MATH 250 TOPIC 11 LIMITS. A. Basic Idea of a Limit and Limit Laws. Answers to Exercises and Problems

Transcription

1 Math 5 T-Limits Page MATH 5 TOPIC LIMITS A. Basic Idea of a Limit and Limit Laws B. Limits of the form,, C. Limits as or as D. Summary for Evaluating Limits Answers to Eercises and Problems

2 Math 5 T-Limits Page A. Basic Idea of a Limit and Limit Laws In Calc I, you undoubtedly had problems similar to the following. Let f() = +. Find f(). If you said the answer is 3, you are correct. How did you get it? You probably plugged in for! However, is never supposed to equal. Instead, you are really interested in what f() is doing as gets closer and closer to. Let s review what is happening in this it. A it typically has the form f(). Notice that there are two things moving or changing. The input variable is moving closer to a. As the input changes, the output f() also changes. So now you have a bunch of outputs f(). Are they getting close to some value? You know is moving arbitrarily close (but never equal) to a. The question is, are the outputs f() moving arbitrarily close to some number? Do you epect the outputs to arrive at some value? If you do, then this value is the it. Once again. You know where is going. Can you say without a doubt where you epect f() to go? If you can, then this value is the it. Eample A.. Find ( + 3). As gets close to (written ) from the right side or the left side, we epect ( + 3) to get close to 5. To see that this is really happening, consider the following table f() Table A. We see that the closer gets to, the closer f() getsto5.thus, ( +3)=5.

3 Math 5 T-Limits Page 3 Notice in Table A. that we substituted values for on both sides of. Remember: (i) (ii) (iii) ( ) means is allowed to assume values on either side of. +( ) means is always greater than and is coming down to from the right. ( ) means is always less than and is coming up to from the left. Important Fact from Calc I: f() eists if and only if f() and f() both eist and are + the same. This means that if f() f() or if one of these its doesn t + eist, then f() doesn t eist. Eample A.. Consider the following graph. 3 3 When the input moves along the -ais, the function f() travels along the graph. As gets close to, you epect that the f() ory values on the graph will tend to. So, f() =. Of course, f() = 3, which f(). However, this is irrelevant as far as the it is concerned. The it is the value you epect f() to hit.

4 Math 5 T-Limits Page 4 Aside: The fact that f() f() is etremely relevant for the concept of continuity. Are any bells ringing? Practice Problem A.. In Eample A., find: a) f(); (b) f(); (c) f(); (d) f() 3 Answers As you probably remember, there are it laws from Calc I. Eample. Find ( +sin). π Answer: π +sin( π) π = +. To do this problem, you (maybe subconsciously!) used the fact that π ( +sin) = π +. π In general, we have the following rules from Calc I. Suppose f() and g() both eist. Then: (f() ± g()) = f() ± g() (A.) (cf()) = c f(), for any constant c. (A.) (f()g()) = f() g() (A.3) ( ) f() f() = g() g(), provided g(). (A.4) Use the above rules to evaluate the following its. Practice Problem A.. Evaluate: + a) ; (b) + π 4 +cos ; (c) (7 )ln Answers

5 Math 5 T-Limits Page 5 B. Limits of the form,, Eample B.. Let f() =. Find f(). The answer is. You epect the output values of f() to get arbitrarily close to. Is there anything wrong with getting? No! Is this a valid answer? Certainly! Eample B.. Let f() = +. Find f(). The answer is still, since now you epect that the values of f() will tend to, which is. However, consider the following eample. Eample B.3.. As gets close to, your first thought is that the it is getting close to, which is undefined! What happens now? Consider another eample. Eample B.4. Find. As, your first thought is that the it is. How do we handle this? Notice that the its in Eamples B., B.3, and B.4 have the form,,and, respectively. Let us consider each form separately.

6 Math 5 T-Limits Page 6 Form Like Eample B., its having the first form zero. are easy. The it is Eample B.5. Find π As π, cos. cos (form ), which is. Eercise B.. Evaluate the following its. a) 3 b) e ln( +) Answers Form Net let s analyze Eample B.3, where the it had the form ( ). The function in this eample is f() = bottom zero but not the top (form ).. The input = makes the This means the function has a vertical asymptote at =. (See Math 5, Review Topic 6, for an eplanation of vertical asymptotes.) Thus, as gets close to from the right or the left, the outputs will become unbounded. (This always happens at a vertical asymptote!). To see what the its are

7 Math 5 T-Limits Page 7 from either side, we draw a quick sketch of f() = around =. Thus, + =, and right and left are different, = =. Since the its from the does not eist. Remember! Eample B.3 is typical of it problems f() which have the form. The answer is either,, or does not eist. Since = a is a vertical asymptote, as a from either side the outputs will become unbounded. You must check both sides to see if the outputs are positive (heading toward ) or negative (heading toward ). Eample B.6. Find tan. π The function tan = sin cos.at = π, this has the form = (form ). Therefore = π is a vertical asymptote for tan, which means the function becomes unbounded as approaches π from the left. To see if the it is + or, consider values of slightly less than π. The sine will be positive and the cosine will also be positive. Therefore tan will be positive. Thus, tan =. π

8 Math 5 T-Limits Page 8 Note: Eample B.6 also follows from looking at the graph of the tangent function. Eercise B.. Evaluate the following its. a) b) π 3 cos Answers Form When a it has the third form, like Eample B.4, this always means more work. You can t conclude anything as the problem stands, and any answer is possible. You ll have to do some manipulation (or pull some other trick out of your math hat!) to change the starting epression into an equivalent form that lets you see what is happening. We will refer you to your calculus book to read the steps involved in showing =. Eample B.7. ( ) Eample B.8. Find. 4 4 = (See your calculus book for details.) This has form, so more manipulation is needed. Here we multiply top and bottom by +. We are performing legal algebra and changing the starting epression into an equivalent one that lets us see what is happening. + 4 = 4 4 = + ( 4)( +) =. + ( So, 4 ) ( ) = =

9 Math 5 T-Limits Page 9 [f( + h) f() ] Eample B.9. Find, where f() =. Well, h h f( + h) =( + h) = +h + h. This means [ ] [ f( + h) f() +h + h ] =. h h 4 h This has the form. Here though, the required algebraic manipulation is straightforward. [ ] [ ] [ ] f( + h) f() h + h h( + h) = = h h h h h h = [ + h] =. h Note: Taking the it in any difference quotient as it stands will yield the form. Manipulating it to find the it is the same as calculating the derivative of f using the formal definition of a derivative (Review Topic A). Practice Problems. Find the following its. B.. B.. cos Answer B.5. cos Answer Hint: multiply by (cos +). cos + Answer B.3. B f( + h) f() Answer B.6. h h Answer, where f() =. Answer

10 Math 5 T-Limits Page C. Limits as or as In many situations in Calc II, you will have to evaluate its where or. Let s do some eamples. Eample C.. Find f(), where f() = +. Remember our intuitive definition of a it. We know is getting large. Is there anything we can say about what we epect for f()? Yes. Since f() = +, we epect f() to get large as gets large. Thus, f() =. Sometimes when processes compete for control, a it may appear indeterminable. Eample C.. Find ( 3 ). This problem has the form. There are two competing processes going on, so to speak. Which one wins? Does the first override the second, does the second override the first, or do they offset each other in some fashion? Once again we must do more work or manipulation to see what is happening (and any answer is possible!). We proceed as follows. ( 3 ) = ( ). Now we have a product of an arbitrarily large positive number and an arbitrarily large negative number (since ( ) is negative). We epect that ( ), which is the answer. Thus, ( 3 ) = ( ) =. Eample C.3. Find + 3. This has the form which (like ) really means two competing processes. Does the numerator win or does the denominator, or do they counterbalance each other in some way? Again, we must manipulate the epression to

11 Math 5 T-Limits Page see what is happening. The following procedure is a good tool for problems of this type. In it problems involving quotients of powers of where or, factor out the highest power of in the numerator and the highest power of in the denominator. (The powers of may not be the same.) Doing this for Eample C.3 yields: + ( + 3 = ) ( + ( 3 = ) ) ( 3 = ). Note: and 3 both as. Factoring is also useful in problems like the following (which you will encounter when studying infinite series). Eample C.4. Find n [ n ]. n Once again, this has the form which means more manipulation is needed to determine the outcome. ( n ( ) ( ) ) n = n n n n ( = n n) = =. n

12 Math 5 T-Limits Page Practice Problems. Evaluate the following its. C C.4. ( 3 +) /3 C Answers C.5. C.3. ( + ) C.6. (e + e ) Answers D. Summary for Evaluating Limits Consider the general epression f().. Remember that you know a. The question is, does f() get arbitrarily close to some value? If it does, then this value is the it. (See Eamples A., B., and Problem A..) p(). Many its involve quotients of the form q() are continuous functions. where p()andq() If the quotient at = a has the form: a) b), it is. (Eamples B., B.5), f() has a vertical asymptote at = a, which means f() becomes unbounded as a. The answer is either,, or does not eist. Evaluate f for values of close to a and on either side of a to see which answer applies. (Eamples B.3, B.6)

13 Math 5 T-Limits Page 3 c) or, you must do more manipulation to get an equivalent epression from which you can interpret the answer. (Eamples B.7, B.8, C.3, C.4) Note: In Calc II you will learn L Hospital s rule which is a great tool for analyzing it problems of the form or. 3. There are indeterminate forms such as where two operations work against each other or compete for control. (You will encounter more indeterminate forms in Calc II.) As above, you must manipulate the starting epression into an equivalent one that allows you to see what is happening. (See Eample C., Problem C.3) Beginning of Topic Review Topics 5 Skills Assessment

14 Math 5 B Eercises Answers Page 4 B.. a) 3 = = = ( ) Form e b) ln( +) = ln = ln = ( ) Form Return to Review Topic B.. a). This has the form ( has a vertical asymptote at = and thus becomes unbounded as approaches from either side. As, is negative. Thus, =. b) π 3. This has the form cos ), which means the function sin π 3 cos π 3 = 3 = 3 ( ) form, which means = π 3 is a vertical asymptote. As π 3, ( cos ) is positive (why?!) and is positive. Thus π cos =. As π +, (cos ) is negative 3 3 and is positive. This implies π + 3 cos the its from the left and right are not equal, π 3 does not eist. =. Since cos Return to Review Topic

15 Math 5 A Problems Answers Page 5 A.. These answers come from studying the graph of f() in Eample A.. a) f() = b) f() = 3 + c) f() =3 3 d) Based on b) and c), 3 f() does not eist. A a) Using A.4, + = + = 3 b) Using A.4 and then A. on the denominator, π 4 +cos = = π/4 ( +cos) π/4 π/4 + cos = π/4 π/4 π 4 +. c) Using A.3, (7 )ln = (7 ) ln =5ln

16 Math 5 B Problems Answers Page 6 Find the following its. B.. cos = sin() cos () B.. has form. Try factoring. = ( )( +) ( ) = ( +) = ( )( +) ( ) = ( +)= B.3. + has form 4 ( ) +. This means the function has a vertical asymptote at =. This further means that + becomes unbounded as + and as. We must check both sides. As + + As + + does not eist. is positive, and so + + =. is negative, and so + =. Thus, B has form. Let s manipulate as follows = = 3 = 3 3 = 9. (3 ) (3)( 3)

17 Math 5 B Problems (Continued) Answers Page 7 B.5. cos has form =. This one is tricky. Use the hint and also remember Eample B.7. cos cos = = sin (cos +) = = cos + cos + = cos (cos +) ( ) cos + = ( ) cos + = =. NOTE: In Calc II you will learn L Hospital s Rule which makes this problem easy. B.6. Let f() =,orf( ) = ( ). Then, f( + h) f() h h = h = h + h h = h h h( + h) = h ( + h) ( + h) h ( + h) =. So, we have just proved the derivative of is.

18 Math 5 C Problems Answers Page 8 C Since, let s factor from the numerator and 3 from the denominator (the highest power possible in each case). Thus, + ( + ) = 3( ) 3 ( ) + = ( ). 3 Note that ( + ) and( ) 4as. This means ( ) + ( ) = 3 ( + ) ( ) =. NOTE: For the quotient of two polynomials, if and the power of in the denominator is greater than the power of in the numerator, the it is. C = 3( + 3 ) ( 3 ) = 3/ + 3 ( 3 ) = Thus, + 3 ( 3) +.Now and( 3 ) as ( 3) =.

19 Math 5 C Problems (Continued) Answers Page 9 C.3. ( + ) has the indeterminate form, which means that we must manipulate. ( + ) = ( + ) ( +) ( ) = = ++ ( ) Since +and both get arbitrarily large as, ( + ( ) ) = =. ++ ( ). ++ ( 3( + 3 ) ) /3 C.4. ( 3 +) /3 = ( + ) /3 = 3 ( ) /3 = + =. 3 C.5. Find. No matter how large becomes, oscillates between and. It does not stay arbitrarily close to any single value. Therefore, does not eist. C.6. As gets large, e = e gets small since e gets large. Therefore (e + e ) gets large as gets large. This means (e + e )=.