2 3 Calculating Limits Using the Limit Laws

2. 3 Calculating Limits Using the Limit Laws: In section 2. 2 we used graphs and tables of values to determine the values of limits. In this section, we use properties of limits to calculate limits analytically. Direct Substitution: Substitute the value that x approaches into the function. 3 -4 4 Basic Limits: b c cn

1. The limit of a sum is the sum of the limits. 2. The limit of a difference is the difference of the limits. 3. The limit of a constant times a function is the constant times the limit of the function. 4. The limit of a product is the product of the limits. 5. The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not 0.

EX #1: Use the Limit Laws and the graph of f and g to evaluate the following limits, if they exist. = = 1 + 5(-1) = - 4 = 2 (DNE) = DNE = = 1. 4 0 = DNE

EX #2: The domain is all real numbers, so direct substitution can be used.

EX #2: The domain is all real numbers, so direct substitution can be used.

When using substitution… * if ____occurs, the limit DOES exist try _________ simplifying re-substituting. and __________ Methods of Simplifying: 1. Factoring 2. Rationalizing * if _____occurs, the limit DOES NOT exist (vertical asymptote at x = c)

What if there is no canceling after you factor? When the numerator and denominator both approach zero, the limit may be finite or infinite value.

Double check. EX #3: DNE Substitution gives us 1 a. Left of 1, 0. 9 (0. 9)3 +1, Positive. 1 a. Right of 1, 1. 1 (1. 1)3 +1, Positive. 2 a. Left of 1, 0. 9 (0. 9) – 1, Negative 2 b. Right of 0, 0. 1 (1. 1) – 1, Positive.

EX #5: substitution: try simplifying numerator:

EX #6: substitution: Multiply by the conjugate. try simplifying numerator:

Recall that Right side. Since | x | = x for x > 0, we have Left side. . For x < 0, we have | x | = – x , so Therefore, by Theorem 1, . .

Using previous facts that | x | = x for x > 0 and | x | = – x we have for x < 0, We can make the following changes… Right side. Left side. Since the left and right sides are different, it follows from Theorem 1 that the limit does not exist.

EX #9: If determine whether exists. Right side. Left side. Since the left and right sides are equal, a limit does exist and .

The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the Pinching Theorem. It says that if g (x) is squeezed 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 L at a.

Let f (x) = 2 x and h (x) = x 4 – x 2 + 2, so f (x) < g (x) < h (x). Since , then .

We can’t use the Product Law, because does not exist. We will have to apply the Squeeze Theorem, and will need a function f smaller than and a function h bigger than g such that both f (x) and h(x) approach 0. The sine of any number must range from -1 to 1, so…

Inequalities remain true if we multiply positive values to all the parts. x 2 > 0, so multiply all parts by x 2. The limits of the outer functions will be zero… …and can be our f and h functions.

Gathering all our function, limits, and inequalities… We have satisfied the Squeeze Theorem, so

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