Calculus Maimus Notes 3.: Properties of Limits 3. Properties of Limits When working with its, you should become adroit and adept at using its of generic functions to find new its of new functions created from combinations and modifications to those generic functions. I ll show you what I mean, but first, some important properties of its that make it all work. Properties of Limits 1. f g f g c c c. f g f g c c c 3. k f k f c c 4. Here s a cool one: f g f g c c c 5. f c g c g c f, g c n 6. f f c c 7. If g c n L, then f ( g( )) f ( ) c L 8. c 3 f ( ) 3 f ( ) c, etc... g ( ) g( ) c c Eample 1: 1. Given that f 8, g c eist, eplain why., c h, find the its that eist. If the it does not c (a) f 3g c Rewriting using the properties, Page 1 of 1
Calculus Maimus f 3 g c c 8 3 16 6 Notes 3.: Properties of Limits The it is the only mathematical operation that can reproduce itself and that can be put anywhere it is needed in order to evaluate an epression. Rewriting a given epression in a more useful, convenient form will always be helpful. After a while, you may become so adept that you will not have to rewrite the epression, but will be able to go directly to the numerical plug-in step. Never combine the numbers in your head and just write down the final answer. Showing step two at the left communicates HOW you are evaluating the epression. When in doubt of how much to show, air on the side of being over-thorough. g (b) c Rewriting using the properties, we get g 4 4 4 16 c Be careful with any constant multiples that also get squared. They can still be pulled out front, but must be done so correctly. Don t forget that squaring a negative number yields a positive number. On an AP Free-Response question, you do not need to simplify your numeric answers, so stopping on line two would be acceptable (as long as the grouping symbols are in place). The Multiple-Choice section, of course, always has simplified, tempting choices. In this course, I epect you to know when you can stop, and then I epect you to keep going to a simplified answer. It s all part of your training for both sections. (c) c 3 f g 4 h When evaluating epressions, obtaining one DNE term or factor pretty much ruins it for all the other terms and factors in the epression. Rewriting using the properties, we get 3 c f g 4 h c c 3 8 4 1 DNE DNE Be careful, remember that is not necessarily DNE. It s indeterminate form that requires more analysis. Page of 1
Calculus Maimus We can also use the properties to evaluate a combination of given functions at different values. Eample : Given the graphs of f and g are given below. Notes 3.: Properties of Limits graph of f graph of g Determine whether the following its eist. If they do, then find the it. (a) f (c) ( ) 3 g( ) Rewriting, we get f ( ) 3 g( ) Pulling values off the graphs, we get 3 8 g( f ( )) 3 (b) 3 g( ) f ( ) Rewriting, we get g( ) 3 3 f ( ) 3 Evaluating, we get 3 1 9 1 (d) g 1 (e) Working from the inside out, we first get f 1, so now we get 3 1 g 1 1 1 f ( ) g( ) 4 (we ll speed this one up a bit) Rewriting, we get g g 1 1 g 1 1 (f) 5 3 f ( ) 1 (this one too!) DNE 5 1 3 DNE (party pooper) Page 3 of 1
Calculus Maimus Notes 3.: Properties of Limits Knowing this handful of properties will enable you to find its of what at first appear to be daunting epressions. They are tools that allow you to cut any problem down to an appropriate size. But wait! There s more. Very often, we must evaluate its involving trig functions, and very often, these its are not easy to evaluate unless we have one of two things: 1) a calculator or ) a couple of trig its memorized. Eample 3: With a calculator, evaluate (a) sin and (b) 1 cos (a) If we tried to use our properties to evaluate this it, we d get, an indeterminate form, which simply means that the it could be 4, -7,, DNE, or any number of other things. We need to figure out graphically and/or numerically. Entering the epression into our calculators and graphing, we get By the evidence, we re convinced that the it is one, that is sin 1. Memorize this!! But what s the reciprocal of one??? Yes one. So 1 also. Know this! sin Page 4 of 1
Calculus Maimus (b) Trying to evaluate 1 cos by direct substitution also yields. Graphing, we get Notes 3.: Properties of Limits We conclude that 1 cos. Memorize this too!! What about? Well the reciprocal of (which is really ) is undefined. So 1 cos 1 DNE. 1 cos The more interesting question is what is epression s numerator, we get cos 1? Well, if we factor out a negative one from the cos 1 1 cos By transformations, this just reflects our graph of 1 cos This does not change the behavior at the origin, so, alas, we can conclude that (the graph shown above) across the -ais. cos 1 In general, if g, then g sin 1 and g g of the behavior of the graphs at undergoing a horizontal dilation. 1 cos. This is due to the preservation g For instance sin 5 1 and 5 1 cos and even 7 sin 7 1 7 We can now use these memorized its in conjunction with our it properties. Page 5 of 1
Calculus Maimus Notes 3.: Properties of Limits Eample 4: Evaluate the following sin (a) Plugging in first, we get. For this one, we need to algebraically split up the numerator in order to be able to use our properties and avoid the same troublesome. We get sin sin 1 sin 1 11 Splitting up terms in a numerator is a common algebraic maneuver. Look for it often. Be careful, though. You cannot split up the denominators (at least not in the same easy manner). The it of a constant is the constant itself, since the graph of a constant is a horizontal line having the same y-value everywhere. You must write the it each and every time you are rewriting the epression UNTIL you get to the step when you plug in, then it s no longer needed. (b) tan Plugging in first, we get. This means now that we need to rewrite tangent in terms of sine and/or cosine. The ratio identity, sin tan, does a nice job of this. cos sin cos cos sin cos sin cos sin 11 1 From step one to step two, we clean up the comple fraction by multiplying by the reciprocal of the fraction in the denominator. This is another common algebraic maneuver. Look for it often. Here our goal was to regroup our factors to resemble one of the two trig forms we memorized. Page 6 of 1
Calculus Maimus Notes 3.: Properties of Limits (c) sec 1 sec The variable here is theta instead of, no problem. Plugging in first, we that we get. Now we must try to write the epression in terms of sine and/or cosine. The reciprocal identity of secant, 1 sec, will do the trick, along with a little algebra. cos 1 1 cos 1 cos 1 1 cos cos 1 cos cos 1 cos Remember the quickest way to clean up a messy comple fraction is to multiply by the LCM of all the miniature denominators over the LCM. The original epression was just the one we had memorized, but in a very clever disguise. (d) 4sin 5 cos Plugging in, we get. Since it s already in terms of sine and cosine, we try to split up the terms and/or factors to resemble one of the two trig it forms we memorized. 4 sin 5 4 sin 5 5 4 sin 5 5 cos cos 5 cos 5 1 415 The angle behind the trig function dictates the form that we need. Since it was 5, we needed a factor of 5 in the denominator. We already had the, so we just multiplied by a clever form of one, namely 5, then associated the 5 we needed 5 with the in the denominator. This is a pretty common maneuver. Look for it. I don t think you can have just one bulleted point, so here s the second one. Page 7 of 1
Calculus Maimus (e) tan 3 sin 4 Notes 3.: Properties of Limits Plugging in, we get. Like before, we first write the epression in terms of sine and/or cosine. Our angles are 3 and 4, so we ll have to multiply by clever forms of one to introduce what factors we need. In this case, we are not only in need of the constant multiples, but the variable as well. tan 3 sin 3 1 3 sin 3 4 sin 4 cos3 sin 4 cos3 3 4 sin 4 1 sin 3 4 3 4 cos3 3 sin 4 111 3 3 4 4 This is a really good one that is a lot of fun. They are pretty common, so you should get used to them. In line four, our divided out. If you want, you only need to introduce a single, but I think it s easier to create what you need one factor at a time, regroup, then simplify what s left over. (f) 3 cos 1 Once again, we get by direct substitution. In this case, if we split up the epression as 3 cos 1, direct substitution gives (from above, the second it gives the reciprocal of zero, 1, which means vertical asymptote which means infinity.) It turns out that is another indeterminate form (zero and infinity are reciprocals of each other!!). In a case like this, we need to try something even more clever than before, like multiplying by cos 1, the conjugate of the denominator cos 1 over itself, a very clever form of one. 3 cos 1 cos 1 cos 1 3 cos 1 cos 1 3 cos 1 sin 3cos 1 sin sin 6 1 1 6 Page 8 of 1
Calculus Maimus Notes 3.: Properties of Limits All that work with trig functions and its is making me hungry. How about you? If you re telling yourself that you could sure go for a sandwich right about now, then....well, then you re out of luck. But I have the net best thing: a sandwich theorem! A Limit Sandwich (called the Sandwich (or Squeeze) Theorem) IF g f h for all, ecept possibly at c, and IF g L h c f L, THEN c c You ve probably lost your appetite just reading that, right! No problem, you ll be eating again in no time. All the theorem really says is that if you have a function that is everywhere contained between two other functions, then at a point where the two outer functions are sandwiched or squeezed together through a single point, then the one in between them must pass through that point as well. Think of it as thousands of concert goers from all over the stadium leaving the concert at the end of the night through a single turnstile. Here s a visual h( ) f ( ) sin 1 In this graph, 1 sin for all ecept possibly at. Since and, we 1 know that sin too! g( ) Another way to think about it is that h and g are the bread, and f is the salami. g( ) h( ) f ( ) I know what you re saying, or thinking: How and the heck do I go about finding such functions that bound the one I m interested in? Where do we get the bread? What if I don t even like bologna? Those are great questions! To answer the first question, sometimes, they can be difficult to find. Other times, they magically appear through algebraic manipulations. In fact, we ve already done this. Here s and eample: Page 9 of 1
Calculus Maimus tan? tan sin sin 1 sin 1 11 1 cos cos cos Notes 3.: Properties of Limits Here, finding the desired it of tan at zero, we stumbled across two other functions that contain our desired function that both have its of one; therefore, we know our desired it must be one. Here s the graph. 1 h( ) cos tan f ( ) sin g( ) Here, we didn t really need to know what was going on graphically behind the scenes. The Squeeze Theorem worked its magic surreptitiously. Generally, though, you ll be asked to find the it of a function at a point, and you will be given information about the ones bounding it. Eample 5: If 3 g( ) 3, find g( ). 1 We are told that some function g lies between two other functions. Since we re interested in the it of our unknown, bounded function at 1, it suffices to show what the its of the two outer functions at 1 are. We will then know that the it we desire must lie between the other two. Hopefully, the its of the outer functions at 1 are the same, and we can then make a definitive determination about g. Because our two outer functions are continuous everywhere, and thus at 1, we can evaluate the its by direct substitution. 1 3 3 1 3 1 3 1 1 3 11 3 3 Therefore g( ) 3 too! 1 For a problem like this, it is important to clearly communicate that you are evaluating the its at the two outer functions, then drawing a conclusion based on those results. The word therefore does a good job of showing this conclusion. Using the word too is not necessary, but it does help communicate your results emphatically. Words such as also and as well also work as well in the place of too. Page 1 of 1