Section 2.5. Evaluating Limits Algebraically
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1 Section 2.5 Evaluating Limits Algebraically (1) Determinate and Indeterminate Forms (2) Limit Calculation Techniques (A) Direct Substitution (B) Simplification (C) Conjugation (D) The Squeeze Theorem (3) Limits of Piecewise and Absolute Value Functions MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 1 / 14
2 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
3 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. The form of a limit is not the same as its value! MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
4 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. The form of a limit is not the same as its value! It is a tool for inspecting the limit. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
5 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. The form of a limit is not the same as its value! It is a tool for inspecting the limit. lim x arctan(x) : form 0 0 x 0 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
6 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. The form of a limit is not the same as its value! It is a tool for inspecting the limit. lim x arctan(x) : form 0 0 x 0 lim (1 + x) 1 x : form 0 x MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
7 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. The form of a limit is not the same as its value! It is a tool for inspecting the limit. lim x arctan(x) : form 0 0 x 0 lim (1 + x) 1 x : form 0 x lim cos(x) 1 x : form 1 x 0 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
8 The Form of a Limit The form of a limit lim x c is the expression resulting from substituting x = c into. The form of a limit is not the same as its value! It is a tool for inspecting the limit. lim x arctan(x) : form 0 0 x 0 lim (1 + x) 1 x : form 0 x lim cos(x) 1 x : form 1 lim ln(x) sin(x) : form 0 x 0 + x 0 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 2 / 14
9 The Form of a Limit Determinate Forms are forms which always represent the same limit. For example, the form 1 always represents a limit which equals 0. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 3 / 14
10 The Form of a Limit Determinate Forms are forms which always represent the same limit. For example, the form 1 always represents a limit which equals 0. Assume c 0, c ± + 0 c 0 ± MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 3 / 14
11 The Form of a Limit Determinate Forms are forms which always represent the same limit. For example, the form 1 always represents a limit which equals 0. Assume c 0, c ± + 0 c 0 ± Indeterminate Forms are called indeterminate because they represent limits which may or may not exist and may be equal to any value. The form itself does not indicate the value of the limit. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 3 / 14
12 The Form of a Limit Determinate Forms are forms which always represent the same limit. For example, the form 1 always represents a limit which equals 0. Assume c 0, c ± + 0 c 0 ± Indeterminate Forms are called indeterminate because they represent limits which may or may not exist and may be equal to any value. The form itself does not indicate the value of the limit. There are 7 indeterminate forms: 0 0 ± ± MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 3 / 14
13 We can use direct substitution to evaluate limits of functions that are continuous (Section 2.4) or have determinate forms. Simplification and Limits If f (x) = g(x) for values near x = a, then lim f (x) = lim g(x) x a x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 4 / 14
14 We can use direct substitution to evaluate limits of functions that are continuous (Section 2.4) or have determinate forms. Simplification and Limits If f (x) = g(x) for values near x = a, then lim f (x) = lim g(x) x a x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 4 / 14
15 We can use direct substitution to evaluate limits of functions that are continuous (Section 2.4) or have determinate forms. Simplification and Limits If f (x) = g(x) for values near x = a, then lim f (x) = lim g(x) x a x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 4 / 14
16 Conjugation The expression a + b is conjugate to the expression a b. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 5 / 14
17 Conjugation The expression a + b is conjugate to the expression a b. a 2 b 2 = (a b)(a + b) MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 5 / 14
18 Conjugation The expression a + b is conjugate to the expression a b. Rationalize the denominator: a 2 b 2 = (a b)(a + b) MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 5 / 14
19 Conjugation The expression a + b is conjugate to the expression a b. a 2 b 2 = (a b)(a + b) Rationalize the denominator: Rationalize the numerator: MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 5 / 14
20 Simplification and Conjugation Examples (Example I) Evaluate the following limits: 2x 2 + 4x + 6 (a) lim x 1 x 2 x 2 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 6 / 14
21 Simplification and Conjugation Examples (Example I) Evaluate the following limits: 2x 2 + 4x + 6 (a) lim x 1 x 2 x 2 (b) lim h 0 h h 2 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 6 / 14
22 Simplification and Conjugation Examples (Example I) Evaluate the following limits: 2x 2 + 4x + 6 (a) lim x 1 x 2 x 2 (c) lim t 0 (t + 3) 2 9 t (b) lim h 0 h h 2 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 6 / 14
23 Simplification and Conjugation Examples (Example I) Evaluate the following limits: 2x 2 + 4x + 6 (a) lim x 1 x 2 x 2 (c) lim t 0 (t + 3) 2 9 t (b) lim h 0 h h (d) lim + 1 x x x MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 6 / 14
24 iclicker Question 1 Evaluate the limit: (A) (B) lim x 16 3 x 9x x 2 (D) (C) (E) Does Not Exist. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 7 / 14
25 iclicker Question 2 Evaluate the limit: lim x 9 3 x 9x x 2 (A) 1 45 (B) 0.02 (C) 1 54 (D) (E) Does Not Exist. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 8 / 14
26 Limits of Comparable Functions If f (x) g(x) for values of x near a then if both limits exist. lim f (x) lim g(x) x a x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 9 / 14
27 Limits of Comparable Functions If f (x) g(x) for values of x near a then if both limits exist. lim f (x) lim g(x) x a x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 9 / 14
28 The Squeeze Theorem If f (x) g(x) h(x) for values of x near a and lim f (x) = L = lim h(x) x a x a then lim g(x) = L. x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 10 / 14
29 The Squeeze Theorem If f (x) g(x) h(x) for values of x near a and lim f (x) = L = lim h(x) x a x a then lim g(x) = L. x a MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 10 / 14
30 Walking a drunk through a door MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 10 / 14 The Squeeze Theorem If f (x) g(x) h(x) for values of x near a and lim f (x) = L = lim h(x) x a x a then lim g(x) = L. x a
31 The Squeeze Theorem If f g h for values near x = a and lim f (x) = L = lim h(x) x a x a then lim x a g(x) = L. 1 sin(x) 1 1 cos(x) 1 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 11 / 14
32 Example II: The Squeeze Theorem Evaluate ( ) 1 lim x 2 sin x 0 x MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 12 / 14
33 Example II: The Squeeze Theorem Evaluate ( ) 1 lim x 2 sin x 0 x MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 12 / 14
34 iclicker Question 3 Evaluate ( lim 17x sin x 0 ( ) ) x (A) (B) 1 (C) 0 (D) 4 (E) Does Not Exist MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 13 / 14
35 iclicker Question 3 Evaluate ( lim 17x sin x 0 ( ) ) x (A) (B) 1 (C) 0 (D) 4 (E) Does Not Exist MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 13 / 14
36 iclicker Question 3 Evaluate ( lim 17x sin x 0 ( ) ) x (A) (B) 1 (C) 0 (D) 4 (E) Does Not Exist MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 13 / 14
37 iclicker Question 3 Evaluate ( lim 17x sin x 0 ( ) ) x (A) (B) 1 (C) 0 (D) 4 (E) Does Not Exist MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 13 / 14
38 Piecewise and Absolute Valued Functions Evaluating the limit of a piecewise function differs from evaluating the limit of elementary functions only when the limiting value is a break point. ({ 2, 0, 2} below) x x < 2 sin(x) 2 x 0 f (x) = x 2 0 < x < 2 cos(x) x 2 MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 14 / 14
39 Piecewise and Absolute Valued Functions Evaluating the limit of a piecewise function differs from evaluating the limit of elementary functions only when the limiting value is a break point. ({ 2, 0, 2} below) x x < 2 sin(x) 2 x 0 f (x) = x 2 0 < x < 2 cos(x) x 2 Absolute value functions are secretly piecewise functions! { (x a) x < a x a = x a x > a When confronted with an absolute valued function, calmly write it as a piecewise function before any other step. MATH 125 (Section 2.5) Evaluating Limits Algebraically The University of Kansas 14 / 14
40 Section 2.7 Limits at Infinity (1) Graphs at Infinity (2) Calculating Limits at Infinity (A) Vertical Asymptotes (B) Horizontal Asymptotes (C) Roots and Infinite Limits (3) Asymptotes of Common Functions MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 1 / 11
41 Infinity in the Input lim f (x) = L x The values of f (x) can be made as close to L as we would like by taking x sufficiently large. MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 2 / 11
42 Infinity in the Output lim f (x) = x a The values of f (x) can be made as large as we want by taking x sufficiently close to a but not equal to a. MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 3 / 11
43 Vertical Asymptotes The line x = a is a vertical asymptote of the curve y = f (x) if at least one of the following statements is true: lim f (x) = ± lim x a f (x) = ± lim x a + f (x) = ± x a MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 4 / 11
44 Vertical Asymptotes The line x = a is a vertical asymptote of the curve y = f (x) if at least one of the following statements is true: lim f (x) = ± lim x a f (x) = ± lim x a + f (x) = ± x a MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 4 / 11
45 Vertical Asymptotes The line x = a is a vertical asymptote of the curve y = f (x) if at least one of the following statements is true: lim f (x) = ± lim x a f (x) = ± lim x a + f (x) = ± x a MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 4 / 11
46 Vertical Asymptotes The line x = a is a vertical asymptote of the curve y = f (x) if at least one of the following statements is true: lim f (x) = ± lim x a f (x) = ± lim x a + f (x) = ± x a MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 4 / 11
47 Horizontal Asymptotes The line y = L is a horizontal asymptote of the curve y = f (x) if either lim f (x) = L or lim f (x) = L. x x MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 5 / 11
48 Horizontal Asymptotes The line y = L is a horizontal asymptote of the curve y = f (x) if either lim f (x) = L or lim f (x) = L. x x If n is a positive rational number, then lim x 1 x n = 0 lim 1 x n = 0 x (if n has odd denominator, otherwise DNE) MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 5 / 11
49 Horizontal Asymptotes The line y = L is a horizontal asymptote of the curve y = f (x) if either lim f (x) = L or lim f (x) = L. x x If n is a positive rational number, then lim x 1 x n = 0 lim 1 x n = 0 x (if n has odd denominator, otherwise DNE) Rational Functions: lim x ± a n x n a 1 x + a 0 b m x m b 1 x + b 0 = 0 if n < m a n b n if n = m ± if n > m MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 5 / 11
50 Example I: Calculating Horizontal Asymptotes 3 + 9x 6 (i) lim x 1 2x 3 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 6 / 11
51 Example I: Calculating Horizontal Asymptotes 3 + 9x 6 (i) lim = 3 x 1 2x 3 2 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 6 / 11
52 Example I: Calculating Horizontal Asymptotes 3 + 9x 6 (i) lim = 3 x 1 2x 3 2 { Reminder: x 2 x if x 0 = x = x if x < 0 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 6 / 11
53 Example I: Calculating Horizontal Asymptotes 3 + 9x 6 (i) lim = 3 x 1 2x 3 2 { Reminder: x 2 x if x 0 = x = x if x < x 6 (ii) lim x 1 2x 3 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 6 / 11
54 Example I: Calculating Horizontal Asymptotes 3 + 9x 6 (i) lim = 3 x 1 2x 3 2 { Reminder: x 2 x if x 0 = x = x if x < x 6 (ii) lim x 1 2x 3 = 3 2 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 6 / 11
55 Example I: Calculating Horizontal Asymptotes 3 + 9x 6 (i) lim = 3 x 1 2x 3 2 { Reminder: x 2 x if x 0 = x = x if x < x 6 (ii) lim x 1 2x 3 = 3 2 Horizontal Asymptotes y = 3 2 y = 3 2 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 6 / 11
56 iclicker Question 1 Evaluate the following limit: lim x 2x 3 16x (A) 1 2 (B) 2 (C) 1 2 (D) 2 (E) Does Not Exist MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 7 / 11
57 The word asymptote always refers to behavior involving infinity. Horizontal asymptotes refer to the end behavior of the graph. Functions can cross the lines of their horizontal asymptotes! MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 8 / 11
58 Example II: Calculating Horizontal Asymptotes Find the Horizontal Asymptotes of the following functions: (i) f (x) = 4x 2 + x 2x MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 9 / 11
59 Example II: Calculating Horizontal Asymptotes Find the Horizontal Asymptotes of the following functions: (i) f (x) = 4x 2 + x 2x (ii) g(x) = x 3 2x 8 MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 9 / 11
60 Example II: Calculating Horizontal Asymptotes Find the Horizontal Asymptotes of the following functions: (i) f (x) = 4x 2 + x 2x (ii) g(x) = x 3 2x 8 (iii) h(x) = arctan(e x ) MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 9 / 11
61 Example II: Calculating Horizontal Asymptotes Find the Horizontal Asymptotes of the following functions: (i) f (x) = 4x 2 + x 2x (ii) g(x) = x 3 2x 8 (iii) h(x) = arctan(e x ) A tank contains 100 L of pure water. Brine that contains 30g of salt per liter of water is pumped into the tank at a rate of 25 L min. What happens to the concentration of salt as time t approaches? MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 9 / 11
62 iclicker Question 2 Evaluate the following limit: lim e 1 x 1 x 1 (A) (B) (C) 1 (D) 0 (E) Does Not Exist MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 10 / 11
63 Summary of Limit Techniques For finite limiting values we have four algebraic techniques: (I) Direct Substitution (Continuity and Determinate Forms) (II) Simplification (III) Conjugation (Roots) (IV) The Squeeze Theorem For infinite limiting values we have two techniques: 1 (i) Manipulations involving lim x ± x n = 0 (ii) Horizontal asymptotes of common functions (e x, ln x, arctan(x),...) MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 11 / 11
64 Summary of Limit Techniques For finite limiting values we have four algebraic techniques: (I) Direct Substitution (Continuity and Determinate Forms) (II) Simplification (III) Conjugation (Roots) (IV) The Squeeze Theorem For infinite limiting values we have two techniques: 1 (i) Manipulations involving lim x ± x n = 0 (ii) Horizontal asymptotes of common functions (e x, ln x, arctan(x),...) Limits are at the heart of the Calculus sequence and will be used in defining every major object of study. Despite their role in defining objects, calculating limits becomes less common as we develop more techniques. MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 11 / 11
65 Summary of Limit Techniques For finite limiting values we have four algebraic techniques: (I) Direct Substitution (Continuity and Determinate Forms) (II) Simplification (III) Conjugation (Roots) (IV) The Squeeze Theorem For infinite limiting values we have two techniques: 1 (i) Manipulations involving lim x ± x n = 0 (ii) Horizontal asymptotes of common functions (e x, ln x, arctan(x),...) Limits are at the heart of the Calculus sequence and will be used in defining every major object of study. Despite their role in defining objects, calculating limits becomes less common as we develop more techniques. MATH125 has two remaining sections focused heavily on limits: Section 2.6/3.6 on trigonometric limits Section 4.5 on evaluating limits with indeterminate forms MATH 125 (Section 2.7) Limits at Infinity The University of Kansas 11 / 11
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