. The number L exists if we can make values of f(x) as close to L as we desire by choosing values of x sufficiently close to c. When the values of f(x) do not approach a single finite value L as x approaches c, we say the limit does not exist. Vera Babenko Math 11-2. ecture
As the definition states, a limit as x * c can exist only if the function approaches a single finite value as x approaches c from both the left and right of c. Vera Babenko Math 11-2. Lecture t.
Example Limits Figure shows three functions For which the limit exists as x approaches 2. Use this figure to find the following. o urn f(x) and f(2) (if it exists) x 2 o urn g(x) and g(2) (if it exists) x-+2 o urn h(x) and h(2) (if it exists) Ver Bbnko Math 11-2. Lecture 1
Solution O From the graph, we see that as x approaches 2 from both the left and the right, the graph approaches the point (2, 3). Thus f(x) approaches the single value 3. That is lim f(x) 3 The value of f(2) is the y-coordinate of the point on the graph at x = 2. Thus f(2) = 3. o Figure shows that as x approaches 2 from both the left and the right, the graph approaches the open circle at (2, 1). Thus lim g(x) = 1 x *2 The figure also shows that at x= 2 there is no point on the graph. Thus g(2) is undefined. Vera Babenko Math 11-2. Lecture 1. u1.
G Last figure shows that urn h(x) = 1 The figure also shows that at x 2 there is a point on the graph at (2,4). Thus h(2) = 4, and we see that lirn h(x) h(2). x 42 Vera Babenko Math 11-2. Lecture
As Example shows the limit of the function as x approaches c may or may not he the same as the value of the function at x c. Also we saw that the limit as x approaches 2 meant the limit as x approaches 2 from both the left and the right. We can also consider limits only from the left or only from the right; these are called one-sided limits. Vera Babeniw Math 11-2. Lecture
One-Sided Limits. Definition. Limit from the Right: urn f(x) L x +c+ means the values of f(x) approach the value L as x * c but x> c. Limit from the Left: urn f(x) = M x *c means the values of f(x) approach the value M as x + c but x < C. Vera Babenko Math 11-2. Lecture
Note that when one or both one-sided limits fail to exist, then the limit does not exist. Also, when the one-sided limits differ, such as if L M above, then the values of f(x) do not approach a single value as x approaches c, and lim f(x) does not exist. x *c Vera Babenico Math 11-2. Lecture
2 Example Using the functions graphed in Figure, determine why the limit as does not exist for x o f(x). o g(x). O h(x). L, Vera Babenko Math 11-2. Lecture
2 Solution 1) As x from the left side and the right side of x = 2, f(x) increases without bound, which we denote by saying that f(x) approaches +oo as x * 2. n this case, lim f(x) does not exist x *2 [denoted by lim f(x) DNE] because f(x) does not approach a x *2 finite value as x * 2. n this case, we write - f(x) +oo as x * 2 The graph has a vertical asymptote at x = 2. jj: f\, ;t Vera Babenku Math 11-2. Lecture
2) As x 4 2 from the left, g(x) approaches cx, and as x * 2 from the right, g(x) approaches +oo, so g(x) does not approach a finite value as x + 2. Therefore, the limit does not exist. The graph of y = g(x) has a vertical asymptote at x = 2. n this case, we summarize by writing lim g(x) DNE or g(x) * oc as x * 2 x 2 urn g(x) DNE or g(x) * +oc as x * 2 x*2± Tim g(x) DNE x *2 Vera Babenko Math 11-2. Lecture 1
3) As x * 2 from the left, the graph approaches the point at (2,), so lim h(x) =. As x * 2 from the right, the graph approaches the open circle at (2,3), so lim h(x) = 3. Because these one-sided limits differ, urn h(x) does not exist. x *2 V Babenka Math 11-2. Lecture 1
The Limit. 1) The limit is said to exist only if the following conditions are satisfied: o The limit L is a finite value (real number) o The limit as x approaches c from the left equals the limit as x approaches c from the right. That is lim f(x) = lim f(x) x_c x_*c+ 2) The limit of a function as x approaches c is independent of the value of the function at c. When lim 1(x) exists, the value of the x c function at c may be () the same as the limit, (ii) undefined, or (iii) defined but different from the limit. Vera Babepka Math 11-2. Lecture
Properties of Limits f k is a constant, urn f(x) = L, and urn g(x) = M, then the x +c x >c following are true. O limk=k x * C o limx=c X *C o m[f(x)±g(x)]=l±m o im[f(x).g(x)] = LM OHm=h, if MZO o is even ji= ç/ii f(x) =.Yi, provided that L> when ii Ve.a Babenko Math 11-2. Lecture
Polynomial function The function f(x) = ax 1 + a_ix + +a1x + 2, where and n is a positive integer, is called a polynomial function of degree n. urn f(x) = f(c) x +c for all values c (by properties 1-4) Vera Babenico Math 11-2. Lecture 1
Rational function The function h(x) = where both f(x) and g(x) are polynomial functions, is called a rational function. when g(c) (by property 5) f(x) f(c) km h(x) = urn = x c x *c g(x) g(c) Vera Bbenkn Math 11-2. Lecture 1
Example Find the foltowing limits, if they are exist. 1) urn (x3 2x) 2) urn X2_4X Veta Babenko Math 11-2. Lecture
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