MTH Calculus I Week 8: Limits at Infinity and Curve Sketching

Transcription

1 MTH 35 - Calculus I Week 8: Limits at Infinity and Curve Sketching Adam Gilbert Northeastern University January 2, 24 Objectives. After reviewing these notes the successful student will be prepared to Determine finite and infinite its at infinity Determine the horizontal asymptotes (if any) of the graph of a function Analyze the graph of a function Combining all of the concepts we have discussed in calculus, sketch an accurate graph of a function

2 3.5 Limits at Infinity In this section we revisit our old friend the it. In all of our previous eperiences, we were taking its as our independent variable approached some finite value. Since the it eplores what happens as we get really really close, but never really get to..., it seems as though the it process is built for eploring infinity. Similar to when we introduced the notion of it, we will have some very technical definitions, however, the idea itself is not etremely sophisticated. Let s introduce the definition of a it at infinity and then break it down into something more tangible. Definition (Limit at Infinity). Let f() be a function defined on R and let L be a real number. The statement f() = L means that for every ε > there eists an N > so that f() L < ε as long as > N The statement f() = L means that for every ε > there eists an N < so that f() L < ε as long as < N Remark. If we lose all of the technical jargon and symbols, here is what the definition is saying: We say that f() = L if as gets really really large (approaching infinity), the corresponding function values are approaching the number L We say that f() = L if as becomes more and more negative (approaching negative infinity), the corresponding function values are approaching the number L Eample. Find. Solution. As, the denominator of is getting larger and larger. If the denominator of a rational epression increases without bound, while the numerator remains constant, then the fraction itself becomes smaller and smaller. That is, the fraction is approaching. Thus, = 2

3 Theorem. If r > and c is any real number, then c r = and, as long as r is such that r is defined for < c r = Eample 2. Evaluate 5. 2 Solution. 5 = 2 = 5 = p() Note. In order to evaluate (where p() and q() are polynomials), we need a q() strategy for making each term look similar to c, as in Theorem. r Strategy (Evaluating Rational Limits at Infinity). In order to evaluate the following steps: p() ± q(), use. Identify the highest power of in the entire rational epression (numerator and denominator). 2. Divide every term in the numerator and denominator by this power of 3. Simplify 4. Evaluate the it using Theorem Eample 3. Evaluate the following its: a) b) c)

4 Solution. a) We first identify that 3 is the highest power of present in the entire epression: So we divide every term by 3 = = Now we use Theorem to evaluate the it = = 2 = b) We first identify that 4 is the highest power of present in the entire epression: So we divide every term by = = Now we use Theorem to evaluate the it =

5 = 3 6 = 2 c) Again, we identify the largest power of, which is 3 in our eample: And we proceed by dividing each term by 3 Then we use Theorem to evaluate = = = = 8 In this case we must be a little bit careful. We know the result will be either or, but how do we choose? We know that the numerator is approaching 8, which is no big deal - we know that all numbers near 8 are positive. The denominator approaching is the problem. Does the denominator approach from below (through the negative numbers) or from above (through the positive numbers)? Notice that for large values of, we have that > 4, since the denominator belonging 3 to the fraction on the right is so much larger than the one on the left. > 4 3 = 4 3 > So we know that our denominator is approaching through the positive numbers. That is, as approaches, the numerator approaches 8, and the denominator approaches, with both numerator and denominator positive. Since a positive number divided 5

6 by a positive number yields a positive number, the it itself is positive. Thus, our choice is made and the it is +. = = Warning. As you can see, the case in which the largest power of sits in the denominator requires some etra care. Be cautious in this case to determine whether the it is or through a similar analysis to the one provided in the eample above. Note 2. Strategy can be applied to its of the form polynomial-like functions. See the eample below. 9 Eample 4. Evaluate ± p(), where p() and q() are q() Solution. Notice that the denominator of the epression within the it is not quite a polynomial. It is, however, sort of close. With some attention to detail we can apply our strategy. The first instance where etra attention is necessary is when identifying the largest power of. The highest power of here is, but why? At initial glance the highest power of seems to be However, take a closer look at the environment in which 4 is living! Now, ignoring the +6 we can see that the denominator will act like 4 = as tends toward. Thus, the largest power of in the denominator is, which is the same as the largest power of in the numerator. Thus, we consider the largest power of to be. We will now divide every term in the numerator and denominator by, but again, this poses a problem because the denominator does not have terms. The square root function is in the way! 9 =

7 We ll use a small trick here. Let s disguise in such a way that he will be allowed into the square root. 9 = Notice that, since 4 =, we haven t changed a thing! = = Using Strategy in conjunction with the fact that the square root function is continuous, so we can take its of the inside pieces, we see = 4 9 = = Note 3. Limits at infinity are used to define and to find horizontal asymptotes for functions. Definition 2 (Horizontal Asymptote). The line y = L is a horizontal asymptote for the function f() if at least one of the following statements is true:. f() = L 2. f() = L Eample 5. Show that f() = 8 2 has a horizontal asymptote at y =. Solution

8 We use our strategy for evaluating rational its, and identify as the highest power of present in our epression. Thus, we divide each term in the numerator and denominator of the epression inside of our it by, simplify, and evaluate. = = = = Thus y = is a horizontal asymptote for f() There is another strategy for evaluating some of the its we have encountered thus far. I feel that Strategy is more efficient when dealing with these its, however another effective method is called L Hopital s Rule (stated below). This rule can also be applied to its where our strategy doesn t seem to work. As with all of our theorems though, make sure all of the hypotheses are satisfied before making use of the theorem! Theorem 2 (L Hopital s Rule). Let f() = p(), where p() and q() are both differentiable q() functions. Then, If p() = ± and q() = ±, then If p() = and q() =, then p() q() = p () q () p() q() = p () q () The same result holds if (or even if c, where c finite). Eample 6. Evaluate the following its using L Hopital s Rule: a)

9 b) c) Solution. a) = Since the it is in an indeterminant form, we can use L Hopital s Rule: b) = = Thus the it is. Notice that this means the function f() = 2 5 has a horizontal 3 + asymptote at y = = Since the it is in an indeterminant form, we can use L Hopital s Rule: = 4 6 = Again, since the it is in indeterminant form, we can use L Hopital s Rule 4 = 6 = 2 3 Thus the it is 2/3. Notice that this means the function f() = has a horizontal asymptote at y = 2/3. c) = 9

10 Since the it is in an indeterminant form, we can use L Hopital s Rule: = 6 6 = Again, since the it is in indeterminant form, we can use L Hopital s Rule 2 = 6 = Analyzing and Sketching the Graph of a Function Remark 2 (Recap). In this section we bring everything together. Here is a recap of some of the most useful concepts we have encountered: Concept Section Intercepts P Symmetry P Domain and Range P3 Continuity.4 Vertical Asymptotes.5 Differentiability 2. Relative Etrema 3. Concavity 3.4 Points of Inflection 3.4 Horizontal Asymptotes 3.5 Infinite Limits at Infinity 3.5 As you can see, this section is truly synergistic. In fact, this section really reinforces the fact that mathematics builds upon itself. In order to progress in mathematics you must master all of the foundational material in order to move forward! Here is a general strategy for drawing an accurate graph of a function:

11 Strategy 2 (Curve Sketching). Consider a function f(). In order to sketch a fairly accurate graph of f():. Find the domain of f() 2. Determine whether the function has vertical asymptotes or holes at each domain restriction by evaluating the it as approaches each domain restriction individually Draw in any vertical asymptotes as dotted vertical lines, and draw in any holes 3. Find any and y-intercepts and plot them 4. Find any horizontal asymptotes by evaluating f() and f() and draw them in as dotted horizontal lines 5. Find all of the critical points of f() 6. Use the first or second derivative test to classify them as minima, maima, or neither 7. Plot the critical points 8. Find any possible points of inflection and plot them 9. Using all that you have discovered about the function and its first and second derivatives to connect the points you have plotted, being sure to respect any domain restrictions and asymptotic behavior. Eample 7. Analyze and sketch the graph of f() = 3 (2 6). 9 Solution. Find the domain of f(): Since f() is a rational function, we know that we can not allow its denominator to be. 9 = 9 = ±3 Determine the character of the function at these -values (are there vertical asymptotes or just holes) 3 ( 6) 3 9

12 From the Left 3 ( 6) 3 9 As 3 from the left, 9 is positive (remember that < 3 = > 9), but approaching, and the numerator is approaching 5, thus the it from the left is. From the Right 3 ( 6) As 3 from the right, 9 is negative, but approaching, and the numerator is approaching 5, thus the it from the left is. Thus, the function has a vertical asymptote at = 3. Similarly, we analyze the function as 3 3 ( 6) 3 9 From the Left From the Right 3 ( 6) 3 9 As 3 from the left, 9 is negative, but approaching, and the numerator is approaching 5, thus the it from the left is. 3 ( 6) As 3 from the right, 9 is positive, but approaching, and the numerator is approaching 5, thus the it from the left is. Now we draw in the vertical asymptotes and note the behavior of the function near them. y Find any and y intercepts: y-intercept 2

13 Let = = (, 6 ) 3 f() = 3 (()2 6) () 2 9 = 3( 6) 9 = 6 3 is the y intercept for the graph of f() -intercept(s) Let y = 3 ( 6) 9 = = 3 ( 6 ) = = 6 = = = 6 = = ±4 = ( 4, ) and (4, ) are the intercepts for the graph of f() Now we plot the intercepts. y Find any horizontal asymptotes 3

14 Evaluate 3 ( 6) 9 Evaluate 3 ( 6) 9 3 ( 6) 9 3 = 3 = = 3 ( ) 6 9 ( ) 6 9 ( = 3() = = y = 3 is a horizontal asymptote of f() ) 3 ( 6) 9 3 = ( ) ( ) 6 = 2 9 ( 2 ) = 3 = 3() = = y = 3 is a horizontal asymptote of f() From our analysis, f() only has one horizontal asymptote and it is at y = 3. 4

15 Now we plot the horizontal asymptote. y Find all critical points for f() f() = 3 (2 6) 9 = f () = (2 9) (3(2)) (3 ( 6)) (2) ( 9) 2 = f () = 6 ((2 9) ( 6)) ( 9) 2 = f () = 6 ( ) ( 9) 2 = f () = 42 ( 9) 2 Notice that f () is undefined at = ±3, and we can find any remaining critical points by solving f () = f () = = 42 ( 9) 2 = = 42 = = = So we have three critical points. They occur at = and = ±3. Classify the critical points as maima, minima, or neither. We can use the first derivative test to classify our critical points: 5

16 f() f () (, 3) negative 3 undefined undefined ( 3, ) negative 6/3 (, 3) positive 3 undefined undefined (3, ) positive From the table above, we can see that = 3 is neither a minimum nor a maimum, since the derivative does not change sign as it passes through = 3. The same is true for = 3. At =, however, we have a minimum since the derivative changes from negative to positive through this point. Now we plot all of the minima and maima (note that these were plotted already since there was only one and it happened to be one of the intercepts - this will not happen in general). y Find any possible points of inflection Recall that for points of inflection, we need to find the second derivative and locate values of for which it is undefined or equal to. f () = 42 ( 9) 2 = f () = (2 9) 2 (42) 42 (2 ( 9) (2)) ( 9) 4 = f () = 42 (2 9) (( 9) 4 ) ( 9) 4 6

17 = f () = 42 ( 32 9) ( 9) 3 = f () = 26 (2 + 3) ( 9) 3 We can see that f () is undefined at = ±3, and we can find any remaining possible points of inflection by solving f () = f () = = 26 (2 + 3) ( 9) 3 = = 26 ( + 3 ) = = + 3 = = = 3 = There are no remaining possible points of inflection since = 3 has no real solutions! There are only two possible points of inflection occurring at = 3 and = 3. If we try to plot them on the graph we notice that f(3) and f( 3) are both undefined, so again we get to plot no new points (again, this will not happen in general - be sure to check for and plot points of inflection when they eist). Plot any points of inflection on the graph (for this eample, we got no new points from this step) y Use all of the information we have gathered about the graph of f() to draw an accurate sketch: 7

18 y Remark 3. The eample we have just completed truly ties together all of the calculus concepts we have discussed so far in this course. Eamples involving curve sketching required you to combine all of your mathematical knowledge, from basics such as simplification of epressions to using advanced tools and techniques such as the derivative and it processes. 8

19 Closing Homework. For homework, please try the following problems from the tetbook: Section Suggested Problems 3.5, 3, 5, 3-37 (odd), 58, (odd), , 5-23 (odd), 49-52, 59 9