Maths 1036 full - Copy - study manual PDF

Preview

Summary

study manual...

Description

Chapter 1 Functions and limits LEARNING OUTCOMES: On completion of this chapter you should 1. understand the concept of a limit of a function at a point intuitively, 2. be able to estimate the value of a limit of a function at a given point by calculating function values close to the given point, 3. be able to evaluate the limit of a function at a given point graphically, 4. understand and be able to calculate one-sided limits, 5. know the behaviour of a function with infinite limits, 6. understand and be able to determine vertical and horizontal asymptotes of a function, 7. be able to compute limits of exponential functions, 8. be able to use the approporiate Limit Laws to find the limit of a function at a given point,

9. know how to prove that the limit of a function at a given point exists using one-sided limits, 10. be able to use factorization and rationalization to calculate limits of rational functions, 11. be able to use the Squeeze Theorem to calculate limits, 12. know, understand and be able to prove the theorems on trigonometric limits given in these notes on this section and 13. know how to solve the examinable tutorial problems and the worked out problems from these notes on this chapter. The single most important concept in all of calculus is that of a limit. Every single notion of calculus is encapsulated in one sense or another to that of a limit. In this course, we will only consider an intuitive and informal discussion of a limit of a function at a point. The precise approach to limits, using the standard ε-δ definition is reserved for the second year Analysis course and will not be considered here.

1.1

The idea of a limit

Consider a number L and a function f defined near the point x = a, not necessarily at a itself. We wish to investigate the behaviour of f for values x in the domain of f that are close to a. We first consider this numerically, then graphically. x−9 . What hapWorked Example 1.1.1. Consider the function f (x) = √ x−3 pens to the values of f (range values of f ) when x (domain values of f ) is close to 9?

Solution. A table of values will conjecture the behaviour of f . We want to Approaching 9 from the right gives the following table of values. approach 9 and can do this in a direction closely from the left (x < 9, i. e. x−9 from values below 9) and a direction closely from the right (x > 9, i. e. from x>9 f (x) = √ x − 3 values above 9). 9.1 x−9 9.05 x 0 have the inverse y = loga x. Particularly, the logarithmic function to the base e, loge x, is called the natural logarithmic function y = loge x which we denote in future as y = ln x (logarithmus naturalis), i. e., ln x = loge x.

f (x) DOES NOT EXIST. lim f (x) , lim+ f (x) ⇒ lim x→a

x→a−

x→a

It is obvious that it is tedious to determine lim− f (x), lim+ f (x) and eventux→a x→a ally the existence of lim f (x) using the tabular approach. More importantly, x→a the tabular approach does not prove the existence of the limit nor does it give the exact value of the limit. The graphical way (if we know how to sketch the graph) is more succinct. A computer algebra program may assist you in graphical sketching.

We see that to five decimal places lim f (x) = +

3. y = ln x has all the logarithmic properties:

x→0

Hence, lim− f (x) = lim+ f (x) so that lim f (x) exists, and to five decimal x→0 x→0 x→0 places 1 lim f (x) = lim (1 + x) x ≈ x→0

x→0

The exact value of this limit is given the symbol e (called Eulers number, accredited to the Swiss mathematician Leonhard Euler (1707-1783) who was among the first to study its properties extensively). We thus have 1 x

lim(1 + x) = e x→0

(a) y = ln x ⇔ x = ey ,

(b) eln x = x and ln e x = x, (c) ln xr = r ln x, (d) loga x =

loge x ln x , = loge a ln a

(e) ln xy = ln x + ln y, (f) ln

Ç å

x y

= ln x − ln y.

You will encounter y = ln x in particular in the chapters on integration. Note The number e is an irrational number (like π) and is used frequently in all the properties above remembering that ln x = log x. e of Mathematics. 2 < e < 3 and to five decimals e ≈ 2.71828. In the sequel, e will refer to this irrational number of Euler. Note: If f does not approach a unique value as x approaches a, then x→a lim f (x) does not exist, i. e., Note. 1. The function y = e x is called the natural exponential function. 2. Recall that the exponential functions given as y = ax with a > 0 have the inverse y = loga x. Particularly, the logarithmic function to the base e, loge x, is called the natural logarithmic function y = loge x which we denote in future as y = ln x (logarithmus naturalis), i. e., ln x = loge x.

f (x) DOES NOT EXIST. lim f (x) , lim+ f (x) ⇒ lim x→a

x→a−

x→a

It is obvious that it is tedious to determine lim− f (x), lim+ f (x) and eventux→a x→a ally the existence of lim f (x) using the tabular approach. More importantly, x→a the tabular approach does not prove the existence of the limit nor does it give the exact value of the limit. The graphical way (if we know how to sketch the graph) is more succinct. A computer algebra program may assist you in graphical sketching.

Worked Example 1.1.3. Sketch the graph of the given function and de- Solution. termine graphically whether the corresponding limit exists; if so, find the limit. √ 1. f (x) = 4 − x2 . Find lim f (x). x→2

Solution.

3. h(x) = Solution.

√  −x   

2. Let g(x) = 3 − x    (x − 3)2

if x < 0, if 0 ≤ x < 3, if x > 3.

Find lim g(x) and lim g(x) if they exist. x→0

x→3

|x| . Find lim h(x) if it exists. x→0 x

  1 

if x ∈ Q, i. e., x is rational, if x ∈ R \ Q, i. e., x is irrational.

5. Consider the greatest integer function defined as

4. Let d(x) = 0 ⌊x⌋ = greatest integer less than or equal to x. Find for any a ∈ R, lim d(x) if it exists. The function d is called the Dirichx→a Find lim ⌊x⌋ if it exists. let function. x→4 Solution. Let a ∈ R. As x → a− or x → a+ , x passes through both rational Solution. and irrational numbers. (The rationals and irrationals are said to be dense in R. Between any two real numbers, there is a rational and an irrational number.) Consequently as x approaches a, the values of the Dirichlet function jump back and forth between 0 and 1 and thus d(x) cannot approach and stay close to a fixed number L. Therefore lim d(x) does not exist. x→a

The graphs of the remaining problems have been sketched with a drawing program. We evaluate the limits of the given functions below accordingly using the graphs. All trigonometric functions are defined in radian measure (refer to the Algebra Lecture Guide).

Ç

1å f (x). (In this 2. Let g(x) = sin x . Find lim g(x). Worked Example 1.1.4. 1. Let f (x) = sin x→0 x→0 x . Find lim example, using the tabular method to conjecture x the value of the limit is misleading.) 1

1 0.8

0.5

0.6 0.4

−1

−0.5

0

0

0.5

0.2

1

−0.5

−1 Solution. As x → 0, f frantically fluctuates between 1 and −1 never ap1 proacing a fixed value. Thus lim− sin does not exist. The case is similar x→0 x 1 1 + for x → 0 so that lim+ sin also does not exist. Hence lim sin does not x→0 x→0 x x exist.

−10

5

−5 −0.2

Solution.

10

3. Let h(x) =

Tutorial 1.1.1. 1. Thomas Exercises 11th ed. pp. 75–76: 1, 2, 4, 7, 8 or 12th ed. 2.2 p. 73: 1, 2, 4, 7, 8. 2. If f is a function such that f (2) = 4, can you conclude anything about lim f (x)? Explain your answer and provide a graph for f to substantiate.

1 − cos2 x. Find lim h(x). x→0 x 0.8

x→2

3. If g is a function such that lim g(x) = 4, can you conclude anything about x→2 g(2)? Explain your answer and provide a graph for g to substantiate. 4. Let f (x) = x − ⌊ x⌋. For each integer n, find lim− f (x) and lim+ f (x) if they x→n x→n exist.

0.6 0.4 0.2 −10

5

−5 −0.2 −0.4 −0.6 −0.8

Solution.

10

In the above examples we showed that either a limit of a function f exists at x = a ( f (x) approaches a unique value L ∈ R) or it does not exist ( f (x) does not approach a unique value).

The limit of a function f can also be unbounded, i. e., as x → a+ or x → a− , the values of f (x) become positively large or negatively large. We use the infinity symbol ∞ to denote unboundedness. +∞, or shortly ∞, will denote positively large values and −∞ will denote negatively large values. ∞ is a symbolic representation of having no bound/limit and is not a real number. You cannot algebraically operate with ∞ as if it were an element of the reals.

1 lim− f (x), lim+ f (x), Worked Example 1.1.5. 1. Let f (x) = . Find x→0 x→0 x lim f (x). x→0 Solution. 1 2. Let g(x) = − 2 . Find lim− g(x), lim+ g(x), lim g(x). x→0 x→0 x→0 x Solution. sec x and limπ + sec x. 3. Find lim π− x→ 2

x→ 2

Solution.

The following infinite limit cases arise 1. lim− f (x) = ∞ = lim+ f (x) ⇒ lim f ( x) = ∞ means that f ( x) is arbitrarily x→a x→a x→a positively large for x sufficiently close to a. lim f (x) in fact does not exist x→a here but the infinity notation describes the behaviour of f close to a. 2. lim− f (x) = −∞ = lim+ f (x) ⇒ lim f (x) = −∞ means that f (x) is x→a x→a x→a arbitrarily negatively large for x sufficiently close to a. lim f (x) in fact does x→a not exist here but the infinity notation describes the behaviour of f close to a. 3. lim− f (x) = ∞ and lim+ f (x) = −∞ (or vice versa). lim f (x) in fact does x→a x→a x→a not exist here but the infinity notation describes the behaviour of f close to a− and a+ .

Vertical asymptotes arise when limits are infinite. These are auxiliary vertical lines that the function values approach (but never attain). We say that the line x = a is a vertical asymptote of f if at least one of the following limits hold lim f (x) = ∞ or

x→a−

(or

x→a+

lim f (x) = ∞ or

x→a−

lim f (x) = ∞ or

x→a

x→a

lim f (x) = −∞ or

lim f (x) =

x→a+

lim f (x) = −∞).

The previous worked example illustrates the concept of a vertical asymptote. Horizontal asymptotes arise when we consider the limit of a function f for unbounded values of x, i. e., lim f (x) and lim f (x). If, for arbitrary x→∞ x→−∞ positively or negatively large values of x, the function values approach a unique value b ∈ R, i. e., if lim f (x) = b or lim f (x) = b, then the line x→∞ x→−∞ y = b is called a horizontal asymptote.

The following examples illustrate infinite (unbounded) limits graphically. Horizontal and vertical asymptotes are useful in sketching graphs, as you will discover in Chapter 6 of this guide. In such cases lim f (x) DOES NOT EXIST. x→a

1 lim f (x) For functions of the form f (x) , we may possibly argue with the behaviour lim f (x) and x→−∞ Worked Example 1.1.6. 1. Let f (x) = 2 + . Find x→∞ x g(x) from the graph of f . f (x) of the numerator and denominator near a to calculate lim . x→a Solution. g(x)

2. Let g(x) =

1 . Find lim g(x) and lim g(x) from the graph of g. x→∞ x→−∞ x2

Solution.

2x2 3. Let h(x) = . From the graph evaluate lim− h(x), lim+ h(x), lim− h(x), x→3 x→3 x→−3 9 − x2 lim + h(x), lim h(x) and lim h(x). What are the horizontal and vertical x→∞ x→−∞ x→−3 asymptotes (if any)? Solution.

Worked Example 1.1.7. Evaluate the following limits without the use of a sketch. 2 2 1. lim+ and lim− . x→3 x − 3 x→3 x − 3 Solution:

2. x→−2 lim+

x−1

2 Solution:x (x + 2)

.

1.2

Limit Laws

We now will make use of the following Limit Laws to calculate a limit of a function rather than the approaches used in Section 1.1. Theorem 1.1 (Limit Laws). Let c ∈ R and suppose that lim f (x) = L and x→a lim g(x) = M both exist. Then x→a

1. lim[ f (x) + g(x)] = lim f (x) + lim g(x) = L + M. x→a

x→a

x→a

2. lim[ f (x) − g( x)] = lim f (x) − lim g(x) = L − M. x→a

x→a

x→a

3. lim[c f (x)] = c lim f (x) = cL. x→a

3. lim− cosec x. x→π Solution:

x→a

ò

òï

ï

4. lim[ f (x)g( x)] = lim f (x) lim g(x) = LM. x→a x→a x→a f (x) L f (x) lim x→a = . = x→a g(x) M g(x) lim x→a

5. if M , 0, lim

6. if L , 0 and M = 0, lim

x→a

+



n

f (x) 7. if n ∈ Z , lim x→a



f (x) does not exist. g(x)

ï

òn

= lim f (x)

8. lim c = c. x→a

Tutorial 1.1.2. 1. Thomas Exercises 11th ed. 2.5 pp. 117–118: 6, 13, 16, 9. lim x = a. x→a 21, 39, 42, 43 or 12th ed. 2.6 pp. 114-115: 13–16, 19, 64, 67, 68. 10. if n ∈ Z+ , lim xn = an . x→a

x→a

= Ln .

√n √ 11. if n ∈ Z+ , x→a lim x = n a. If n is even, we assume that a > 0. q √ √ 12. if n ∈ Z+ , lim n f (x) = n lim f (x) = n L. x→a

x→a

13. if lim |h(x)| = 0, then lim h(x) = 0. x→a

x→a

Worked Example 1.2.1. Use the above Limit Laws to calculate the following limits if they exist. 1. lim(5x2 − 9x − 8). x→4

Solution.

2 x 3 + 3 √x . 2. lim x→8 4 − 16 x Solution.

3. x→2 lim

4x − 3

x2 − 4 Solution.

Worked Example 1.2.2. Evaluate lim x→2

.

Solution.

2x2 + 1 x2 + 6x − 4

.

p(x) where both the numerator and denomiq(x) 0 nator tend to zero (said to be of the indeterminate form ), we use factor0 ization to simplify the limit and then use the Limit Laws.

For rational functions f (x) =

Recall that a polynomial function is of the form

Worked Example 1.2.3. Evaluate the following limits if they exist. x2 − 4 f (x) = bn x + bn−1 x + · · · + b2 x + b1 x + b0 . 1. lim x→2 x − 2 with bi ∈ R for i = 1, 2, . . . , n and n any non-negative integer. A rational Solution. p(x) function is of the form f (x) = with p(x) and q(x) polynomials. We q(x) then have the following as a consequence of Theorem 1.1 (1,8,10,5). n

n−1

2

Theorem 1.2. If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a). x→a

3 2. x→−1 lim x + 1 . x+1 Solution.

3. lim

x→−1 (2 x2

Solution.

x+1 . + 7x + 5)2

4. lim s

. xx43−−16 8 Solution. x→2

x 2. lim . 0 x→0 √ where the Quotient Limit Law 1 + 3x − 1 Certain quotients are also of the form 0 rationalizing the numerator or Solution. cannot be applied immediately. However, denominator may allow us to proceed with the Limit Laws. Worked Example 1.2.4. Evaluate the following limits. √ x+1−1 . 1. lim x→0 x Solution.

Tutorial 1.2.1. 1. Thomas Exercises 11th ed. 2.2 pp. 83–85: 2, 4, 5, 8, 12, 15–18, 24, 28, 31, 36, 36, 39, 42, 55, 58 or 12th ed. 2.2 pp. 74–76: 11–13, 18, 21, 22, 24, 34, 37, 42, 53, 56, 79, 82. 2. Given that lim f (x) = 2, lim g(x) = 0 and lim h(x) = −3 find the limits x→a x→a x→a that exist. If the limit does not exist, explain why. √ ( f (x))3 , c. lim 3 2 f (x) − 4h(x), a. lim [ f (x) − h( x)], b. lim x→a x→a x→a f (x) g(x) . , e. lim d. lim x→a g(x) x→a h(x) 3. Find the limit, if it exists. √ √ Ç å 1 1 2+4− + x 5 1 1 x 3 √ , b. lim √ , c. lim . − a. lim x→−3 3 + x x→1 x→0 x+3−2 x 1+x x

Worked Example 1.3.2. Evaluate the following limits. 2x − 3 1. lim . x→∞ 2 x + 7x − 1 Solution.

2.

lim x→∞

√

Solution.

9x2 + 2

3 − 4x

.

Theorem 1.4 (Vertical asymptote: x = a). 1. If n is an even positive integer, then lim

1

= ∞. (x − a)n 1 1 = ∞ and lim− = 2. If n is an odd positive integer, then lim+ x→a (x − a)n x→a (x − a)n −∞. x→a

Worked Example 1.3.3. Evaluate lim+ x→3

Solution.

1 1 and lim− . 3 x→3 (x − 3) (x − 3)3

Theorem 1.5 (Limits of exponential functions). For the exponential function ax , 1. if a > 1, then lim ax = ∞

x→∞

and

lim ax = 0.

x→−∞

2. if 0 < a < 1, then lim ax = 0

x→∞

and

lim ax = ∞.

x→−∞

Worked Example 1.3.4. Evaluate the following exponential limits. 1. lim e x and lim e x . x→∞

Solution.

x→−∞

x −x e x − e−x 2. x→∞ lim e − e x→−∞ . and lim e x + e−x e x + e−x Solution.

2. Determine algebraically whether x→−4 lim |x + 4| exists. x + 4 The following theorem allows us to prove algebraically the existence (or Solution. non-existence) of limits and to calculate their value. We used this theorem graphically in section 1.1.

1.4

Limits through One-sided Limits

f (x) = L ⇔ lim− f (x) = L = lim+ f (x). Theorem 1.6. lim x→a x→a

x→a

Worked Example 1.4.1. 1. Prove that lim |x| = 0. x→0

Proof.

Tutorial 1.4.1. 1. Thomas Exercises 11th ed. 2.4 p. 108: 13, 18–20 or Worked Example 1.5.1. 1. Use the Sandwich Theorem to prove that 12th ed. 2.4 pp. 91-92: 12–15, 17–20, 52. lim 2 (|x| + 1) = 1. (x + 1) = 1 using the fact that lim x→0 2. For the following functions f (x), determine if the one-sided limits at the x→0 indicated point a exist, find whether the limit at a exists, and compare with the value of f (a) if f is defined at a. 1 x2 − 1 1 a. f (x) = − at a = 1, c. f (x) = at a = 0, b. f (x) = |x − 1| x |x| ⌊x⌋ + ⌊−x⌋ at a = −3.

1.5

The Sandwich Theorem & Trigonometric Limits

Proof. 2. Use the Sandwich theorem to prove that lim sin θ = 0 and hence conclude θ→0 that lim cos θ = 1. (Note that θ is in radian measure.) θ→0

The next important theorem allows us to compute the limits of functions that are bounded (squeezed, pinched or sandwiched) between two other Proof. Consider the sector OAC with central angle θ in the unit circle for θ > 0 (first quadrant) and θ < 0 (fourth quadrant). functions that have the same limit.

x→a

x→a

O

1

D

C

θ | sin θ|

Theorem 1.7 (Sandwich Theorem). If f (x) ≤ g( x) ≤ h(x) when x is near a (except possibly at a) and lim f (x) = L = lim h(x), then lim g(x) = L.

A

x→a

|θ|

| sin θ|

Note. The Sandwich Theorem is also called Squeeze Theorem. |θ| A

θ O

D θ>0

C

−1 θ0

lim sin θ = 0 θ→0

and

which will be used in Theorem 1.8.

lim cos θ = 1 θ→0

tan θ

A

We see that

C

By the above sin A =1 A→0 A lim

or

lim

A→0

A =1 sin A

Together with the above theorem and the Limit Laws we also have the fol- Worked Example 1.5.2. Evaluate the following trigonometric limits. sin 5x lowing important limit. 1. lim . x→0 x 1 − cos θ Solution. = 0. Theorem 1.9. lim θ→0 θ Proof.

sin2 x . 2. lim x→0 x Solution.

sin( x2 ) . x→0 x Solution. 3. lim

Worked Example 1.5.2. Evaluate the following trigonometric limits. sin 5x 1. lim ....