3.3 Limits and Infinity

Transcription

1 Calculus Maimus. Limits Infinity Infinity is not a concrete number, but an abstract idea. It s not a destination, but a really long, never-ending journey. It s one of those mind-warping ideas that is difficult to fathom, sometimes challenging to work with, often filled with paradoical qualities. As the anonymous schoolboy once observed, It s the place where things happen that don t. We ve already seen infinity in our evaluation with its. Remember that when direct substitution of c 0 yields, the graph has a vertical asymptote at c. The only option on either side of the vertical 0 asymptote is for the graph to increase without bound (to positive infinity) or decrease without bound (to negative infinity). The it at c does not eist. Here s the definition of a vertical asymptote once again. If f ( ) or f ( ), then there eists a vertical asymptote at c c Knowing where a particular function has vertical asymptotes will allow you to quickly evaluate its at these -values. Some of your parent functions have familiar vertical asymptotes. Eample 1: Evaluate the following quickly without a calculator. We know this parent function 1/ is reflected across the -ais, stretched vertically by a factor of two, then shifted left units. We know the vertical asymptote is at, so the it technically does not eist. However, since it is a one-sided it, we try to answer with more information, namely with either or. Sketching the graph either mentally or on paper will help. We now know that c. Additionally DNE Page 1 of 11

2 Calculus Maimus Sometimes you ll have to find a it at a point on a function you don t recognize when you re not permitted to use your calculator or you calculator s batteries are dead. Eample : 4 Direct substitution here gives 4. The 0 0 means undefined because of some type of infinity, either positive or negative. In either case, we have a finite number plus or minus infinity, which will still be some type of infinity, therefore noneistent. The safest easiest answer then is 4 DNE What could we do, though, if the question was slightly different, of the form 4 If we were answering this one-sided it question on a free-response portion of an eam, DNE would still suffice, but things can not eist for many reasons. A multiple choice question would require you to decide between. Without knowing what the graph looks like, we re left with plugging in a few numbers. Since we re asked to approach from the left, we ll choose two convenient values left of, but very near see what the y-values are doing. 1: : As we moved from left to right (from 1 to 1.5) very near, the y-values increased. This is enough evidence to conclude that the y-values are increasing without bound left of the vertical asymptote at. Therefore 4 Similar calculations can be done to determine the behavior approaching from the right side. The graph at right confirms our calculations. Page of 11

3 Calculus Maimus Into the Great Beyond: LIMITS AT INFINITY We have seen how to take its as approached a specific, finite value, how those its can be infinite, but we can also take its as grows infinitely large. When evaluating its at infinity, we are essentially analyzing what the end behavior of a function is as either or. The end behavior of a graph is very important, especially when trying to graph it by h or answering test questions related to end behavior. Of course, its at infinity must be one-sided. How awkwardly impossible it would be if you, on your way to infinity, met someone coming the other way! Here s the notation we use. f ( ) or f ( ) Assuming a graph DOES have end behavior (unlike f ( ) at ), there are only a couple of things that can happen. 1. The graph can increase without bound (like f ( ) or f ( ). The graph can decrease without bound (like f ( ) or f ( ) f ( ) f ( ) ) ). The graph can oscillate between two fied values (like f ( ) sin ) f ( ) DNE or f ( ) DNE 4. The graph will taper off towards a specific, finite y-value f ( ) L or f ( ) L The last case is of particular interest to all math teachers, most math students, very few chimpanzees. Here s why: Definition: f ( ) L or f ( ) L if only if there eists a Horizontal Asymptote (HA) at y L Horizontal Asymptotes are NOT, repeat, NOT discontinuities. A graph can cross its HA any number of times (although they don t have to). The HA just tells us where the graph will eventually end up. Horizontal Asymptotes are a special quality of many Rational Functions (a polynomial over a polynomial). There are cases based on the comparison of the degrees of the leading terms of the numerator denominator. Remember that the leading terms are not necessarily the ones written in front, but rather the terms with the largest power of. When taking its at infinity, we re essentially looking for the eistence or non-eistence of any HAs. We can do this by analyzing the effective growth rates of the numerator denominator. Here are the three cases: Page of 11

4 Calculus Maimus Case 1: The degree of denominator is greater by any amount than the numerator s. In this case, for etremely large values of, even larger than that, the denominator is getting bigger faster than the numerator, so the fractions are getting smaller, smaller, smaller, smaller, approaching nothing, or ZERO. All terms other than the leading terms are relatively small for very large that they don t significantly contribute to the growth, are only there to assure us that the values will only APPROACH not EQUAL zero. In this case, there will ALWAYS be an HA at y 0. E) (from above) (from below) It doesn t matter how much larger the denominator s degree is than the numerator s, only that it is. The greater the degree difference, the faster the values go to zero. But it s no race, they all infinity to get there. Case : The degree of the numerator denominator are eactly the same. In this case, the top bottom are essentially growing at the same rate. Again the trailing terms are insignificantly small compared to the leading terms. This means that the fractions or ratios are approaching the ratio of the leading terms, or the leading terms coefficients!! In this case, there will always be an HA at leading coefficient y. leading coefficient E) Case : The degree of the numerator is greater by any amount than the denominator. In this case, the numerator s growth is outpacing the denominator s. The fractions or ratios are growing, growing, growing, forever, either in the positive or negative directions (increasing or decreasing without bound). The it will not eist there will be no HA. However, there will still be an asymptote that will take the equation of the quotient. This asymptote becomes the end behavior model for the function. For instance, if the degree difference is one, there will be a Slant (Linear) Asymptote. If the degree difference is two, there will be a Quadratic (Parabolic) Asymptote. If the degree difference is three, there will be a Cubic Asymptote, so on. In this course, we re primarily concerned with finding the equations of only the Slant Asymptotes (SAs). E) To determine what type of infinity (positive or negative), plug in any representative huge -value in the direction we re approaching into the leading terms look what the resulting sign will be. or? Page 4 of 11

5 Calculus Maimus E) We know the answer is some type of infinity. In the second case, were approaching negative infinity, so plugging a negative value into the leading terms, we get 7 negative who cares neg 5 negative who cares 5 neg positive Slant Asymptotes When the degree of the numerator is one larger than the denominator, the quotient when the division is carried out will have a degree of one. In this case there is an SA at y quotient as long as the remainder is NOT ZERO (what have we then? a hole a graph that looks identical to the quotient, ecept at the hole.) Eample : 6 7 If f, find f ( ), f ( ), find any all asymptotes. 1 We should first consider the domain, find any Vertical Asymptotes (VAs) or holes. We know that , so. Plugging into the numerator, we get a non-zero value. This means the graph 1 has a VA at. Net we notice that the degree of the numerator is one larger than the denominator s. We know there will be a Slant Asymptote. We can either find its equation first, then use it to answer the it questions (since the SA will be the end behavior model for f ) or we can plug in representative values to find the sign of infinity in each case. Let s do it the latter way confirm with the SA To find the equation of the SA, we simply do long division (synthetic division only works with dividing by a linear factor). As long as the remainder is non-zero (we know ours will NOT be zero already), the quotient will be the equation of the SA R9 Page 5 of 11

6 Calculus Maimus So there is a HA at y. We already knew there was a VA at 1 Notice that the line of the SA y is a positive sloping line with the following end behavior.. The graph of f confirms our calculations observations. Sometimes, the degrees of the numerator denominator are not as eplicit. This often occurs when either the numerator or denominator are under a radical. For these types of problems, we can use a similar analysis. Eample 4: Find The first thing to notice is that is approaching infinity. The net thing to notice is that we have a quotient. These two things mean that we can compare growth rates of the numerator denominator. The degree of the numerator is one. The degree of the denominator depends on the leading term. To find this, simply take the square root of the leading term of the radic. In this case it would be 9. This means the effective degree of the denominator is one also, the effective leading coefficient of the denominator is. Since the effective degrees are the same, the it will be the quotient of the leading 4 coefficients,. For ratios with radicals, you MUST CHECK THE SIGN of the horizontal asymptote by plugging in representative values for the type of infinity you re approaching. For, plugging in positive values, we get 4 4 So positive who cares 9 positive who cares positive. Page 6 of 11

7 Calculus Maimus For, plugging in negative values, we get 4 4 So negative who cares 9 negative who cares negative. This function has two horizontal asymptotes, each affecting the graph on one end. The graph below confirms our results. We can also find its at infinity on the calculator by plugging in very large -values into the table, values such as Eample 5: Limits at infinity: Bonus Problems! Different functions grow at different rates for large values: log functions grow slowly, polynomials grow faster by order of degree, eponential growth functions grow faster than a polynomial of any degree, factorials, like 4! 4 1, are net, the king of growth is f ( ) If you have a ratio of these types of functions, you can find its at infinity by analyzing respective growth rates. Try these out (a) ? 1.01 (b)? 5 cos ln (c)? (d) (g) 100? (e) 5? (h) 4 4 1!? (f)?! 5? 4 1 (i) 4 16? (j)? 1 1 (k) sin? (l) sin? Which of the above functions have Horizontal Asymptotes??? Page 7 of 11

8 Calculus Maimus There one last special type of it problem that s neither an infinite it nor a it at infinity. Nonetheless, they show up enough to receive special treatment. Recall your parent function jump discontinuity at 0 More importantly, f, shown at right. There is a f 0 is undefined. 1, 0 1, 0 0 DNE Knowing the transformations of this function, knowing the -value where the jump occurs (always the value that yields 0 ), will allow you to answer it questions involving this function quickly 0 accurately. Eample 6: Evaluate the following: (a) Direct substitution yields 0, which usually means that the it still eists finding it requires more 0 mental sweat elbow grease. In this case, however, since we recognize this as a relative of the f family, we know there is jump discontinuity at 5. Since were asked to find the general it, not a one-sided it, we can go straight to the answer... (b) DNE This is almost the same question as above, ecept we re asked to find a one-sided it, which we know will eist. There are a couple of ways to find that value. Page 8 of 11

9 Calculus Maimus We can sketch a graph using transformations, which first requires us to rewrite the epression in stard transformation form. 5, a vertical stretch by a fact of two, shifted 5 5 five units to the right. We can now sketch the graph 10 conclude that 5 5 Another way to answer this question is to realize that this is a graph of a piecewise function with both pieces being horizontal lines. A horizontal line has the same it value at any -value. We just need to know what that value is whether it is positive or negative. In this case, since we re interested in the it from the right of five, we can simply plug ANY -value greater than five into the epression then simplify. This will be our answer. Let s try choosing (c) (it works the same way if the absolute value is in the denominator) Let s try out our new method by plugging in an -value less than five, say That s both fun AND easy. Let s finish up with a couple of piece-wise functions. They are always interesting. Page 9 of 11

10 Calculus Maimus Eample 7: For (a), 1 g( ), find the following: 4, 1 g( ) (b) g( ) (c) 1 1 g( ) 1 (d) g( 1) Notice the two pieces making up the function are polynomials, which are continuous everywhere. That means anywhere we wish to evaluate the it, we may do so by direct substitution, we just need to be sure to plug into the correct piece. (a) We can plug 1 into the top piece to get g( ) 1 (b) We plug 1 into the bottom piece to get g( ) 1 (c) Since the its from each side of 1 are the same, we say the it eists is, or mathematically, g( ). 1 (d) Here, we plug into the top piece (since it is less than or EQUAL to) to get g( 1) Since the it function value at 1 both eist are the same, we can further claim that the function is continuous at 1, thus continuous for all -values. Page 10 of 11

11 Calculus Maimus Eample 8: 4, 1 For g( ), find the following: 4, 1 (a) g( ) (b) g( ) (c) 1 1 g( ) 1 (d) g( 1) Notice this resembles the problem above, ecept that the left piece of the function is slightly different. (a) (b) (c) g( ) 5 1 g( ) 1 g( ) DNE since parts (a) (b) are 1 different values. There is a non-removable jump discontinuity at 1. (d) g( 1) 5. Here the function value eists, while the it does not. Remember that the function value the it value don t necessarily have anything to do with each other, can co-eist without the other. It is the special case when they ARE the same that the function is continuous at that point. Can you draw an eample where neither the it nor the function values eist at a point? Need help? Page 11 of 11