Bob Brown Math 251 Calculus 1 Chapter 2, Section 3 Completed 1
CCBC Dundalk
Properties of Limits
Theorem:
1) Suppose that the limit of f(x) exists at x = c and that b is a constant.
Then the limit of exists at x = c and equals (limit of f(x) at x = c).
2) Suppose that the limit of f(x) exists at x = c and that the limit of g(x) exists at x = c.
Then the limit of the sum (or difference) of f(x) and g(x) exists at x = c and equals the sum (or difference) of the limits.
3) Suppose that the limit of f(x) exists at x = c and that the limit of g(x) exists at x = c. Then the limit of the product of f(x) and g(x) exists at x = c and equals the product of the limits.
4) Suppose that the limit of f(x) exists at x = c and that the limit of g(x) exists at x = c (and does not equal 0.) Then the limit of the quotient of f(x) and g(x) exists at x = c and equals the quotient of the limits.
5) Suppose that the limit of f(x) exists at x = c and that n is a positive integer. Then the limit of exists at x = c and equals the limit of f(x) at x = c raised to the n power.
Numerical Illustration of Property 3
We will illustrate Property 3 with a table. (An example that verifies a property or theorem is not a proof.) We could illustrate any of the other properties in a similar way.
Let f(x) = x + 1 and let g(x) = x2. We will consider the limit as x approaches 3. (Note that some y-values may be rounded.)
f(x) = x + 1 / / f(x)g(x) =2.9 / / /
2.99 / / /
2.999 / / /
3.1 / / /
3.01 / / /
3.001 / / /
As x à 3,
f(x) à / As x à 3,
g(x) à / As x à 3,
f(x)g(x) à
= = =
Two Basic Limit Properties
Theorem:
Let k be a constant. Then = / =Example: = / Example: =
Exercise 1: Explain how the properties of limits are used to justify the following calculation: = 2. (First, use a table to verify the limit value.)
= by Property 4
= by Property 2
= by Property and Property
= using both Basic Limit Properties
= = =
Observation: It seems like we could have arrived at the value of the limit more easily, by directly substituting 4 in place of x. Verify that this substitution does render the correct value of the limit.
=
In general, direct substitution will be our first strategy for evaluating a limit. We will see that direct substitution will produce one of three possible outcomes: (i) the actual limit value, (ii) an answer suggesting that the limit does not exist, or (iii) an answer, called an indeterminate form, that does not reveal whether or not the limit exists, telling you to try some other strategy for evaluating the limit.
Exercise 2: Evaluate , a problem that we first encountered in Handout 2.2.
First Strategy: Direct Substitution
= =
Second Strategy: Algebra
= = = = h(x)
Question: What are the “differences” between and h(x) = 3x + 6?
g(x) / h(x)domain / /
graph /
/
The Point of Exercise 2: These two functions are indeed different functions, as evidenced by their different domains and different graphs. However, they have the same behavior near x = 2.
(*) YET =
This observation suggests an algebraic strategy for evaluating some limits.
Exercise 3: Evaluate algebraically, and check your answer with a table.
Question: Why are we justified in dividing out (“cancelling”) the (x + 5) factors?
a. It’s a right protected by the Constitution of the United States of America.
b. All of the world’s major religions agree that cancellation is a moral duty (and that Spam is not edible meat.)
c. You paid your hard-earned money for this class, so you get to say who or what can or can’t be cancelled.
d. Math has so many rules—this just has to be one of them.
e. Since the limit involves values of x near -5 but not equal to -5, we are therefore not dividing by -5 + 5 = 0.
Exercise 4: Evaluate algebraically, and check your answer with a table.
Three “Special” Limits
The limits calculated in Exercises 5, 6, and 7 should be committed to memory.
Exercise 5: Use a table to evaluate the following limit: =
Exercise 6: Use a table to evaluate the following limit: =
Exercise 7: Use a table to evaluate the following limit: =
Note: Students sometimes assume that no matter what c is. This is not true; don’t make that assumption. See Exercise 8 for an example.
Exercise 8: Use direct substitution to evaluate .
Exercise 9: Use the result of Exercise 5 to compute .
= =
= = = =
Exercise 10: Use the result of Exercise 5 to compute .
= = =
Let u = 5x. We see that as x à 0 , u à
Then = = = =
Squeeze Theorem
Theorem: If for all x in an open interval containing c, except possibly at c itself, and if = L = , then = .
Assume that = L and that and = L
Then if , it follows that = L also.