Calculus (Math 1A) Lecture 6
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1 Calculus (Math 1A) Lecture 6 Vivek Shende September 5, 2017
2 Hello and welcome to class!
3 Hello and welcome to class! Last time
4 Hello and welcome to class! Last time We introduced limits, and discussed slopes of tangent lines.
5 Hello and welcome to class! Last time We introduced limits, and discussed slopes of tangent lines. Today
6 Hello and welcome to class! Last time We introduced limits, and discussed slopes of tangent lines. Today We give rules for computing limits, and use them in examples.
7 Hello and welcome to class! Last time We introduced limits, and discussed slopes of tangent lines. Today We give rules for computing limits, and use them in examples. These rules will be justified later, in terms of the precise ɛ δ definition of limits.
8 Limits For a function f (x), we say lim x x 0 f (x) = y 0 if, by restricting our attention to x very close but not equal to x 0, we can guarantee that f (x) is very close to y 0.
9 The limit of a constant function
10 The limit of a constant function For c a constant, lim c = c (7) x a
11 The limit of f (x) = x lim x = a (8) x a
12 Arithmetic of limits If c is a constant, and lim x a f (x) and lim x a g(x) exist and are finite,
13 Arithmetic of limits If c is a constant, and lim x a f (x) and lim x a g(x) exist and are finite, then: lim [f (x) + g(x)] = [ lim f (x)] + [ lim g(x)] (1) x a x a x a lim [f (x) g(x)] = [ lim f (x)] [ lim g(x)] (2) x a x a x a lim [cf (x)] = c[ lim f (x)] (3) x a x a lim [f (x)g(x)] = [ lim f (x)] [ lim g(x)] (4) x a x a x a f (x) lim x a g(x) = lim x a f (x) lim x a g(x) if lim x a g(x) 0 (5)
14 Arithmetic of limits In words:
15 Arithmetic of limits In words: a sum, difference, constant multiple, product, or quotient
16 Arithmetic of limits In words: a sum, difference, constant multiple, product, or quotient of limits which exist
17 Arithmetic of limits In words: a sum, difference, constant multiple, product, or quotient of limits which exist is the limit of the corresponding sum, difference, constant multiple, product, or quotient.
18 Arithmetic of limits If c is a constant, and lim x a f (x) and lim x a g(x) exist and are finite, then: lim [f (x) + g(x)] = [ lim f (x)] + [ lim g(x)] (1) x a x a x a lim [f (x) g(x)] = [ lim f (x)] [ lim g(x)] (2) x a x a x a lim [cf (x)] = c[ lim f (x)] (3) x a x a lim [f (x)g(x)] = [ lim f (x)] [ lim g(x)] (4) x a x a x a f (x) lim x a g(x) = lim x a f (x) lim x a g(x) if lim x a g(x) 0 (5)
19 Arithmetic of limits Example: lim x 5 x + 1 x + 2
20 Arithmetic of limits Example: lim x 5 x + 1 x + 2 By the quotient law we have lim x 5 x + 1 lim (x + 1) x + 2 = x 5 lim (x + 2) x 5
21 Arithmetic of limits Example: lim x 5 x + 1 x + 2 By the quotient law we have lim x 5 x + 1 lim (x + 1) x + 2 = x 5 lim (x + 2) x 5 assuming these two limits exist
22 Arithmetic of limits Example: lim x 5 x + 1 x + 2 By the quotient law we have lim x 5 x + 1 lim (x + 1) x + 2 = x 5 lim (x + 2) x 5 assuming these two limits exist and the one in the denominator is nonzero.
23 Arithmetic of limits We continue with lim x 5 (x + 1)
24 Arithmetic of limits We continue with lim x 5 (x + 1) By the sum law, lim (x + 1) = lim x + lim 1 x 5 x 5 x 5
25 Arithmetic of limits We continue with lim x 5 (x + 1) By the sum law, lim (x + 1) = lim x + lim 1 x 5 x 5 x 5 assuming these two limits exist.
26 Arithmetic of limits We continue with lim x 5 (x + 1) By the sum law, lim (x + 1) = lim x + lim 1 x 5 x 5 x 5 assuming these two limits exist. They do, and we know how to compute them: lim (x + 1) = lim x + lim 1 = = 6 x 5 x 5 x 5
27 Arithmetic of limits We have learned that lim (x + 1) = lim x + lim 1 = = 6 x 5 x 5 x 5 and in particular, that the limit exists.
28 Arithmetic of limits We have learned that lim (x + 1) = lim x + lim 1 = = 6 x 5 x 5 x 5 and in particular, that the limit exists. Similarly, lim (x + 2) = lim x + lim 2 = = 7 x 5 x 5 x 5
29 Arithmetic of limits We have learned that lim (x + 1) = lim x + lim 1 = = 6 x 5 x 5 x 5 and in particular, that the limit exists. Similarly, lim (x + 2) = lim x + lim 2 = = 7 x 5 x 5 x 5 Putting these together, we see that lim x 5 x + 1 lim (x + 1) x + 2 = x 5 = lim (x + 2) x 5 6 7
30 Try it yourself! Using only the laws we have seen so far i.e., the sum, difference, scalar multiple, product, quotient rules and the rules for limits of the constant and identity functions compute: lim (x + 1)(x + 2)(x + 3) x 4
31 Try it yourself!
32 Try it yourself! lim x 4 (x + 1)(x + 2)(x + 3) = ( lim (x + 1))( lim (x + 2))( lim (x + 3)) x 4 x 4 x 4
33 Try it yourself! lim x 4 (x + 1)(x + 2)(x + 3) = ( lim (x + 1))( lim (x + 2))( lim (x + 3)) x 4 x 4 x 4 = ( lim x 4 x + lim x 4 1)( lim x 4 x + lim x 4 2)( lim x 4 x + lim x 4 3)
34 Try it yourself! lim x 4 (x + 1)(x + 2)(x + 3) = ( lim (x + 1))( lim (x + 2))( lim (x + 3)) x 4 x 4 x 4 = ( lim x 4 x + lim x 4 1)( lim x 4 x + lim x 4 2)( lim x 4 x + lim x 4 3) = (4 + 1)(4 + 2)(4 + 3) = = 210
35 Try it yourself! lim x 4 (x + 1)(x + 2)(x + 3) = ( lim (x + 1))( lim (x + 2))( lim (x + 3)) x 4 x 4 x 4 = ( lim x 4 x + lim x 4 1)( lim x 4 x + lim x 4 2)( lim x 4 x + lim x 4 3) = (4 + 1)(4 + 2)(4 + 3) = = 210 Technical note: the first equalities are justified only because we learn from subsequent further calculations that the limits on the right hand side of each actually exist.
36 Roots and powers
37 Roots and powers Repeatedly applying the product law, we learn lim x a [f (x)]n = [ lim x a f (x)] n (6)
38 Roots and powers Repeatedly applying the product law, we learn lim [f x a (x)]n = [ lim f (x)] n (6) x a In particular, taking f (x) = x: lim x a x n = [ lim x a x] n = a n (9)
39 Roots and powers
40 Roots and powers Applying this to f (x) = n g(x) for g(x) 0 lim g(x) = lim [n g(x)] n = [ lim n g(x)] n x a x a x a Taking n th roots on both sides gives n lim x a g(x) = lim x a n g(x) (11)
41 Roots and powers Applying this to f (x) = n g(x) for g(x) 0 lim g(x) = lim [n g(x)] n = [ lim n g(x)] n x a x a x a Taking n th roots on both sides gives n lim x a g(x) = lim x a n g(x) (11) In particular, taking g(x) = x: n lim x = n a (10) x a
42 Continuity
43 Continuity If f is any polynomial, rational, or algebraic function,
44 Continuity If f is any polynomial, rational, or algebraic function, and a is in its domain,
45 Continuity If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim f (x) = f (a) x a
46 Continuity If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim f (x) = f (a) x a Why this is true: such functions are built from arithmetic operations and taking roots; so repeatedly use the limit laws we just learned.
47 Continuity If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim f (x) = f (a) x a Why this is true: such functions are built from arithmetic operations and taking roots; so repeatedly use the limit laws we just learned. In section 2.3, your book calls this the direct substitution property.
48 Continuity If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim f (x) = f (a) x a Why this is true: such functions are built from arithmetic operations and taking roots; so repeatedly use the limit laws we just learned. In section 2.3, your book calls this the direct substitution property. However from section 2.5 onwards, this will become the definition of continuity:
49 Continuity If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim f (x) = f (a) x a Why this is true: such functions are built from arithmetic operations and taking roots; so repeatedly use the limit laws we just learned. In section 2.3, your book calls this the direct substitution property. However from section 2.5 onwards, this will become the definition of continuity: a function with the above property is continuous.
50 Continuity x + x 1 Example. Compute lim x 5 x x
51 Continuity x + x 1 Example. Compute lim x 5 x x After checking that 5 is in the domain of the expression, we just plug it in: lim x = 7 35
52 Try it yourself! 3x + 3 x 1 Compute lim. x 2 2x 2
53 Try it yourself! 3x + 3 x 1 Compute lim. x 2 2x 2 After checking that 2 is in the domain of the expression, we just plug it in: lim = 7 x
54 Continuity Example. Compute lim x 5 x 2 5x x 5.
55 Continuity Example. Compute lim x 5 x 2 5x x 5. After checking that 5 is in the domain of the expression,
56 Continuity Example. Compute lim x 5 x 2 5x x 5. After checking that 5 is in the domain of the expression, We discover that it s not!
57 Continuity Example. Compute lim x 5 x 2 5x x 5. After checking that 5 is in the domain of the expression, We discover that it s not! So we have to do something other than just plugging it in.
58 Independence of the value
59 Independence of the value If f (x) = g(x) except possibly at x = a
60 Independence of the value If f (x) = g(x) except possibly at x = a then lim f (x) = lim g(x) x a x a
61 Independence of the value If f (x) = g(x) except possibly at x = a then lim f (x) = lim g(x) x a x a More precisely, if one limit is defined, so is the other, and in this case they are equal.
62 Independence of the value If f (x) = g(x) except possibly at x = a then lim f (x) = lim g(x) x a x a More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a.
63 Independence of the value If f (x) = g(x) except possibly at x = a then lim f (x) = lim g(x) x a x a More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a. Note: f nor g need not have the same domain, and need not be defined at a.
64 Independence of the value If f (x) = g(x) except possibly at x = a then lim f (x) = lim g(x) x a x a More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a. Note: f nor g need not have the same domain, and need not be defined at a. One needs only f (x) = g(x) over some [b, a) (a, c].
65 Independence of the value Example. Compute lim x 5 x 2 5x x 5.
66 Independence of the value Example. Compute lim x 5 x 2 5x x 5. lim x 5 x 2 5x x 5
67 Independence of the value Example. Compute lim x 5 x 2 5x x 5. x 2 5x lim x 5 x 5 = lim x 5 x(x 5) x 5
68 Independence of the value Example. Compute lim x 5 x 2 5x x 5. x 2 5x lim x 5 x 5 = lim x(x 5) x 5 x 5 = lim x 5 x
69 Independence of the value Example. Compute lim x 5 x 2 5x x 5. x 2 5x lim x 5 x 5 = lim x(x 5) x 5 x 5 = lim x 5 x = 5
70 Independence of the value Example. Compute lim x 5 x 2 5x x 5. x 2 5x lim x 5 x 5 = lim x(x 5) x 5 x 5 = lim x 5 x = 5 Here in the second equality, we used that x(x 5) x 5 of x other than 5. = x for any value
71 Independence of the value x 2 Example. Compute lim x 2 x
72 Independence of the value x 2 Example. Compute lim x 2 x Note we cannot just plug in x = 2,
73 Independence of the value x 2 Example. Compute lim x 2 x Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist,
74 Independence of the value x 2 Example. Compute lim x 2 x Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist, the limit of the denominator is 0.
75 Independence of the value x 2 Example. Compute lim x 2 x Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist, the limit of the denominator is 0. Remark: If the limit of the numerator were not zero,
76 Independence of the value x 2 Example. Compute lim x 2 x Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist, the limit of the denominator is 0. Remark: If the limit of the numerator were not zero, the limit of the quotient would be infinite or undefined. But it is zero.
77
78 We want to cancel some zeroes from the numerator and denominator.
79 We want to cancel some zeroes from the numerator and denominator. We start by removing the square-root from the denominator: lim x 2 x 2 x x 2 x = lim x 2 x x (x 2)( x = lim ) x 2 (x 2 3) 1 (x 2)( x = lim ) x 2 x 2 4 (x 2)( x = lim ) x 2 (x 2)(x + 2)
80 Now we are able to cancel:
81 Now we are able to cancel: (x 2)( x lim ) x = lim x 2 (x 2)(x + 2) x 2 x + 2
82 Now we are able to cancel: (x 2)( x lim ) x = lim x 2 (x 2)(x + 2) x 2 x + 2 Finally, we can plug in 2.
83 Now we are able to cancel: (x 2)( x lim ) x = lim x 2 (x 2)(x + 2) x 2 x + 2 Finally, we can plug in 2. lim x 2 x x + 2 = =
84 Try it yourself! Compute: lim x 3 x x 3
85 Limits from one-sided limits
86 Limits from one-sided limits If lim x x 0 f (x) and lim x x + 0 f (x) exist
87 Limits from one-sided limits If lim x x 0 f (x) and lim x x + 0 f (x) exist and are equal
88 Limits from one-sided limits If lim x x 0 f (x) and lim x x + 0 f (x) exist and are equal then also lim x x0 f (x) exists
89 Limits from one-sided limits If lim x x 0 f (x) and lim x x + 0 f (x) exist and are equal then also lim x x0 f (x) exists and lim x x 0 f (x) = lim x x0 f (x) = lim x x + 0 f (x)
90 Limits from one-sided limits If lim x x 0 f (x) and lim x x + 0 f (x) exist and are equal then also lim x x0 f (x) exists and lim x x 0 f (x) = lim x x0 f (x) = lim x x + 0 f (x) All the limit laws hold for one-sided limits as well.
91 Limits from one-sided limits We find the one-sided limits direct substitution. lim f (x) = 0 and lim f (x) = 0 by x 0+ x 0
92 Limits from one-sided limits We find the one-sided limits lim f (x) = 0 and lim x 0+ direct substitution. Thus lim x 0 f (x) = 0. f (x) = 0 by x 0
93 Try it yourself Find lim x 1 f (x), where f (x) = { x 2 1 x 1 x > 1 x x < 1
94 Squeeze theorem
95 Squeeze theorem If f (x) g(x) h(x) on some (a, b) (b, c)
96 Squeeze theorem If f (x) g(x) h(x) on some (a, b) (b, c) and lim f (x) = lim h(x) x b x b
97 Squeeze theorem If f (x) g(x) h(x) on some (a, b) (b, c) and lim f (x) = lim h(x) x b x b Then lim x b g(x) exists and lim f (x) = lim g(x) = lim h(x) x b x b x b Also sometimes called the sandwich theorem
98 Squeeze theorem If f (x) g(x) h(x) on some (a, b) (b, c) and lim f (x) = lim h(x) x b x b Then lim x b g(x) exists and lim f (x) = lim g(x) = lim h(x) x b x b x b Also sometimes called the sandwich theorem or the two policemen and one drunk theorem.
99
100 If a police officer stands on each side of a drunk, and both police go to the jail, then no matter how the drunk wobbles about, he will also end up in jail.
101 sin(1/x)
102 Squeeze theorem lim x 2 sin(1/x) =? x 0
103 x 2 sin(1/x)
104 Squeeze theorem
105 Squeeze theorem 1 sin(1/x) 1
106 Squeeze theorem 1 sin(1/x) 1 x 2 x 2 sin(1/x) x 2
107 Squeeze theorem 1 sin(1/x) 1 x 2 x 2 sin(1/x) x 2 lim x 2 = 0 = lim x 2 x 0 x 0
108 Squeeze theorem 1 sin(1/x) 1 x 2 x 2 sin(1/x) x 2 lim x 2 = 0 = lim x 2 x 0 x 0 So by the squeeze theorem lim x 2 sin(1/x) = 0 x 0
109 sin(x)/x Graphs of sin(x)/x, x, sin(x)
110 Some trigonometric limits from the squeeze theorem
111 Some trigonometric limits from the squeeze theorem
112 Some trigonometric limits from the squeeze theorem Area(Triangle ADB) Area(Green sector) Area(Triangle ADF )
113 Some trigonometric limits from the squeeze theorem Area(Triangle ADB) Area(Green sector) Area(Triangle ADF ) 1 2 sin(x) x 2π π 1 2 tan(x)
114 Some trigonometric limits from the squeeze theorem 1 2 sin(x) x 2π π 1 2 tan(x)
115 Some trigonometric limits from the squeeze theorem 1 2 sin(x) x 2π π 1 2 tan(x) sin(x) x sin(x) cos(x)
116 Some trigonometric limits from the squeeze theorem 1 2 sin(x) x 2π π 1 2 tan(x) sin(x) x sin(x) cos(x) x cos(x) sin(x) x
117 Some trigonometric limits from the squeeze theorem 1 2 sin(x) x 2π π 1 2 tan(x) sin(x) x sin(x) cos(x) x cos(x) sin(x) x cos(x) sin(x) x 1
118 Some trigonometric limits from the squeeze theorem 1 2 sin(x) x 2π π 1 2 tan(x) sin(x) x sin(x) cos(x) x cos(x) sin(x) x cos(x) sin(x) x 1 By the squeeze theorem, lim x 0 sin(x) x = 1
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