# ch2 sec 2.6 limits and continuity james stewart pdf online

## Mathematics 265 with Unknown at Arizona State University - Tempe *

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- ch2 sec 2.6 limits and continuity james stewart pdf online

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Calculus: Early Transcendentals (Stewart's Calculus Series)
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62. If and are positive numbers, prove that the equation has at least one solution in the interval . 63. Show that the function is continuous on . 64. (a) Show that the absolute value function is contin- uous everywhere. (b) Prove that if is a continuous function on an interval, then so is . (c) Is the converse of the statement in part (b) also true? In other words, if is continuous, does it follow that is continuous? If so, prove it. If not, find a counterexample. 65. A Tibetan monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 P M. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 P M. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days. f H11341 f H11341 H11341 f H11341 f FH20849xH20850 H33527 H11341 x H11341 H20849H11002H11009, H11009H20850 f H20849xH20850 H33527 H20877 x 4 sinH208491H20862xH20850 0 if x HS33527 0 if x H33527 0 H20849H110021, 1H20850 a x 3 H11001 2x 2 H11002 1 H11001 b x 3 H11001 x H11002 2 H33527 0 ba 55. Prove that is continuous at if and only if 56. To prove that sine is continuous, we need to show that for every real number . By Exercise 55 an equivalent statement is that Use (6) to show that this is true. 57. Prove that cosine is a continuous function. 58. (a) Prove Theorem 4, part 3. (b) Prove Theorem 4, part 5. 59. For what values of is continuous? 60. For what values of is continuous? Is there a number that is exactly 1 more than its cube?61. tH20849xH20850 H33527 H20877 0 x if x is rational if x is irrational tx f H20849xH20850 H33527 H20877 0 1 if x is rational if x is irrational fx lim hl0 sinH20849a H11001 hH20850 H33527 sin a alim xla sin x H33527 sin a lim hl0 f H20849a H11001 hH20850 H33527 f H20849aH20850 af LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES In Sections 2.2 and 2.4 we investigated infinite limits and vertical asymptotes. There we let approach a number and the result was that the values of became arbitrarily large (positive or negative). In this section we let become arbitrarily large (positive or nega- tive) and see what happens to . Let’s begin by investigating the behavior of the function defined by as becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of has been drawn by a computer in Figure 1. As grows larger and larger you can see that the values of get closer and closer to 1. In fact, it seems that we can make the values of as close as we like to 1 by taking sufficiently large. This situation is expressed symbolically by writing lim xlH11009 x 2 H11002 1 x 2 H11001 1 H33527 1 x f H20849xH20850 f H20849xH20850x x 1 0 y y=1 y= ≈-1 ≈+1 FIGURE 1 f x f H20849xH20850 H33527 x 2 H11002 1 x 2 H11001 1 f y x yx 2.6 130 |||| CHAPTER 2 LIMITS AND DERIVATIVES x 0 H110021 0 0.600000 0.800000 0.882353 0.923077 0.980198 0.999200 0.999800 0.999998H110061000 H11006100 H1100650 H1100610 H110065 H110064 H110063 H110062 H110061 f H20849xH20850 SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES |||| 131 In general, we use the notation to indicate that the values of become closer and closer to as becomes larger and larger. DEFINITION Let be a function defined on some interval . Then means that the values of can be made arbitrarily close to by taking suf- ficiently large. Another notation for is as The symbol does not represent a number. Nonetheless, the expression is often read as “the limit of , as approaches infinity, is ” or “the limit of , as becomes infinite, is ” or “the limit of , as increases without bound, is ” The meaning of such phrases is given by Definition 1. A more precise definition, similar to the definition of Section 2.4, is given at the end of this section. Geometric illustrations of Definition 1 are shown in Figure 2. Notice that there are many ways for the graph of to approach the line (which is called a horizontal asymptote) as we look to the far right of each graph. Referring back to Figure 1, we see that for numerically large negative values of , the values of are close to 1. By letting decrease through negative values without bound, we can make as close as we like to 1. This is expressed by writing The general definition is as follows. DEFINITION Let be a function defined on some interval . Then means that the values of can be made arbitrarily close to by taking suf- ficiently large negative. xLf H20849xH20850 lim xlH11002H11009 f H20849xH20850 H33527 L H20849H11002H11009, aH20850f2 lim xlH11002H11009 x 2 H11002 1 x 2 H11001 1 H33527 1 f H20849xH20850 xf H20849xH20850 x x y 0 y=ƒ y=L 0 x y y=ƒ y=L x y 0 y=ƒ y=L y H33527 Lf H9255, H9254 Lxf H20849xH20850 Lxf H20849xH20850 Lxf H20849xH20850 lim xlH11009 f H20849xH20850 H33527 LH11009 x lH11009f H20849xH20850 l L lim xlH11009 f H20849xH20850 H33527 L xLf H20849xH20850 lim xlH11009 f H20849xH20850 H33527 L H20849a, H11009H20850f1 xLf H20849xH20850 lim xlH11009 f H20849xH20850 H33527 L x ` FIGURE 2 Examples illustrating lim ƒ=L Again, the symbol does not represent a number, but the expression is often read as “the limit of , as x approaches negative infinity, is L” Definition 2 is illustrated in Figure 3. Notice that the graph approaches the line as we look to the far left of each graph. DEFINITION The line is called a horizontal asymptote of the curve if either For instance, the curve illustrated in Figure 1 has the line as a horizontal asymp- tote because An example of a curve with two horizontal asymptotes is . (See Figure 4.) In fact, so both of the lines and are horizontal asymptotes. (This follows from the fact that the lines are vertical asymptotes of the graph of tan.) EXAMPLE 1 Find the infinite limits, limits at infinity, and asymptotes for the function f whose graph is shown in Figure 5. SOLUTION We see that the values of become large as from both sides, so Notice that becomes large negative as x approaches 2 from the left, but large posi- tive as x approaches 2 from the right. So Thus both of the lines and are vertical asymptotes. As x becomes large, it appears that approaches 4. But as x decreases through negative values, approaches 2. So This means that both y H33527 4 and y H33527 2 are horizontal asymptotes. M lim xlH11002H11009 f H20849xH20850 H33527 2andlim xlH11009 f H20849xH20850 H33527 4 f H20849xH20850 f H20849xH20850 x H33527 2x H33527H110021 lim x l 2 H11001 f H20849xH20850 H33527H11009andlim x l 2 H11002 f H20849xH20850 H33527H11002H11009 f H20849xH20850 lim xlH110021 f H20849xH20850 H33527H11009 x lH110021f H20849xH20850 x H33527H11006H9266H208622 y H33527H9266H208622y H33527H11002H9266H208622 lim xlH11009 tan H110021 x H33527 H9266 2 lim xlH11002H11009 tan H110021 x H33527H11002 H9266 2 4 y H33527 tan H110021 x lim xlH11009 x 2 H11002 1 x 2 H11001 1 H33527 1 y H33527 1 lim xlH11002H11009 f H20849xH20850 H33527 Lorlim xlH11009 f H20849xH20850 H33527 L y H33527 f H20849xH20850 y H33527 L3 y H33527 L f H20849xH20850 lim x l H11002H11009 f H20849xH20850 H33527 LH11002H11009 132 |||| CHAPTER 2 LIMITS AND DERIVATIVES x _` FIGURE 3 Examples illustrating lim ƒ=L 0 y x y=ƒ y=L x0 y y=ƒ y=L FIGURE 4 y=tan–!x y 0 x pi 2 _ pi 2 FIGURE 5 0 x y 2 2 SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 133 EXAMPLE 2 Find and . SOLUTION Observe that when is large, is small. For instance, In fact, by taking large enough, we can make as close to 0 as we please. Therefore, according to Definition 1, we have Similar reasoning shows that when is large negative, is small negative, so we also have It follows that the line (the -axis) is a horizontal asymptote of the curve . (This is an equilateral hyperbola; see Figure 6.) M Most of the Limit Laws that were given in Section 2.3 also hold for limits at infinity. It can be proved that the Limit Laws listed in Section 2.3 (with the exception of Laws 9 and 10) are also valid if “ ” is replaced by “ ” or “ .” In particular, if we combine Laws 6 and 11 with the results of Example 2, we obtain the following important rule for calculating limits. THEOREM If is a rational number, then If is a rational number such that is defined for all x, then EXAMPLE 3 Evaluate and indicate which properties of limits are used at each stage. SOLUTION As becomes large, both numerator and denominator become large, so it isn’t obvious what happens to their ratio. We need to do some preliminary algebra. To evaluate the limit at infinity of any rational function, we first divide both the numerator and denominator by the highest power of that occurs in the denominator. x x lim xlH11009 3x 2 H11002 x H11002 2 5x 2 H11001 4x H11001 1 V lim xlH11002H11009 1 x r H33527 0 x r r H11022 0 lim xlH11009 1 x r H33527 0 r H11022 05 x lH11002H11009x lH11009x l a y H33527 1H20862xxy H33527 0 lim xlH11002H11009 1 x H33527 0 1H20862xx lim xlH11009 1 x H33527 0 1H20862xx 1 1,000,000 H33527 0.000001 1 10,000 H33527 0.0001 1 100 H33527 0.01 1H20862xx lim xlH11002H11009 1 x lim xlH11009 1 x x ` x _` 1 x 1 x 0 y x y=∆ FIGURE 6 lim =0, lim =0 (We may assume that , since we are interested only in large values of .) In this case the highest power of in the denominator is , so we have (by Limit Law 5) (by 1, 2, and 3) (by 7 and Theorem 5) A similar calculation shows that the limit as is also . Figure 7 illustrates the results of these calculations by showing how the graph of the given rational function approaches the horizontal asymptote . M EXAMPLE 4 Find the horizontal and vertical asymptotes of the graph of the function SOLUTION Dividing both numerator and denominator by and using the properties of lim- its, we have (since for ) Therefore the line is a horizontal asymptote of the graph of . In computing the limit as , we must remember that for , we have . So when we divide the numerator by , for we get 1 x s2x 2 H11001 1 H33527H11002 1 sx 2 s2x 2 H11001 1 H33527H11002 H20881 2 H11001 1 x 2 x H11021 0xsx 2 H33527 H11341 x H11341 H33527H11002x x H11021 0x lH11002H11009 fy H33527s2 H208623 H33527 s2 H11001 0 3 H11002 5 H11554 0 H33527 s2 3 H33527 lim xlH11009 H20881 2 H11001 1 x 2 lim xlH11009 H20873 3 H11002 5 x H20874 H33527 H20881 lim xlH11009 2 H11001 lim xlH11009 1 x 2 lim xlH11009 3 H11002 5 lim xlH11009 1 x x H11022 0sx 2 H33527 x lim xlH11009 s2x 2 H11001 1 3x H11002 5 H33527 lim xlH11009 H20881 2 H11001 1 x 2 3 H11002 5 x x f H20849xH20850 H33527 s2x 2 H11001 1 3x H11002 5 y H33527 3 5 3 5 x lH11002H11009 H33527 3 5 H33527 3 H11002 0 H11002 0 5 H11001 0 H11001 0 H33527 lim x l H11009 3 H11002 lim x l H11009 1 x H11002 2 lim x l H11009 1 x 2 lim x l H11009 5 H11001 4 lim x l H11009 1 x H11001 lim x l H11009 1 x 2 H33527 lim x l H11009 H20873 3 H11002 1 x H11002 2 x 2 H20874 lim x l H11009 H20873 5 H11001 4 x H11001 1 x 2 H20874 lim x l H11009 3x 2 H11002 x H11002 2 5x 2 H11001 4x H11001 1 H33527 lim x l H11009 3x 2 H11002 x H11002 2 x 2 5x 2 H11001 4x H11001 1 x 2 H33527 lim x l H11009 3 H11002 1 x H11002 2 x 2 5 H11001 4 x H11001 1 x 2 x 2 x xx HS33527 0 134 |||| CHAPTER 2 LIMITS AND DERIVATIVES 1 y=0.6 x y 0 y= 3≈-x-2 5≈+4x+1 SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 135 Therefore Thus the line is also a horizontal asymptote. A vertical asymptote is likely to occur when the denominator, , is 0, that is, when . If is close to and , then the denominator is close to 0 and is positive. The numerator is always positive, so is positive. Therefore If is close to but , then and so is large negative. Thus The vertical asymptote is . All three asymptotes are shown in Figure 8. M EXAMPLE 5 Compute . SOLUTION Because both and x are large when x is large, it’s difficult to see what happens to their difference, so we use algebra to rewrite the function. We first multiply numerator and denominator by the conjugate radical: The Squeeze Theorem could be used to show that this limit is 0. But an easier method is to divide numerator and denominator by . Doing this and using the Limit Laws, we obtain Figure 9 illustrates this result. M The graph of the natural exponential function has the line (the x-axis) as a horizontal asymptote. (The same is true of any exponential function with base .) In a H11022 1 y H33527 0y H33527 e x H33527 lim x l H11009 1 x H20881 1 H11001 1 x 2 H11001 1 H33527 0 s1 H11001 0 H11001 1 H33527 0 lim x l H11009 (sx 2 H11001 1 H11002 x) H33527 lim x l H11009 1 sx 2 H11001 1 H11001 x H33527 lim x l H11009 1 x sx 2 H11001 1 H11001 x x x H33527 lim x l H11009 H20849x 2 H11001 1H20850 H11002 x 2 sx 2 H11001 1 H11001 x H33527 lim x l H11009 1 sx 2 H11001 1 H11001 x lim x l H11009 (sx 2 H11001 1 H11002 x) H33527 lim x l H11009 (sx 2 H11001 1 H11002 x) sx 2 H11001 1 H11001 x sx 2 H11001 1 H11001 x sx 2 H11001 1 lim xlH11009 (sx 2 H11001 1 H11002 x) x H33527 5 3 lim x l H208495H208623H20850 H11002 s2x 2 H11001 1 3x H11002 5 H33527H11002H11009 f H20849xH208503x H11002 5 H11021 0x H11021 5 3 5 3 x lim xlH208495H208623H20850 H11001 s2x 2 H11001 1 3x H11002 5 H33527H11009 f H20849xH20850s2x 2 H11001 1 3x H11002 5x H11022 5 3 5 3 xx H33527 5 3 3x H11002 5 y H33527H11002s2H208623 H33527 H11002 H20881 2 H11001 lim xlH11002H11009 1 x 2 3 H11002 5 lim xlH11002H11009 1 x H33527H11002 s2 3 lim xlH11002H11009 s2x 2 H11001 1 3x H11002 5 H33527 lim xlH11002H11009 H11002 H20881 2 H11001 1 x 2 3 H11002 5 x FIGURE 8 y= œ„„„„„„ 3x-5 2≈+1 x y y= œ„2 3 y=_ œ„2 3 x= 5 3 FIGURE 9 y= ≈+1œ„„„„„-x x y 0 1 1 N We can think of the given function as having a denominator of 1. fact, from the graph in Figure 10 and the corresponding table of values, we see that Notice that the values of approach 0 very rapidly. EXAMPLE 6 Evaluate . SOLUTION If we let , we know that as . Therefore, by (6), (See Exercise 71.) M EXAMPLE 7 Evaluate . SOLUTION As x increases, the values of sin x oscillate between 1 and H110021 infinitely often and so they don’t approach any definite number. Thus does not exist. M INFINITE LIMITS AT INFINITY The notation is used to indicate that the values of become large as becomes large. Similar mean- ings are attached to the following symbols: EXAMPLE 8 Find and . SOLUTION When becomes large, also becomes large. For instance, In fact, we can make as big as we like by taking large enough. Therefore we can write Similarly, when is large negative, so is . Thus These limit statements can also be seen from the graph of in Figure 11. My H33527 x 3 lim xlH11002H11009 x 3 H33527H11002H11009 x 3 x lim xlH11009 x 3 H33527H11009 xx 3 1000 3 H33527 1,000,000,000100 3 H33527 1,000,00010 3 H33527 1000 x 3 x lim xlH11002H11009 x 3 lim xlH11009 x 3 lim xlH11002H11009 f H20849xH20850 H33527H11002H11009lim xlH11009 f H20849xH20850 H33527H11002H11009lim xlH11002H11009 f H20849xH20850 H33527H11009 xf H20849xH20850 lim xlH11009 f H20849xH20850 H33527H11009 lim xlH11009 sin x lim x l H11009 sin x lim x l 0 H11002 e 1H20862x H33527 lim t l H11002H11009 e t H33527 0 x l 0 H11002 t lH11002H11009t H33527 1H20862x lim x l 0 H11002 e 1H20862x V y=´ x0 1 y 1 FIGURE 10 e x lim xlH11002H11009 e x H33527 06 136 |||| CHAPTER 2 LIMITS AND DERIVATIVES x 0 1.00000 H110021 0.36788 H110022 0.13534 H110023 0.04979 H110025 0.00674 H110028 0.00034 H1100210 0.00005 e x N The problem-solving strategy for Example 6 is introducing something extra (see page 76). Here, the something extra, the auxiliary aid, is the new variable .t FIGURE 11 lim x#=`, lim x#=_` x ` x _` x y 0 y=˛ SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 137 Looking at Figure 10 we see that but, as Figure 12 demonstrates, becomes large as at a much faster rate than . EXAMPLE 9 Find . | SOLUTION It would be wrong to write The Limit Laws can’t be applied to infinite limits because is not a number ( can’t be defined). However, we can write because both and become arbitrarily large and so their product does too. M EXAMPLE 10 Find . SOLUTION As in Example 3, we divide the numerator and denominator by the highest power of in the denominator, which is just x: because and as . M The next example shows that by using infinite limits at infinity, together with intercepts, we can get a rough idea of the graph of a polynomial without having to plot a large num- ber of points. EXAMPLE 11 Sketch the graph of by finding its inter- cepts and its limits as and as . SOLUTION The -intercept is and the -intercepts are found by setting : . Notice that since is positive, the function doesn’t change sign at ; thus the graph doesn’t cross the -axis at . The graph crosses the axis at and . When is large positive, all three factors are large, so When is large negative, the first factor is large positive and the second and third factors are both large negative, so Combining this information, we give a rough sketch of the graph in Figure 13. M lim xlH11002H11009 H20849x H11002 2H20850 4 H20849x H11001 1H20850 3 H20849x H11002 1H20850 H33527H11009 x lim xlH11009 H20849x H11002 2H20850 4 H20849x H11001 1H20850 3 H20849x H11002 1H20850 H33527H11009 x 1H110021 2x2 H20849x H11002 2H20850 4 x H33527 2, H110021, 1y H33527 0 xf H208490H20850 H33527 H20849H110022H20850 4 H208491H20850 3 H20849H110021H20850 H33527H1100216y x lH11002H11009x lH11009 y H33527 H20849x H11002 2H20850 4 H20849x H11001 1H20850 3 H20849x H11002 1H20850V x lH110093H20862x H11002 1 lH110021x H11001 1 lH11009 lim xlH11009 x 2 H11001 x 3 H11002 x H33527 lim xlH11009 x H11001 1 3 x H11002 1 H33527H11002H11009 x lim xlH11009 x 2 H11001 x 3 H11002 x x H11002 1x lim xlH11009 H20849x 2 H11002 xH20850 H33527 lim xlH11009 xH20849x H11002 1H20850 H33527H11009 H11009H11002H11009H11009 H33527H11009H11002H11009 lim xlH11009 H20849x 2 H11002 xH20850 H33527 lim xlH11009 x 2 H11002 lim xlH11009 x lim xlH11009 H20849x 2 H11002 xH20850 y H33527 x 3 x lH11009y H33527 e x lim xlH11009 e x H33527H11009 y y=(x-2)$(x +1)#(x-1) FIGURE 13 0 x _1 _16 21 x0 100 y 1 y=˛ y=´ FIGURE 12 ´ is much larger than ˛ when x is large. PRECISE DEFINITIONS Definition 1 can be stated precisely as follows. DEFINITION Let be a function defined on some interval . Then means that for every there is a corresponding number such that then In words, this says that the values of can be made arbitrarily close to (within a distance , where is any positive number) by taking sufficiently large (larger than , where depends on ). Graphically it says that by choosing large enough (larger than some number ) we can make the graph of lie between the given horizontal lines and as in Figure 14. This must be true no matter how small we choose . Figure 15 shows that if a smaller value of is chosen, then a larger value of may be required. Similarly, a precise version of Definition 2 is given by Definition 8, which is illustrated in Figure 16. DEFINITION Let be a function defined on some interval . Then means that for every there is a corresponding number such that then H11341 f H20849xH20850 H11002 L H11341 H11021H9255x H11021 N if NH9255 H11022 0 lim xlH11002H11009 f H20849xH20850 H33527 L H20849H11002H11009, aH20850f8 FIGURE 14 lim ƒ=L x ` FIGURE 15 lim ƒ=L x ` y 0 x N L y=ƒ y=L-∑ y=L+∑ 0 y x N L when x is in here ƒ is in here y=L-∑ y=L+∑ ∑ ∑ y=ƒ NH9255H9255 y H33527 L H11001H9255y H33527 L H11002H9255 fN xH9255N NxH9255H9255 Lf H20849xH20850 H11341 f H20849xH20850 H11002 L H11341 H11021H9255x H11022 Nif NH9255 H11022 0 lim xlH11009 f H20849xH20850 H33527 L H20849a, H11009H20850f7 138 |||| CHAPTER 2 LIMITS AND DERIVATIVES SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 139 In Example 3 we calculated that In the next example we use a graphing device to relate this statement to Definition 7 with and . EXAMPLE 12 Use a graph to find a number such that then SOLUTION We rewrite the given inequality as We need to determine the values of for which the given curve lies between the horizon- tal lines and . So we graph the curve and these lines in Figure 17. Then we use the cursor to estimate that the curve crosses the line when . To the right of this number the curve stays between the lines and . Round- ing to be safe, we can say that then In other words, for we can choose (or any larger number) in Definition 7. M EXAMPLE 13 Use Definition 7 to prove that . SOLUTION Given , we want to find such that if then In computing the limit we may assume that . Then . Let’s choose . So if then H20895 1 x H11002 0 H20895 H33527 1 x H11021H9255x H11022 N H33527 1 H9255 N H33527 1H20862H9255 1H20862x H11021H9255 &? x H11022 1H20862H9255x H11022 0 H20895 1 x H11002 0 H20895 H11021H9255x H11022 N NH9255 H11022 0 lim xlH11009 1 x H33527 0 N H33527 7H9255H33527 0.1 H20895 3x 2 H11002 x H11002 2 5x 2 H11001 4x H11001 1 H11002 0.6 H20895 H11021 0.1x H11022 7if y H33527 0.7y H33527 0.5 x H11015 6.7y H33527 0.5 y H33527 0.7y H33527 0.5 x 0.5 H11021 3x 2 H11002 x H11002 2 5x 2 H11001 4x H11001 1 H11021 0.7 H20895 3x 2 H11002 x H11002 2 5x 2 H11001 4x H11001 1 H11002 0.6 H20895 H11021 0.1x H11022 Nif N H9255H33527 0.1L H33527 3 5 lim xlH11009 3x 2 H11002 x H11002 2 5x 2 H11001 4x H11001 1 H33527 3 5 x _` FIGURE 16 lim ƒ=L x N y L y=L-∑ y=L+∑ y=ƒ 0 FIGURE 17 1 0 15 y=0.7 y=0.5 y= 3≈-x-2 5≈+4x+1 Therefore, by Definition 7, Figure 18 illustrates the proof by showing some values of and the corresponding values of . M Finally we note that an infinite limit at infinity can be defined as follows. The geomet- ric illustration is given in Figure 19. DEFINITION Let be a function defined on some interval . Then means that for every positive number there is a corresponding positive number N such that then Similar definitions apply when the symbol is replaced by . (See Exercise 70.)H11002H11009H11009 f H20849xH20850 H11022 Mx H11022 Nif M lim xlH11009 f H20849xH20850 H33527H11009 H20849a, H11009H20850f9 N H9255 lim xlH11009 1 x H33527 0 140 |||| CHAPTER 2 LIMITS AND DERIVATIVES x y 0 N=5 ∑ =0.2 FIGURE 18 x y 0 N=1 ∑ =1 x y 0 N=10 ∑ =0.1 FIGURE 19 lim ƒ=` x ` 0 x y N M y=M (c) (d) (e) (f) The equations of the asymptotes x y 1 1 lim x l H11002H11009 f H20849xH20850 lim x l H11009 f H20849xH20850lim x l H110021 H11001 f H20849xH208501. Explain in your own words the meaning of each of the following. (a) (b) (a) Can the graph of intersect a vertical asymptote? Can it intersect a horizontal asymptote? Illustrate by sketching graphs. (b) How many horizontal asymptotes can the graph of have? Sketch graphs to illustrate the possibilities. 3. For the function whose graph is given, state the following. (a) (b) lim x l H110021 H11002 f H20849xH20850lim x l 2 f H20849xH20850 f y H33527 f H20849xH20850 y H33527 f H20849xH208502. lim xlH11002H11009 f H20849xH20850 H33527 3lim xlH11009 f H20849xH20850 H33527 5 EXERCISES2.6 SECTION 2.6 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES |||| 141 13–14 Evaluate the limit and justify each step by indicating the appropriate properties of limits. 13. 14. 15–36 Find the limit. 15. 16. 17. 18. 20. 21. 22. 23. 24. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. ; 37. (a) Estimate the value of by graphing the function . (b) Use a table of values of to guess the value of the limit. (c) Prove that your guess is correct. ; 38. (a) Use a graph of to estimate the value of to one decimal place. (b) Use a table of values of to estimate the limit to four decimal places. (c) Find the exact value of the limit. f H20849xH20850 lim xlH11009 f H20849xH20850 f H20849xH20850 H33527s3x 2 H11001 8x H11001 6 H11002s3x 2 H11001 3x H11001 1 f H20849xH20850 f H20849xH20850 H33527sx 2 H11001 x H11001 1 H11001 x lim x l H11002H11009 (sx 2 H11001 x H11001 1 H11001 x) lim x l H20849H9266H208622H20850 H11001 e tan x lim x l H11009 H20849e H110022x cos xH20850 lim x l H11009 tan H110021 H20849x 2 H11002 x 4 H20850lim x l H11009 1 H11002 e x 1 H11001 2e x lim x l H11009 x 3 H11002 2x H11001 3 5 H11002 2x 2 lim x l H11002H11009 H20849x 4 H11001 x 5 H20850 lim x l H11009 sx 2 H11001 1lim x l H11009 x H11001 x 3 H11001 x 5 1 H11002 x 2 H11001 x 4 lim x l H11009 cos x lim x l H11009 (sx 2 H11001 ax H11002sx 2 H11001 bx ) lim xlH11002H11009 (x H11001sx 2 H11001 2x )lim x l H11009 (s9x 2 H11001 x H11002 3x) 25. lim x l H11002H11009 s9x 6 H11002 x x 3 H11001 1 lim x l H11009 s9x 6 H11002 x x 3 H11001 1 lim x l H11009 x H11001 2 s9x 2 H11001 1 lim u l H11009 4u 4 H11001 5 H20849u 2 H11002 2H20850H208492u 2 H11002 1H20850 lim t l H11002H11009 t 2 H11001 2 t 3 H11001 t 2 H11002 1 lim x l H11009 x 3 H11001 5x 2x 3 H11002 x 2 H11001 4 19. lim y l H11009 2 H11002 3y 2 5y 2 H11001 4y lim x l H11002H11009 1 H11002 x H11002 x 2 2x 2 H11002 7 lim x l H11009 3x H11001 5 x H11002 4 lim x l H11009 1 2x H11001 3 lim x l H11009 H20881 12x 3 H11002 5x H11001 2 1 H11001 4x 2 H11001 3x 3 lim x l H11009 3x 2 H11002 x H11001 4 2x 2 H11001 5x H11002 8 4. For the function whose graph is given, state the following. (a) (b) (c) (d) (e) (f) The equations of the asymptotes 5–10 Sketch the graph of an example of a function that satisfies all of the given conditions. 5. is odd 6. 8. 9. 10. is even ; 11. Guess the value of the limit by evaluating the function for and . Then use a graph of to support your guess. ; 12. (a) Use a graph of to estimate the value of correct to two decimal places. (b) Use a table of values of to estimate the limit to four decimal places. f H20849xH20850 lim xlH11009 f H20849xH20850 f H20849xH20850 H33527 H20873 1 H11002 2 x H20874 x f1004, 5, 6, 7, 8, 9, 10, 20, 50, x H33527 0, 1, 2, 3,f H20849xH20850 H33527 x 2 H208622 x lim x l H11009 x 2 2 x ff H208490H20850 H33527 0,lim x l H11009 f H20849xH20850 H33527 2, lim x l 3 f H20849xH20850 H33527H11002H11009, lim x l H11009 f H20849xH20850 H33527 3 lim x l 4 H11001 f H20849xH20850 H33527H11009,lim x l 4 H11002 f H20849xH20850 H33527H11002H11009,lim x l H11002H11009 f H20849xH20850 H33527H11002H11009, lim x l 0 H11001 f H20849xH20850 H33527 2,lim x l 0 H11002 f H20849xH20850 H33527 4,f H208490H20850 H33527 3, lim x l H11009 f H20849xH20850 H33527H110023 lim x l H11002H11009 f H20849xH20850 H33527 3,lim x l H110022 f H20849xH20850 H33527H11009, lim x l 0 H11002 f H20849xH20850 H33527H11002H11009lim x l 0 H11001 f H20849xH20850 H33527H11009, lim x l H11002H11009 f H20849xH20850 H33527 0,lim x l H11009 f H20849xH20850 H33527H11009, lim x l 2 f H20849xH20850 H33527H11002H11009, 7. lim x l H11002H11009 f H20849xH20850 H33527 1 lim x l H11009 f H20849xH20850 H33527 1,lim x l 0 H11002 f H20849xH20850 H33527H11002H11009, lim x l 0 H11001 f H20849xH20850 H33527H11009, flim x l H11009 f H20849xH20850 H33527 0, f H208491H20850 H33527 1, f H208490H20850 H33527 0, f 2 0 x y 1 lim x l H110022 H11001 tH20849xH20850 lim x l 0 tH20849xH20850lim x l 3 tH20849xH20850 lim x l H11002H11009 tH20849xH20850lim x l H11009 tH20849xH20850 t 142 |||| CHAPTER 2 LIMITS AND DERIVATIVES 53. (a) Use the Squeeze Theorem to evaluate . ; (b) Graph . How many times does the graph cross the asymptote? ; 54. By the end behavior of a function we mean the behavior of its values as and as . (a) Describe and compare the end behavior of the functions by graphing both functions in the viewing rectangles by and by . (b) Two functions are said to have the same end behavior if their ratio approaches 1 as . Show that P and Q have the same end behavior. Let and be polynomials. Find if the degree of is (a) less than the degree of and (b) greater than the degree of . 56. Make a rough sketch of the curve ( an integer) for the following five cases: (i) (ii) , odd (iii) , even (iv) , odd (v) , even Then use these sketches to find the following limits. (a) (b) (c) (d) Find if, for all , 58. (a) A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the tank at a rate of 25 LH20862min. Show that the concentration of salt after minutes (in grams per liter) is (b) What happens to the concentration as ? 59. In Chapter 9 we will be able to show, under certain assump- tions, that the velocity of a falling raindrop at time t is where t is the acceleration due to gravity and is the terminal velocity of the raindrop. (a) Find .lim t l H11009 vH20849tH20850 v* vH20849tH20850 H33527 v*H208491 H11002 e H11002ttH20862v* H20850 vH20849tH20850 t lH11009 CH20849tH20850 H33527 30t 200 H11001 t t 10e x H11002 21 2e x H11021 f H20849xH20850 H11021 5sx sx H11002 1 x H11022 1lim x l H11009 f H20849xH2085057. lim x l H11002H11009 x n lim x l H11009 x n lim x l 0 H11002 x n lim x l 0 H11001 x n nn H11021 0 nn H11021 0nn H11022 0 nn H11022 0n H33527 0 ny H33527 x n Q QP lim xlH11009 PH20849xH20850 QH20849xH20850 QP55. x lH11009 H20851H1100210,000, 10,000H20852H20851H1100210, 10H20852H20851H110022, 2H20852H20851H110022, 2H20852 QH20849xH20850 H33527 3x 5 PH20849xH20850 H33527 3x 5 H11002 5x 3 H11001 2x x lH11002H11009x lH11009 f H20849xH20850 H33527 H20849sin xH20850H20862x lim x l H11009 sin x x 39–44 Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes. 39. 40. 42. 43. 44. ; 45. Estimate the horizontal asymptote of the function by graphing for . Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy? ; 46. (a) Graph the function How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits and (b) By calculating values of , give numerical estimates of the limits in part (a). (c) Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.] 47. Find a formula for a function that satisfies the following conditions: , , , , 48. Find a formula for a function that has vertical asymptotes and and horizontal asymptote . 49–52 Find the limits as and as . Use this infor- mation, together with intercepts, to give a rough sketch of the graph as in Example 11. 49. 50. 51. 52. y H33527 x 2 H20849x 2 H11002 1H20850 2 H20849x H11001 2H20850 y H33527 H208493 H11002 xH20850H208491 H11001 xH20850 2 H208491 H11002 xH20850 4 y H33527 x 3 H20849x H11001 2H20850 2 H20849x H11002 1H20850y H33527 x 4 H11002 x 6 x lH11002H11009x lH11009 y H33527 1x H33527 3x H33527 1 lim x l 3 H11001 f H20849xH20850 H33527H11002H11009lim x l 3 H11002 f H20849xH20850 H33527H11009 f H208492H20850 H33527 0lim x l 0 f H20849xH20850 H33527H11002H11009lim x l H11006H11009 f H20849xH20850 H33527 0 f f H20849xH20850 lim x l H11002H11009 s2x 2 H11001 1 3x H11002 5 lim x l H11009 s2x 2 H11001 1 3x H11002 5 f H20849xH20850 H33527 s2x 2 H11001 1 3x H11002 5 H1100210 H33355 x H33355 10f f H20849xH20850 H33527 3x 3 H11001 500x 2 x 3 H11001 500x 2 H11001 100x H11001 2000 y H33527 2e x e x H11002 5 y H33527 x 3 H11002 x x 2 H11002 6x H11001 5 y H33527 1 H11001 x 4 x 2 H11002 x 4 y H33527 2x 2 H11001 x H11002 1 x 2 H11001 x H11002 2 41. y H33527 x 2 H11001 1 2x 2 H11002 3x H11002 2 y H33527 2x H11001 1 x H11002 2 SECTION 2.7 DERIVATIVES AND RATES OF CHANGE |||| 143 DERIVATIVES AND RATES OF CHANGE The problem of finding the tangent line to a curve and the problem of finding the velocity of an object both involve finding the same type of limit, as we saw in Section 2.1. This spe- cial type of limit is called a derivative and we will see that it can be interpreted as a rate of change in any of the sciences or engineering. TANGENTS If a curve has equation and we want to find the tangent line to at the point , then we consider a nearby point , where , and compute the slope of the secant line : Then we let approach along the curve by letting approach . If approaches a number , then we define the tangent t to be the line through with slope . (This mPm m PQ axCPQ m PQ H33527 f H20849xH20850 H11002 f H20849aH20850 x H11002 a PQ x HS33527 aQH20849x, f H20849xH20850H20850PH20849a, f H20849aH20850H20850 Cy H33527 f H20849xH20850C 2.7 (a) How large do we have to take so that ? (b) Taking in Theorem 5, we have the statement Prove this directly using Definition 7. 66. (a) How large do we have to take so that ? (b) Taking in Theorem 5, we have the statement Prove this directly using Definition 7. 67. Use Definition 8 to prove that . 68. Prove, using Definition 9, that . 69. Use Definition 9 to prove that . 70. Formulate a precise definition of Then use your definition to prove that 71. Prove that and if these limits exist. lim x l H11002H11009 f H20849xH20850 H33527 lim t l 0 H11002 f H208491H20862tH20850 lim x l H11009 f H20849xH20850 H33527 lim t l 0 H11001 f H208491H20862tH20850 lim xlH11002H11009 H208491 H11001 x 3 H20850 H33527H11002H11009 lim xlH11002H11009 f H20849xH20850 H33527H11002H11009 lim xlH11009 e x H33527H11009 lim xlH11009 x 3 H33527H11009 lim xlH11002H11009 1 x H33527 0 lim xlH11009 1 sx H33527 0 r H33527 1 2 1H20862sx H11021 0.0001x lim xlH11009 1 x 2 H33527 0 r H33527 2 1H20862x 2 H11021 0.0001x65. ; (b) Graph if and . How long does it take for the velocity of the raindrop to reach 99% of its terminal velocity? ; 60. (a) By graphing and y H33527 0.1 on a common screen, discover how large you need to make x so that . (b) Can you solve part (a) without using a graphing device? ; 61. Use a graph to find a number such that if ; 62. For the limit illustrate Definition 7 by finding values of that correspond to and . ; 63. For the limit illustrate Definition 8 by finding values of that correspond to and . ; 64. For the limit illustrate Definition 9 by finding a value of that corres- ponds to .M H33527 100 N lim xlH11009 2x H11001 1 sx H11001 1 H33527H11009 H9255H33527 0.1H9255H33527 0.5 N lim xlH11002H11009 s4x 2 H11001 1 x H11001 1 H33527H110022 H9255H33527 0.1H9255H33527 0.5 N lim xlH11009 s4x 2 H11001 1 x H11001 1 H33527 2 H20895 3x 2 H11001 1 2x 2 H11001 x H11001 1 H11002 1.5 H20895 H11021 0.05thenx H11022 N N e H11002xH2086210 H11021 0.1 y H33527 e H11002xH2086210 tH33527 9.8 mH20862s 2 v* H33527 1 mH20862svH20849tH20850 James Stewart Stewart - Calculus - Early Transcedentals 6e calculus

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