Mathematics

4.1. Background

Topics: limit of f(x) as x approaches a, limit of f(x) as x approaches infinity, left- and right-hand limits.

4.1.1. Definition. Suppose that f is a real valued function of a real variable, a is an accumulation point of the domain of f , and ` ∈ R. We say that ` is the limit of f(x) as x approaches a if for every neighborhood V of ` there exists a corresponding deleted neighborhood U of a which satisfies the following condition:

for every point x in the domain of f which lies in U the point f(x) lies in V .

Once we have convinced ourselves that in this definition it doesn’t matter if we work only with symmetric neighborhoods of points, we can rephrase the definition in a more conventional algebraic fashion: ` is the limit of f(x) as x approaches a provided that for every � > 0 there exists δ > 0 such that if 0 < |x− a| < δ and x ∈ dom f , then |f(x)− `| < �.

4.1.2. Notation. To indicate that a number ` is the limit of f(x) as x approaches a, we may write either

lim x→a

f(x) = l or f(x)→ ` as x→ a.

(See problem 2.)

21

22 4. LIMITS

4.2. Exercises

(1) lim x→3

x3 − 13×2 + 51x− 63 x3 − 4×2 − 3x+ 18

= a

5 where a = .

(2) lim x→0

√ x2 + 9x+ 9− 3

x = a

2 where a = .

(3) lim x→1

x3 − x2 + 2x− 2 x3 + 3×2 − 4x

= 3

a where a = .

(4) lim t→0

t√ 4− t− 2

= .

(5) lim x→0

√ x+ 9− 3

x =

1

a where a = .

(6) lim x→2

x3 − 3×2 + x+ 2 x3 − x− 6

= 1

a where a = .

(7) lim x→2

x3 − x2 − 8x+ 12 x3 − 10×2 + 28x− 24

= −a 4

where a = .

(8) lim x→0

√ x2 − x+ 4− 2 x2 + 3x

= −1 a

where a = .

(9) lim x→1

x3 + x2 − 5x+ 3 x3 − 4×2 + 5x− 2

= .

(10) lim x→3

x3 − 4×2 − 3x+ 18 x3 − 8×2 + 21x− 18

= .

(11) lim x→−1

x3 − x2 − 5x− 3 x3 + 6×2 + 9x+ 4

= −4 a

where a = .

(12) lim x→0

2x sinx

1− cosx = .

(13) lim x→0

1− cosx 3x sinx

= 1

a where a = .

(14) lim x→0

tan 3x− sin 3x x3

= a

2 where a = .

(15) lim h→0

sin 2h

5h2 + 7h = .

(16) lim h→0

cot 7h

cot 5h = .

(17) lim x→0

secx− cosx 3×2

= 1

a where a = .

(18) lim x→∞

(9×8 − 6×5 + 4)1/2

(64×12 + 14×7 − 7)1/3 = a

4 where a = .

(19) lim x→∞

√ x( √ x+ 3−

√ x− 2) = a

2 where a = .

(20) lim x→∞

7− x+ 2×2 − 3×3 − 5×4

4 + 3x− x2 + x3 + 2×4 = a

2 where a = .

(21) lim x→∞

(2×4 − 137)5

(x2 + 429)10 = .

4.2. EXERCISES 23

(22) lim x→∞

(5×10 + 32)3

(1− 2×6)5 = − a

32 where a = .

(23) lim x→∞

(√ x2 + x− x

) =

1

a where a = .

(24) lim x→∞

x(256×4 + 81×2 + 49)−1/4 = 1

a where a = .

(25) lim x→∞

x

(√ 3×2 + 22−

√ 3×2 + 4

) = a √ a where a = .

(26) lim x→∞

x 2 3 ( (x+ 1)

1 3 − x

1 3 )

= 1

a where a = .

(27) lim x→∞

(√ x+ √ x−

√ x− √ x

) = .

(28) Let f(x) =

 2x− 1, if x < 2;x2 + 1, if x > 2. Then limx→2− f(x) = and limx→2+ f(x) = . (29) Let f(x) =

|x− 1| x− 1

. Then lim x→1−

f(x) = and lim x→1+

f(x) = .

(30) Let f(x) =

 5x− 3, if x < 1;x2, if x ≥ 1. Then limx→1− f(x) = and limx→1+ f(x) = . (31) Let f(x) =

 3x+ 2, if x < −2;x2 + 3x− 1, if x ≥ −2. Then limx→−2− f(x) = and limx→−2+ f(x) = . (32) Suppose y = f(x) is the equation of a curve which always lies between the parabola

x2 = y − 1 and the hyperbola yx+ y − 1 = 0. Then lim x→0

f(x) = .

24 4. LIMITS

4.3. Problems

(1) Find lim x→0+

( e−1/x sin(1/x)− (x+ 2)3

) (if it exists) and give a careful argument showing

that your answer is correct.

(2) The notation limx→a f(x) = ` that we use for limits is somewhat optimistic. It assumes the uniqueness of limits. Prove that limits, if they exist, are indeed unique. That is, suppose that f is a real valued function of a real variable, a is an accumulation point of the domain of f , and `, m ∈ R. Prove that if f(x) → ` as x → a and f(x) → m as x → a, then l = m. (Explain carefully why it was important that we require a to be an accumulation point of the domain of f .)

(3) Let f(x) = sinπx

x+ 1 for all x 6= −1. The following information is known about a function g

defined for all real numbers x 6= 1:

(i) g = p

q where p(x) = ax2 + bx+ c and q(x) = dx+ e for some constants a, b, c, d, e;

(ii) the only x-intercept of the curve y = g(x) occurs at the origin; (iii) g(x) ≥ 0 on the interval [0, 1) and is negative elsewhere on its domain; (iv) g has a vertical asymptote at x = 1; and (v) g(1/2) = 3.

Either find lim x→1

g(x)f(x) or else show that this limit does not exist.

Hints. Write an explicit formula for g by determining the constants a . . . e. Use (ii) to find c; use (ii) and (iii) to find a; use (iv) to find a relationship between d and e; then use (v) to obtain an explicit form for g. Finally look at f(x)g(x); replace sinπx by sin(π(x− 1) + π) and use the formula for the sine of the sum of two numbers.

(4) Evaluate lim x→0

√ |x| cos (π1/x2) 2 + √ x2 + 3

(if it exists). Give a careful proof that your conclusion is

correct.

4.4. ANSWERS TO ODD-NUMBERED EXERCISES 25

4.4. Answers to Odd-Numbered Exercises

(1) −4 (3) 5

(5) 6

(7) 5

(9) −4 (11) 3

(13) 6

(15) 2

7

(17) 3

(19) 5

(21) 32

(23) 2

(25) 3

(27) 1

(29) −1, 1 (31) −4, −3

CHAPTER 5

CONTINUITY

5.1. Background

Topics: continuous functions, intermediate value theorem. extreme value theorem.

There are many ways of stating the intermediate value theorem. The simplest says that con- tinuous functions take intervals to intervals.

5.1.1. Definition. A subset J of the real line R is an interval if z ∈ J whenever a, b ∈ J and a < z < b.

5.1.2. Theorem (Intermediate Value Theorem). Let J be an interval in R and f : J → R be continuous. Then the range of f is an interval.

5.1.3. Definition. A real-valued function f on a set A is said to have a maximum at a point a in A if f(a) ≥ f(x) for every x in A; the number f(a) is the maximum value of f . The function has a minimum at a if f(a) ≤ f(x) for every x in A; and in this case f(a) is the minimum value of f . A number is an extreme value of f if it is either a maximum or a minimum value. It is clear that a function may fail to have maximum or minimum values. For example, on the open interval (0, 1) the function f : x 7→ x assumes neither a maximum nor a minimum.

The concepts we have just defined are frequently called global (or absolute) maximum and global (or absolute) minimum.

5.1.4. Definition. Let f : A → R where A ⊆ R. The function f has a local (or relative) maximum at a point a ∈ A if there exists a neighborhood J of a such that f(a) ≥ f(x) whenever x ∈ J and x ∈ dom f . It has a local (or relative) minimum at a point a ∈ A if there exists a neighborhood J of a such that f(a) ≤ f(x) whenever x ∈ J and x ∈ dom f .

5.1.5. Theorem (Extreme Value Theorem). Every continuous real valued function on a closed and bounded interval in R achieves its (global) maximum and minimum value at some points in the interval.

5.1.6. Definition. A number p is a fixed point of a function f : R→ R if f(p) = p.

5.1.7. Example. If f(x) = x2 − 6 for all x ∈ R, then 3 is a fixed point of f .

27

28 5. CONTINUITY

5.2. Exercises

(1) Let f(x) = x3 − 2×2 − 2x− 3 x3 − 4×2 + 4x− 3

for x 6= 3. How should f be defined at x = 3 so that it becomes a continuous function on all of R?

Answer: f(3) = a

7 where a = .

(2) Let f(x) =

 1 if x < 0

x if 0 < x < 1

2− x if 1 < x < 3 x− 4 if x > 3

.

(a) Is it possible to define f at x = 0 in such a way that f becomes continuous at x = 0? Answer: . If so, then we should set f(0) = .

(b) Is it possible to define f at x = 1 in such a way that f becomes continuous at x = 1? Answer: . If so, then we should set f(1) = .

(c) Is it possible to define f at x = 3 in such a way that f becomes continuous at x = 3? Answer: . If so, then we should set f(3) = .

(3) Let f(x) =

 x+ 4 if x < −2 −x if −2 < x < 1 x2 − 2x+ 1 if 1 < x < 3 10− 2x if x > 3

.

(a) Is it possible to define f at x = −2 in such a way that f becomes continuous at x = −2? Answer: . If so, then we should set f(−2) = .

(b) Is it possible to define f at x = 1 in such a way that f becomes continuous at x = 1? Answer: . If so, then we should set f(1) = .

(c) Is it possible to define f at x = 3 in such a way that f becomes continuous at x = 3? Answer: . If so, then we should set f(3) = .

(4) The equation x5 + x3 + 2x = 2×4 + 3×2 + 4 has a solution in the open interval (n, n+ 1) where n is the positive integer .

(5) The equation x4−6×2−53 = 22x−2×3 has a solution in the open interval (n, n+1) where n is the positive integer .

(6) The equation x4 + x + 1 = 3×3 + x2 has solutions in the open intervals (m,m + 1) and (n, n+ 1) where m and n are the distinct positive integers and .

(7) The equation x5 + 8x = 2×4 + 6×2 has solutions in the open intervals (m,m + 1) and (n, n+ 1) where m and n are the distinct positive integers and .

5.3. PROBLEMS 29

5.3. Problems

(1) Prove that the equation

x180 + 84

1 + x2 + cos2 x = 119

has at least two solutions.

(2) (a) Find all the fixed points of the function f defined in example 5.1.7.

Theorem: Every continuous function f : [0, 1]→ [0, 1] has a fixed point. (b) Prove the preceding theorem. Hint. Let g(x) = x − f(x) for 0 ≤ x ≤ 1. Apply the

intermediate value theorem 5.1.2 to g.

(c) Let g(x) = 0.1×3 + 0.2 for all x ∈ R, and h be the restriction of g to [0, 1]. Show that h satisfies the hypotheses of the theorem.

(d) For the function h defined in (c) find an approximate value for at least one fixed point with an error of less than 10−6. Give a careful justification of your answer.

(e) Let g be as in (c). Are there other fixed points (that is, points not in the unit square where the curve y = g(x) crosses the line y = x)? If so, find an approximation to each such point with an error of less than 10−6. Again provide careful justification.

(3) Define f on [0, 4] by f(x) = x + 1 for 0 ≤ x < 2 and f(x) = 1 for 2 ≤ x ≤ 4. Use the extreme value theorem 5.1.5 to show that f is not continuous.

(4) Give an example of a function defined on [0, 1] which has no maximum and no minimum on the interval. Explain why the existence of such a function does not contradict the extreme value theorem 5.1.5.

(5) Give an example of a continuous function defined on the interval (1, 2] which does not achieve a maximum value on the interval. Explain why the existence of such a function does not contradict the extreme value theorem 5.1.5.

(6) Give an example of a continuous function on the closed interval [3,∞) which does not achieve a minimum value on the interval. Explain why the existence of such a function does not contradict the extreme value theorem 5.1.5.

(7) Define f on [−2, 0] by f(x) = −1 (x+ 1)2

for −2 ≤ x < −1 and −1 < x ≤ 0, and f(−1) = −3.

Use the extreme value theorem 5.1.5 to show that f is not continuous.

(8) Let f(x) = 1

x for 0 < x ≤ 1 and f(0) = 0. Use the extreme value theorem 5.1.5 to show

that f is not continuous on [0, 1].

30 5. CONTINUITY

5.4. Answers to Odd-Numbered Exercises

(1) 13

(3) (a) yes, 2 (b) no, — (c) yes, 4

(5) 3

(7) 1, 2

Part 3

DIFFERENTIATION OF FUNCTIONS OF A SINGLE VARIABLE

CHAPTER 6

DEFINITION OF THE DERIVATIVE

6.1. Background

Topics: definition of the derivative of a real valued function of a real variable at a point

6.1.1. Notation. Let f be a real valued function of a real variable which is differentiable at a point a in its domain. When thinking of a function in terms of its graph, we often write y = f(x), call x the independent variable, and call y the dependent variable. There are many notations for the derivative of f at a. Among the most common are

Df(a), f ′(a), df

dx

∣∣∣∣ a

, y′(a), ẏ(a), and dy

dx

∣∣∣∣ a

.

33

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