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Why [0/0] and [ – ] Are Indeterminate Forms Whereas k/0 Isn’t?
Page 20 [0/0] is an indeterminate form because a limit of this form is unpredictable. If you go back to the 8 problems on last two pages, each one is [0/0] and each one has a different answer. That means, it can be anything from a constant (zero or non-zero) to infinity (positive or negative) and it can be DNE also. Whereas, the limit of the form k/0 (k  0) is very predictable since it only can be ___________ (and can’t be anything else). Therefore, the (true) meaning of a limit in an indeterminate form is: A limit that can’t be done by direction substitution nor can it be predicted using direction substitution. Now let’s look at why [ – ] is also an indeterminate form. Of course, one might wonder: Is it just 0? So let’s discuss it from Set Theory point of view. What is the difference between the following sets? How is it related to limits? 1. {1, 2, 3, 4, 5, …} – {1, 2, 3, …} = 2. {1, 2, 3, 4, 5, …} – {4, 5, 6, …} = 3. {1, 2, 3, 4, 5, …} – {2, 4, 6, …} = As you can see from the above, [ – ] is also as unpredictable as [0/0]. We will present you some examples in limits:

More on the Indeterminate Form of [∞ – ∞]
Note: Most limit problems of the form [ – ] really come from [k/0 – k/0] (recall k/0 is either + or – on page 10). On the other hand, [k/0 – k/0] does not necessarily yield [ – ]. Let’s consider the following (where +k denotes a positive nonzero constant and –k denotes a negative nonzero constant): [k/0 – k/0]  [ – ]? Comment

Calculating Limits Using the Limit Laws
Limit Laws Suppose that c is a constant and the limits lim f(x) and lim g(x) exist. Then 1. 2. 3. 4. 5. provided that lim g(x)  0 xa xa Constant Multiple Law Sum Law Difference Law Product Law Quotient Law xa Applications: Given: limx1 f(x) = 2, limx1 g(x) = –3, limx2 f(x) = 4 and limx2 g(x) = 0. Evaluate the following: 1. limx1 [f(x) + g(x)] 2. limx1 [2f(x)  3g(x)] 3. limx2 [f(x)g(x)] 4. limx2 [f(x)/g(x)] 5. limx2 [g(x)/f(x)] Note: These problems are conceptual and intangible since neither the graphs nor the equations of the functions are given. Nevertheless, they are easy. Even without any of these limit laws, when you are given limx1 f(x) = 2 and limx1 f(x) = –3, of course you are going to evaluate limx1 [f(x) + g(x)] as 2 + (3) = 1 (what else can it be?)

Calculating Limits Using the Limit Laws (cont’d)
Limit Laws Suppose that the limits lim f(x) and lim g(x) exist. Then 6. 7. if n is even, lim f(x) 0 xa xa Power Law Root Law xa Applications: I. Given: limx1 f(x) = 2, limx2 f(x) = 8 and limx3 f(x) = 4. Evaluate the following: 1. limx1 [f(x)]4 2. limx2 3. limx3 II. Evaluate: limx3 (2x + 4)5 Use the graph and limit laws to evaluate the following limits: y x O 1. limx1 [f(x) + g(x)] 2. limx3 [2f(x)  3g(x)] 3. limx5 [f(x)g(x)] 4. limx3 [f(x)/g(x)] 5. limx4 [f(x) + g(x)] g f

If f(x)  g(x)  h(x) when x is near a (except possibly at a) and then
The Squeeze Theorem The Squeeze Theorem If f(x)  g(x)  h(x) when x is near a (except possibly at a) and then g f L h a Figure 9 In layman’s terms, the Squeeze Theorem says if g(x) is squeezed in between f(x) and h(x) near a, and if f and h have the same limit L at a, then g is forced to have the same limit at a also (see Figure 9). The Squeeze Theorem, a.k.a. the Sandwich Theorem or the Pinching Principle, allows us to evaluate limits that are not in any previous prescriptions: Example: Use the Squeeze Theorem to show that y = x2 y = x2 sin (1/x) Note: Notice that limx0 sin (1/x) = sin () = DNE (see graph of y = sin x below), therefore, limx0 x2 sin (1/x) = limx0 x2limx0 sin (1/x) would be 0DNE, which is a special format not mentioned previously. y = –x2 y = sin x 1 –1

Limits Involving Absolute Value
We first saw a problem involving absolute value on page 12 (problem 3 to be exact). Since every problem on that page is evaluated using the tabular method. The question is: How do we evaluate limit problems involving absolute value without using the tabular method? The answer is: Algebraically, with a split consideration. Recall the basic algebraic definition of |x|: The fact is: where c is called the cutoff number. To determine this cutoff number, we set the expression = 0 and solve for x. Example: since 2x + 5 = 0  x = Examples: (See above) Q: How does this relate to limit problems involving absolute values? A: When the function involves absolute value, try to rewrite it without absolute value. Note: Not all limit problems involving absolute values require the split consideration. For example, we do not need to (and we don’t) split the absolute-value expression into two non-absolute-value expressions when direct substitution is sufficient or when it yields k/0 (on the other hand, we do need to split it when direct substitution yields 0/0). Examples: 1. limx–2 |2x + 5| = 2. limx–2 3/|2x + 4| =