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Limit 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. x y h/d x y h/d x y h/d y x O y x O x y h/d x y h/d limx2 f(x) = ___ Is f continuous at 2? ___ limx–2 f(x) = ___ Is f continuous at –2? ___ limx1 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.

Limit 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 determine whether 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.

Limits 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 limx0 (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 –.001 0.001 0.01 0.1 sin x One 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 form and a, the number x is approaching, makes the expression = 0. This is what I mean: How is derived from ? If we have where expression 1 and expression 2 are not the same 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.

Limits Involving Trigonometry—cont’d
If we have where expression 1 and expression 2 are not the same but Recall: if the value of a makes both expressions = 0 (e.g., limx0 (sin x)/(2x) and limx0 (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 the Make 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.

Application of limx0 (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: and Keeping the above limit in mind, it allows us to evaluate the following limits: