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26Limit and Continuity of Piecewise-defined Functions When we write f(x) = |2x + 5| as f(x) = , the new format is called a piecewise-defined function because it has two (or more) pieces instead of one. And usually, when we have a piecewise-defined function, we will be asked to find the limit at the cutoff numbers (from one piece to another piece) and to determine whether the function is continuous at those numbers. For example,Ex 1. Ex 2.xyh/dxyh/dxyh/dyxOyxOxyh/dxyh/dlimx2 f(x) = ___Is f continuous at 2? ___limx–2 f(x) = ___Is f continuous at –2? ___limx1 f(x) = ___Is f continuous at 1? ___Note:When we sketch the graph of a piecewise-defined function first, of course we can tell the limit and continuity at a particular number. However, this approach often takes longer than it really should. We also can tell the limit and continuity of a piecewise-defined function at any number without any graphing—just do it algebraically.
27Limit and Continuity of Piecewise-defined Functions (cont’d) Recall that to determine the limit at a number a, we must consider both __________________ at a. To determine whether the function is continuous at a, not only the limit at a must exist, but also i. _____________ and ii. _____________. This is how we are going to determine the limit and continuity of a piecewise-defined function algebraically.Ex 1. Given . Find the limits at x = –1 and 3 if they exist. Also determinewhether f is continuous at –1 and 3 or not.Note: Most piecewise-defined functions have the property that each piece is nice, i.e., each piece itself is continuous. Therefore, we can use direct substitution to find the one-sided limits at a cutoff number as well as its function-value (of course, we have to determine which piece should be used for the “plug-it-in”). Direct substitution is also feasible when the number in question is not a cutoff number and if the piece contains that number is nice. For example, use the function f above to find the limit at x = 1 and determine its continuity:_______________________________________.This brings us to a classic problem when the piecewise-defined function contains a piece which is not so nice: Find the value of k so that f(x) is continuous at x = 2.
28Limits Involving Trigonometry We haven’t done a lot of limit problems involving trigonometry, since we need to recall some of the basic trig. identities and the two limits on the right. The second one really can be derived from the first one, therefore you only need to know limx0 (sin x)/x = 1. Let’s see why the limit is 1 (using the tabular method and a calculator of course, however, make sure your calculator is in radian mode).1.2.If you compare the 1st row (x) and the 2nd row (sin x), you can see the values are really close to each other, and the ratio between two quantities approximately the same is approximately equal to __.x–.1–.01–.0010.0010.010.1sin xOne thing we have to be careful is that limit of (sin x)/x is not always 1, because it also depends on the number x is approaching. For example, the two limits on the right will not be 1.1.2.On the other hand, many cases are equal to 1, provided that it is of the formand a, the number x is approaching, makes the expression = 0. This is what I mean:How isderived from ?If we have where expression 1 and expression 2 are not thesame but if the value of a makes both expressions = 0 (for example, ),the limit will almost never be 1. Nonetheless, most of the time, the limit does exist.
29Limits Involving Trigonometry—cont’d If we have where expression 1 and expression 2 are not the same butRecall:if the value of a makes both expressions = 0 (e.g., limx0 (sin x)/(2x) and limx0 (sin 2x)/x),we will, with proper algebra, make expression 1 and expression 2 exactly the same. This is what we mean:1. 2.Another way to evaluate:3. 4.As we can see from example 2 above, we can do either way—either make x (in denominator) to become 2x or expand sin 2x as 2 sin x cos x. On the other hand, though the expansions for sin 3x and sin 4x (as in examples 3 and 4) exist, they are not familiar to most of us (and they are also much harder to derive). Therefore, the expansion for sine of an expression is almost never a good way to evaluate these limits since the better way should be: Use proper algebra to make theMake expression 2 the same as expression 1 if you see:Make expression 1 the same as expression 2 if you see:expression without the sine to be exactly the same as the expression with the sine (see above). Using this method, it allows us to tackle some much harder limits:5. 6.
30Application of limx0 (sin x)/x = 1 Q: If , what is ?A: __ (and in general, if , then ___)Knowing this, it allows us to evaluate limits with sine of an expression in the denominator.1. 2.Recall:Make expression 2 the same as expression 1 if you see:Make expression 1 the same as expression 2 if you see:In previous slide, we are only saying that don’t try to expand sine of expression using trig. identities. We are not saying that we never need to use trig. identities because sometimes we do need them, especially for this one:Why not just keep in mind that:andKeeping the above limit in mind, it allows us to evaluate the following limits:
31Know Your Basic Trig. Limits: The box on the right shows the six basic trig. limits we should know:The limits in the 2nd and the 4th one are obtained by reciprocating the limits in the 1st and the 3rd one, respectively. And reciprocating 1 is still 1!The limit in the last one is obtained by negating the limit in the 5th one. And negating 0 is still 0!On the other hand, what are the following limit:The 6 trig. limits we should keep in mind:Note: If the sign of the infinity is of no concern, we can say the 3rd limit is __ instead of ___ (After all, 1/0 is either + or –).Examples:Four more ExamplesThe four examples below are not previously mentioned. The first three limits have to do with trig. powers (we can even make a generalization of these limits), while the last one requires a totally different method (well, we kind of mention it—hint: it’s on this slide).