1.3 Limits and Continuity

Transcription

1 .3 Limits and Continuity.3. Limits Problem 8. What will happen to the functional values of as x gets closer and closer to 2? f(x) = Solution. We can evaluate f(x) using x values nearer and nearer to 2 (but not equal to 2). x f(x) = The table above shows x approaching 2 from the left through values less than 2. The it operation repeats this process an infinite number of times by imputing numbers nearer and nearer to 2, but never 2 itself. Now let x approach 2 from the right side through values greater than 2. x f(x) = In this example, both sets of outputs from f(x) move toward the same number: 4, asx approaches 2 from both sides. Hence, the it of the function f(x) = as x approaches 2, is 4. Mathematically, we write x2 =4. Remark. For f(x) in the above example, f(2) = 4 and so, f(x) =f(2). Sotheitof the function, in this case, is equal to the plug-in value. Is this always true? No, as we will see in the next problem. Problem 9. What will happen to the functional values of as x gets closer and closer to 3? f(x) = x2 9 x +3 38

2 Solution. We evaluate f(x) using x values nearer and nearer to 3 (but not equal to 3). x f(x) = The table above shows x approaching 3 from the left through values less than 3. Now let x approach 3 from the right side through values greater than 3. x f(x) = In this example, both sets of outputs from f(x) move toward the same number: 6, asx approaches 3 from both sides. Hence, the it of the function f(x) = x2 9 as x approaches x+3 3, is 6. Mathematically, we write 9 x+3 9 x+3 9 x! 3 x +3 = 6. Remark. For f(x) in the above example, f(x) 6= f( 3). Why? So the it of the x! 3 function, in this case, is not equal to the plug-in value. Remark. Limits can fail to exist for many reasons. Three common reasons are: f(x) approaches a different number from the right side of a than it approaches from the left side of a. f(x) increases, or decreases, without bound as x approaches a. f(x) oscillates between two fixed values as x approaches a. 39

3 .3.2 One-Sided Limits Definition.3.. One-sided its are its that have inputs which approach the specified number from one side as opposed to both sides like regular its. Limit from the Right: f(x) =L. This is read The it of f(x) as x approaches + c from the right equals L. Limit from the Left: c from the left equals L. f(x) =L. This is read The it of f(x) as x approaches The next theorem shows that for the it of a function to exist, both the it from the right and the it from the left must exist and produce the SAME value. Theorem.3.2. Let f be a function and let c and L be real numbers. Then f(x) =L if and only if f(x) = f(x) =L Limits at Infinity; Horizontal Asymptotes Limits at infinity have the general form: f(x) and x! the end behavior of a function on an infinite interval. Problem 0. Evaluate x! Solution. As x gets very large,, and x!. also gets very close to 0. Thus, x! x gets very large in magnitude, x! gets very close to 0. Wesay gets very close to 0. f(x). This section discusses x! =0.Notethat =0.Asx approaches never equals 0. As Definition.3.3 (Horizontal Asymptote). The line y = L is a horizontal asymptote of the graph of f if f(x) =L or f(x) =L. x! x! Figure 2: The lines y = 3 2 and y = 3 2 are horizontal asymptotes for f in above graph Remark. The graph of a function f(x) can have at most two horizontal asymptotes. 40

4 .3.4 Infinite Limits; Vertical Asymptotes Infinite its can occur when evaluating f(x) if f(x) increases, or decreases, without bound as x approaches c. Definition.3.4. For any function f, we write f(x) =+ to mean that as x gets closer and closer to a (from both sides of a), then f(x) becomes arbitrarily large in the positive sense. Definition.3.5. For any function f, we write f(x) = to mean that as x gets closer and closer to a (from both sides of a), then f(x) becomes arbitrarily large in the negative sense. Problem. Find. x!0 Solution. We evaluate x!0 + (a) (b) x!0 + x!0 Thus, =+. x!0 and x!0 as follows: =+, the function increases without bound as x approaches 0 from the right. =+, the function increases without bound as x approaches 0 from the left. Definition.3.6 (Vertical Asymptote). If f(x) approaches positive or negative infinity as x approaches c from the right or the left, then the line x = c is a vertical asymptote of the graph of f. Figure 22: The lines x = 6 and x =5are vertical asymptotes for f in above graph Problem 2. Find the vertical asymptotes of f(x) = 2 x 3. 2 Solution. Since =+, the function increases without bound as x approaches 3 x!3 + x 3 from the right. Thus, x =3is a vertical asymptote. 4

5 .3.5 Limit Theorems Theorem.3.7 (Limit Laws). Let f(x) and g(x) be functions. Then we have the following:. c = c 2. x = a 3. [c f(x)] = c [ f(x)] 4. [f(x)+g(x)] = f(x)+ g(x) 5. [f(x) g(x)] = f(x) g(x) 6. [f(x) g(x)] = [ f(x)] [ g(x)] f(x) 7. = f(x) g(x) 8. [f(x) n ]= 9. n p f(x) = n g(x), when g(x) 6= 0 h i n f(x) whenever n. q f(x), where f(x) 0 if n 2 is even. Problem 3. Compute x!3 (x 3 +). Solution. (x 3 +)=x 3 += =9. x!3 x!3 x!3 x!3 Problem 4. Compute 5x 3 +4 x 3 Solution. Since (x 3) 6= 0, using (7) of Theorem.3.7, we obtain. 5x 3 +4 = x 3 x Problem 5. Compute 2 4. (5x3 +4) (x 3) = 5(2) = 44. Solution. Note that () = 0 and so we cannot use (7) of Theorem.3.7. However, (x2 4) = 0 and so we obtain 0 (an indeterminate form ). Thus, we can try to factor 0 the top, cancel, and then plug in. Note that x2 4 x 2 4 =(x +2)=2+2=4. = ()(x+2) = x +2for x 6= 2. Therefore, 42

6 Definition.3.8. Afunctionf is called a polynomial function if it has the form f(x) =a n x n + a n x n + a x + a 0, where n is a natural number. The constants a n,a n,...,a 0 are called the coefficients of the polynomial function. Definition.3.9. Afunctionf is called a quadratic function if it has the form f(x) =a 2 + a x + a 0. Definition.3.0. Afunctionf is called a cubic function if it has the form f(x) =a 3 x 3 + a 2 + a x + a 0. Definition.3.. Afunctionf is called a rational function if it has the form f(x) = p(x) q(x) where p(x) and q(x) are polynomial functions. Theorem.3.2 (Limits of Polynomial and Rational Functions). If p is a polynomial function and c is a real number, then p(x) =p(c). If r is a rational function, say r(x) = p(x),andc is a real number such that q(c) 6= 0, then q(x) r(x) =r(c) =p(c) q(c). Theorem.3.3. Let c be a constant and r be a positive real number. Then c. =0 x! x r 2. x! c =0. x r p(x) To evaluate, where p(x) and q(x) are polynomials, divide each term in the x!± q(x) numerator and denominator by the highest power of x that appears in the denominator q(x). Problem 6. Evaluate Solution. We evaluate 5x x!+ 3 x x!+ 5 +7, using Theorems.3.3 and.3.7, as follows: 3 x 5 +7 x!+ 3 x = 5+ 7 x!+ 3 x = 5+ x!+ 3 x!+ 7 x = =

7 .3.6 Finding Limits Algebraically Using Theorem.3.7, we can often (but not always) evaluate f(x) algebraically (when f(x) is defined by a single formula) as follows: Substitute the specified number a into the function f(x) and evaluate the value f(a):. If you obtain a number, then this number is the it of the function. 2. If you get a none-zero constant divided by zero, the function has no it for the specified number, that is, f(x) does not exist. 3. If you get 0 0 (an indeterminate form), try one of the following: (a) factor the top and/or bottom, cancel, plug in and repeat the process. (b) rationalize the numerator, cancel, plug in and repeat the process. (c) rationalize the denominator, cancel, plug in and repeat the process. 4. If the above methods fail, you may have to use the numeric method. Problem 7. Evaluate the following its:. (7 x). 2. 5x 3 +4 x x px 4 x!4. p x!0 p x!0 Problem 8. Evaluate x!3 f(x), where f(x) = 2 8 x 3 if x 6= 3 2 if x =3 Remark. The it of a function at a point does not depend on the value of the function at the point. Theorem.3.4. Suppose that h(x) apple g(x) for all x in an open interval containing c, except possibly c itself, and suppose that h(x) and g(x) both exist. Then h(x) apple g(x). 44

8 .3.7 The Pinching Theorem Theorem.3.5 (The Pinching Theorem). Suppose that h(x) apple f(x) apple g(x) for all x in an open interval containing c, exceptpossiblyc itself, and if then f(x) exists and is equal to L. h(x) =L = g(x) Problem 9. Suppose that we have shown that 6 apple sin(x) x apple for all x near 0. What can we conclude from Theorem.3.5? x Solution. Since ( 2 )==,weconcludethat sin x =. x!0 6 x!0 x!0 x.3.8 A Precise Definition of the Limit Definition.3.6 (, Definition of Limit). Let f : I! R be a function and let c be an point in the open interval I. We shall say that L is the it of f at c, denotedby f(x) =L, to mean that the following holds: For every ">0 there exists a >0 such that for all I, if0 < x c <,then f(x) L <"..3.9 Continuity Definition.3.7. Afunctionf is said to be continuous at a point c if f(x) =f(c). Remark. Afunctionf is continuous at a point c if the following three conditions hold:. f(c) is defined. 2. f(x) exists. 3. f(x) =f(c). If one or more of the above conditions fail to hold, then f is said to be discontinuous at c and c is called a point of discontinuity of f. Problem 20. Find the points of discontinuity of the following functions:. f(x) = x2 2. f(x) =. 3. h(x) = x2 4 5x+4. (, if x 6= 2; if x =2. 45

9 Problem 2. Let h(x) = kx 2 9 if x + k if x<2 Determine k so that h is continuous at the point 2. Solution. By Definition.3.7, we need to find k so that h(x) =h(2). By Theorem.3.2, we need k so that h(x) = h(x) =h(2). + Note that h(x) =k 22 9 = h(2), and + h(x) =2+k. To make h continuous at 2, wemusthavek = 2 + k. Solving for k, wegetk =7. Definition.3.8. Afunctionissaidtobecontinuous on an open interval (a, b) if f is continuous at all points c in the interval (a, b). Remark. Afunctioniscontinuouson(a, b) if there are no breaks or jumps in its graph. If you can trace the graph without lifting your pen from the page, it is continuous. Definition.3.9. Let f be a function.. f is said to be continuous from the right at c if f(x) =f(c) f is said to be continuous from the left at c if f(x) =f(c). Definition Afunctionissaidtobecontinuous on an closed interval [a, b] if. f is continuous at all points c in the interval (a, b); 2. f(x) =f(a); + 3. x!b f(x) =f(b). Theorem.3.2. If f(x) and g(x) are continuous functions at the point c, then. f + g is continuous at c; 2. f g is continuous at c; 3. kf is continuous at c, where k is any constant; 4. f g is continuous at c; 5. f g is continuous at c, ifg(c) 6= 0. Theorem The following functions are continuous (on their domains); that is:. Polynomial functions are continuous everywhere, that is, on (, +). 2. Rational, radical, and trig. functions are continuous at every point in their domain. 46

10 Theorem If g is continuous at c, and f is continuous at the point g(c), then (f g) =f(g(x)) is continuous at c. The above theorem shows that the composition of continuous functions is a continuous function. Theorem.3.24 (Intermediate Value Theorem). If f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), thenthereisatleastonenumberc in (a, b) such that f(c) =k. Definition Afunctionf : I! R is said to be one-to-one (or an injection), if distinct elements in I get mapped to distinct elements in R; morespecifically or equivalently, for all x 2 I and 2 I, ifx 6=,thenf(x ) 6= f( ) for all x 2 I and 2 I, iff(x )=f( ),thenx =. Definition Afunctionf has an inverse function, denoted by f,if. f(f (x)) = x for all x in the domain of f 2. f (f(x)) = x for all x in the domain of f. The domain of f is the range of f. Anotherwaytosaythatf is the inverse of f is when for all real numbers a and b, f (a) =b if and only if f(b) =a. (N) When dealing with the name f of a function, the name f is reserved to identify the inverse function of f. However, one may write (f(x)) for.ingeneral,f (x) 6=. f(x) f(x).3.0 Finding a formula for the inverse of a one-to-one function Remark Given a function f(x) which is one-to-one. To find a formula for f (y) try the following steps:. Write the equation y = f(x). 2. Solve the equation y = f(x) for the unique x, obtainingx = g(y). 3. In g(y) replace the variable y with x. 4. Then f (x) =g(x). Problem 22. One can prove that the function f : R! R defined by f(x) =(x one-to-one and onto. Find a formula for the inverse function f. 4) 3 +2is Theorem Let f : I! J where I is an interval and the range of f is J. If f is continuous and is one-to-one, then f : J! I is also continuous. 47