A Basic Principle Consider the it f(x) x a If you have a formula for the function f and direct substitution gives the indeterminate form 0, you may be able to evaluate the it algebraically. 0 Principle for Algebraic Evaluation of Limits (page 103 of the text) Clint Lee Math 112 Lecture 3: Evaluation of Limits II 2/26 Example 9 Algebraic Evaluation for Example 1 Consider one of the its from Example 1 Note that if x = 1 f(x) = x + 1 x 2 1 x 1 x + 1 x 2 1 Clint Lee Math 112 Lecture 3: Evaluation of Limits II 3/26
Continuing Example 9 Thus x 1 x + 1 x 2 1 Clint Lee Math 112 Lecture 3: Evaluation of Limits II 4/26 Factor and Cancel As seen in Example 9 one technique for evaluating a its is to factor and cancel. Factoring and Canceling in Rational Functions For a rational function, if direct substitution gives the indeterminate form 0 0, it is guaranteed that you will be able to factor the numerator and denominator so that the factor that gives rise to the zeroes in the numerator and denominator will cancel. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 5/26
Rationalizing Rationalizing in Algebraic Functions For an algebraic function involving roots, especially square roots, it may be useful to rationalize either the numerator or the denominator as a preinary to factoring and canceling. This is illustrated in the next example. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 6/26 Example 10 Rationalizing the Numerator Compute the value of the it x2 + 9 5 First note that direct substitution gives 0, but the function is not a rational 0 function, so we cannot factor and cancel immediately. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 7/26
Solution: Example 10 Since the expression contains a square root, it is possible that rationalizing will help. Here we rationalize the numerator by multiplying the numerator and denominator by, called the of the numerator. This gives: x2 + 9 5 = Clint Lee Math 112 Lecture 3: Evaluation of Limits II 8/26 Solution: Example 10 continued Now factor and cancel x2 + 9 5 = The first and last expressions above are equal for all x except x = 4. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 9/26
Solution: Example 10 continued Thus x2 + 9 5 = x + 4 x2 + 9 + 5 The it on the right can be evaluated by direct substitution to give x2 + 9 5 = The technique used in this example can be used in certain its of algebraic functions involving square roots, but not all. Do not over use it. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 10/26 Change of Variables in Limits Change of Variables If we know the two its then g(x) = b and f(u) = L x a u b f (g(x)) = f(u) = L x a u b This is a change of variables in the it of f (g(x)) at x = a to a new it in the new variable u = g(x) at u = b. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 11/26
Example 11 An Exponential Limit Given that evaluate the it x 0 2 x 1 x 1 x 2 x 2 x 1 = ln 2 Recall that we evaluated this it numerically in Example 7. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 12/26 Solution: Example 11 First note that 2 x 2 x 1 Then make the change of variables u = x 1 x 1 2 x 2 x 1 This agrees with the numerical calculation in Example 7. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 13/26
Example 12 Three Different Methods Consider the it x 2 As usual, direct substitution yields 0. This is a simple it, but we will 0 evaluate it using three different approaches. (a) (b) (c) rationalizing the numerator factoring and canceling changing variables Clint Lee Math 112 Lecture 3: Evaluation of Limits II 14/26 Solution: Example 12(a) The root conjugate is x + 2. Multiplying top and bottom by this expression gives x 2 for all x in [0, 4) (4, ). Thus x 2 = Clint Lee Math 112 Lecture 3: Evaluation of Limits II 15/26
Solution: Example 12(b) Note that the denominator can be written as a difference of squares and factored as so that x 2 Clint Lee Math 112 Lecture 3: Evaluation of Limits II 16/26 Solution: Example 12(c) Make the change of variable u = x. x 2 The three methods used in this example illustrate the approaches that can be taken in most its involving algebraic expressions. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 17/26
The Squeeze Theorem Theorem (The Squeeze Theorem) If f, g, and h are functions for which for all x in some open interval containing a, and then Clint Lee Math 112 Lecture 3: Evaluation of Limits II 18/26 The Squeeze Theorem Comments on the Squeeze Theorem The Squeeze Theorem is a generalization of the principle stated at the beginning of this lecture. It states that if the function g is squeezed between an upper function h and a lower function f, and, if h and f have a common it value at a, then g must have the same it value at a as h and f. It is also true for one-sided its. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 19/26
A Basic Trigonometric Limit Using the Squeeze Theorem We will use the Squeeze Theorem to prove that θ 0 sin θ θ = 1 Proof: First note that the function f(θ) = sin θ θ f( θ) is, since Clint Lee Math 112 Lecture 3: Evaluation of Limits II 20/26 A Basic Trigonometric Limit (proof continued) Thus and So that and sin θ θ 0 + θ θ 0 θ 0 exists sin θ θ θ 0 sin θ θ sin θ θ Clint Lee Math 112 Lecture 3: Evaluation of Limits II 21/26
A Basic Trigonometric Limit (proof continued) For the remainder of the proof consider the angle θ in a unit circle defining two triangles AOC and DOC, and the circular sector AOC, shown below. It is clear from the diagram that Area of AOC < Area of sector AOC and Area of sector AOC < Area of DOC θ 1 A D O B C Clint Lee Math 112 Lecture 3: Evaluation of Limits II 22/26 A Basic Trigonometric Limit (proof continued) From the observations above we have 1 2 sin θ < 1 2 θ < 1 2 tan θ Cancelling the 2 s and noting that tan θ = sin θ cos θ gives sin θ < θ < sin θ cos θ Clint Lee Math 112 Lecture 3: Evaluation of Limits II 23/26
A Basic Trigonometric Limit (proof completed) The pair of inequalities above give an opportunity to use the Squeeze Theorem with f(θ) = cos θ, g(θ) = sin θ θ, h(θ) = 1 Now we have two easy its: Clint Lee Math 112 Lecture 3: Evaluation of Limits II 24/26 Example 13 Applying the Basic Trig Limit Evaluate the it Solution: θ 0 1 cos θ θ 2 We use the same idea as rationalizing the numerator. Multiply top and bottom by 1 + cos θ to give 1 cos θ θ 2 Clint Lee Math 112 Lecture 3: Evaluation of Limits II 25/26
Solution: Example 13 continued Thus θ 0 1 cos θ θ 2 since θ 0 cos θ = 1. This agrees with our result in Example 6. Clint Lee Math 112 Lecture 3: Evaluation of Limits II 26/26