Chapter 2 Limits and Continuity (極限與連續性)

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Chapter 2 Limits and Continuity (極限與連續性)

Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

Fall 2022

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 1/85

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本章預定授課範圍

2.2 Limit of a Function and Limit Laws 2.4 One-Sided Limits

2.5 Continuity

2.6 Limits Involving Infinity; Asymptotes of Graphs

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Section 2.2

Limit of a Function and Limit Laws (函數的極限值與極限法則)

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Informal Definition of a Limit

Def (函數極限值的非正式定義)

Let f be a real-valued function defined on X⊆ R with c∈\X. If f(x) becomesarbitrarily closeto a unique number L∈ R as x

approaches cfrom either side, then the limit of f is L as x approaches c, denoted by lim

x→cf(x) = L.

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Example 2 (在 x = 1 處的極限值均為 2)

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Nonexistence of a Limit (1/3)

Type I (在 c 點兩邊的極限值不相等)

Iff(x)→ L1 andf(x)→ L2, whereL₁ ̸= L2, as x approaches cfrom either side, then lim

x→cf(x) @.

In this case, we say that f has a jump discontinuity at x = c.

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Nonexistence of a Limit (2/3)

Type II (函數值在 c 點附近無上下界)

If f(x)increases (遞增) or decreases (遞減) without bound as x approaches c from either side, then lim

x→cf(x) @.

In this case, we say that f has an infinite discontinuity at x = c.

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Nonexistence of a Limit (3/3)

Type III (函數值在 c 點附近震盪)

If f(x)oscillates (震盪) between two fixed valuesas x approaches c from either side, then lim

x→cf(x) @.

In this case, we say that f has an oscillating discontinuity at x = c.

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Example 4 (在 x = 0 處的極限值不存在)

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Precise Definition of a Limit (p. 88, Section 2.3)

Def (函數極限值的正式定義)

Let f be a real-valued function defined on X⊆ R with c∈\X. Then

xlim→cf(x) = L.

⇐⇒∀ε > 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ (and x ∈ X), then|f(x) − L| < ε.

Note: this is also called the ε-δ definition of a limit.

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極限的正式定義

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Thm (Basic Limit Laws; 1/2)

Let b, c∈ R and let f and g be real-valued functions with

xlim→cf(x) = L, lim

x→cg(x) = K.

(1) lim

x→cb = b and lim

x→c|x| = |c|.

(2) lim

x→cxⁿ= cⁿ and lim

x→c

[ f(x)]n

= Lⁿ ∀ n ∈ N.

(3) lim

x→c[b· f(x)] = b ·[

limx→cf(x)]

= b· L.

(4) lim

x→c

[

f(x)± g(x)]

= [

limx→cf(x)]

±[

xlim→cg(x)]

= L± K.

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Thm (Basic Limit Laws; 2/2)

(5) lim

x→c

[f(x)· g(x)]

= [lim

x→cf(x)]

·[

limx→cg(x)]

= L· K.

(6) lim

x→c

f(x) g(x) =

xlim→cf(x)

xlim→cg(x) = LK if K̸= 0. (7) lim

x→c

√n

f(x) =√ⁿ

L ∀ n ∈ N, where L ≥ 0 if n is even.

(8) The Limit of f◦ g: (合成函數的極限值) If lim

x→cf(x) = f(K), then lim

x→cf(g(x)) = f(

limx→cg(x))

= f(K).

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Example 5 (極限法則的例子)

(a) lim

x→c(x³+ 4x²− 3) = lim

x→cx³+ 4 (lim

x→cx²)

− lim

x→c3 = c³+ 4c²− 3.

(b) lim

x→c

x⁴+x²−1

x²+5 = ^x^lim^→c^(x

4+x²−1)

xlim→c(x²+5) = ^c⁴_c^+c2+5²⁻¹. (c) lim

x→(−2)

√4x²+ 3 =√

4(−2)²+ 3 =√ 19.

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Thm (Limits of Elementary Functions; 1/2)

Let c be a real number in the domain of the given function.

(1) If p(x) is a polynomial, then lim

x→cp(x) = p(c).

(2) If r(x) = p(x)/q(x) is a rational function with q(c)̸= 0, then

xlim→cr(x) = r(c) = p(c)/q(c).

(3) lim

x→c

√n

x =√ⁿ

c ∀ n ∈ N, where c ≥ 0 when n is even and c∈ R when n is odd.

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Thm (Limits of Elementary Functions; 2/2)

(4) Limits of 6 trigonometric functions are given by

xlim→csin x = sin c, lim

x→ccos x = cos c, lim

x→ctan x = tan c,

xlim→ccot x = cot c, lim

x→csec x = sec c, lim

x→ccsc x = csc c.

(5) lim

x→ca^x= a^c for a > 0and c∈ R.

(6) lim

x→cln x = ln c for c > 0.

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Thm (化簡函數後求極限值)

If∃ δ > 0 s.t.f(x) = g(x) ∀ x ∈ (c − δ, c) ∪ (c.c + δ), then

xlim→cf(x) = lim

x→cg(x).

Note: 將 f(x) 簡化為 g(x) 後，兩者在 c 點附近的極限值相等!

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Example 7 (分子和分⺟消去公因式)

Evaluate lim

x→1

x²+ x− 2 x²− x .

Sol: Note that the given function can be simplified as

f(x) = x²+ x− 2

x²− x = (x + 2)(x− 1)

x(x− 1) = x + 2

x = g(x) for x̸= 1.

Thus, from above Thm, we see that

xlim→1

x²+ x− 2 x²− x = lim

x→1f(x) = lim

x→1g(x) = 1 + 2 1 = 3.

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Example 7 的示意圖

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Example 9 (以有理化技巧求極限)

Evaluate lim

x→0

√x²+ 100− 10

x² .

Sol: Applying the rationalizing technique, we obtain

xlim→0

√x²+ 100− 10

x² = lim

x→0

(√x²+ 100− 10

x² ·

√x²+ 100 + 10

√x²+ 100 + 10 )

= lim

x→0

x² x²(√

x²+ 100 + 10) = lim

x→0

√ 1

x²+ 100 + 10

= 1

√0²+ 100 + 10 = 1

20 = 0.05.

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Thm 4 (The Sandwich Theorem; 三明治定理)

If∃ δ > 0 s.t.h(x)≤ f(x) ≤ g(x) ∀ x ∈ (c − δ, c) ∪ (c, c + δ), and

xlim→ch(x) = L = lim

x→cg(x), then lim

x→cf(x) = L.

Note: this theorem is also called the Squeeze Theorem (夾擠定理

或夾擊定理).

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Thm 4 的示意圖

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Example 11 (三明治定理的例子)

Prove the following statements using the Sandwich Theorem:

(a) lim

θ→0sin θ = 0.

(b) lim

θ→0cos θ = 1.

(c) For any real-valued function f, lim

x→c|f(x)| = 0 =⇒ lim

x→cf(x) = 0.

In fact, it is also true that lim

x→c|f(x)| = 0 ⇐⇒ lim_x

→cf(x) = 0.

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Proof of Example 11

(a) Since −|θ| ≤ sin θ ≤ |θ| for θ ∈ R (ssee Section 1.3), and lim

θ→0(−|θ|) = lim

θ→0|θ| = 0, it follows from the Sandwich Thm that lim

θ→0sin θ = 0.

(b) From the Sandwich Thm and 0≤ 1 − cos θ ≤ |θ| for θ ∈ R (see Section 1.3), we obtain lim

θ→0(1− cos θ) = 0 and hence

θlim→0cos θ = lim

θ→0

[

1− (1 − cos θ)]

= 1− 0 = 1.

(c) The result follows from−|f(x)| ≤ f(x) ≤ |f(x)| for x ̸= c and the Sandwich Thm.

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Example 11 的示意圖

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Section 2.4 One-Sided Limits

(單邊極限)

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One-Sided Limits

Def (單邊極限值的定義)

(1) f has the limit L from the right (or the right-hand limit L; 右 極限值) at c, denoted by lim

x→c⁺f(x) = L, if f(x)→ L as x → c from the right.

(2) f has the limit L from the left (or the left-hand limit L; 左極限 值) at c, denoted by lim

x→c⁻f(x) = L, if f(x)→ L as x → c from the left.

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單邊極限值的示意圖

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Existence of a Limit

Thm 6 (函數極限值存在的等價條件)

Suppose that f is defined on an open interval containing c, except possibly at c itself. Then

xlim→cf(x) = L⇐⇒ lim

x→c⁻f(x) = L = lim

x→c⁺f(x).

(f 在 c 點的極限值為 L⇐⇒ 左右極限值均為 L)

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Example 4 (Thm 6 的例子)

Notice that y = f(x) = sin¹_x is defined on the open intervals (−∞, 0) and (0, ∞), respectively, and f(0) is NOT

well-defined.

Since the graph of f oscillates between −1 and 1 on both sides of x = 0, we know that

xlim→0⁻f(x) and lim

x→0⁺f(x) @.

Therefore, lim

x→0f(x) = lim

x→0sin¹_x @.

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Example 4 的示意圖

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Limits at Endpoints of an Interval

Def (區間端點的極限)

(1) If f is defined on an open intervaal (c, b) and not defined on on an open interval (a, c), then dfine

limx→cf(x) = lim

x→c⁺f(x).

(2) If f is defined on an open intervaal (a, c) and not defined on on an open interval (c, b), then dfine

xlim→cf(x) = lim

x→c⁻f(x).

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Example 1 (端點極限的例子)

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Solution of Example 1

At x = 0 : lim

x→0⁺f(x) = 1 = f(0). So, lim

x→0f(x) = 1 by Def.

At x = 1 : lim

x→1⁻f(x) = 0̸= 1 = lim

x→1⁺f(x). So, lim

x→1f(x) @ by Thm 6.

At x = 2 : lim

x→2⁻f(x) = 1 = lim

x→2⁺f(x). So, f(2) = 2̸= 1 = lim

x→2f(x)∃ by Thm 6. In this case, we say that f has a removable

discontinuity (可移除不連續點) at x = 2.

At x = 3 : From Thm 6, we see that lim

x→3f(x) = 2 = f(3) and hence we say that f has a continuity (連續點) at x = 3.

At x = 4 : lim

→4⁻f(x) = 1̸= f(4). So, lim

x→4f(x) = 1 by Def.

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Example 2 (端點極限的例子)

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Solution of Example 2

For the function f(x) = arcsec x, we see that (i) lim

x→cf(c) = f(c) for any c∈ (−∞, −1) ∪ (1, ∞).

(ii) Since f is NOT defined on (−1, 1), it follows that the endpoint limits of f at x =±1 are

x→(−1)lim f(x) = lim

x→(−1)⁻f(x) = arcsec(−1) = π,

xlim→1f(x) = lim

x→1⁺f(x) = arcsec(1) = 0.

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Thm 7 (Some Special Limits)

(1) lim

θ→0

sin θ

θ = 1. (see Theorem 7) (2) lim

θ→0

1− cos θ

θ = 0. (see Example 5a) (3) lim

x→0(1 + x)^1/x= e.

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Example 5b (Theorem 7 的例子)

Use Theorem 7 to show that lim

x→0 sin 2x

5x = ²₅.

Sol: We fist observe that

sin 2x

5x = sin 2x 2x ·2x

5x for x̸= 0.

If we letθ = 5x, then θ→ 0 as x→ 0 and thus

xlim→0

sin 2x 5x =

(lim

θ→0

sin θ θ

)(lim

x→0

2x 5x )

= (1) (2

5 )

= 2 5.

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Section 2.5 Continuity

(連續性)

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Continuity of a Function

Def (實值函數的連續性)

Let f be a real-valued function defined on I = (a, b) withc∈ I. (1) f is continuous (連續的; 簡寫為 conti.) at c if lim

x→cf(x) = f(c).

(2) f is conti. on I if it is conti. at each c∈ I.

(3) f is everywhere conti. (處處連續) if it is conti. on R = (−∞, ∞).

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Discontinuity of a Function

Def (函數的不連續性)

Let f be a real-valued function defined on I = (a, b) with c∈ I.

(1) f has a discontinuity (不連續點; 簡寫為 disconti.) at c if it is NOT conti. at c.

(2) A disconti. of f at c is called removable (可移除的)if f can be made conti. at cby redefining f(c). Otherwise, the disconti. at c is called nonremovable (不可移除的).

Note: the jump disconti., infinite disconti. and oscillating disconti.

are nonremovable.

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One-Sided Continuity

Def (單邊連續的定義)

(1) f is right-conti. or conti. from the right (佑蓮續) at c if

xlim→c⁺f(x) = f(c).

(2) f is left-conti. or conti. from the left (左蓮續) at c if

xlim→c⁻f(x) = f(c).

Remark

f is conti. at c⇐⇒ f is right-conti. and left-conti. at c.

(f 在 c 點連續⇐⇒ f 在 c 點右蓮續且左連續)

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Example 1 (判斷單點的連續性)

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Solution of Example 1

At x = 0 : f is right-conti. at x = 0 because lim

x→0⁺f(x) = 1 = f(0).

At x = 1 : f has a jump disconti. at x = 1 because lim

x→1⁻f(x) = 0̸= 1 = f(1) = lim

x→1⁺f(x). In fact, f is right-conti.

at x = 1.

At x = 2 : f has a removable disconti. at x = 2 because it can be made conti. at x = 2 by redefining f(2) := lim

x→2f(x) = 1.

At x = 3 : f is conti. at x = 3 because lim

x→3f(x) = 2 = f(3).

At x = 4 : f is NOT left-conti. at x = 4 because lim

x→4⁻f(x) = 1̸= f(4), but it can be made left-conti. there by redefining

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Example 4 (最大整數函數)

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Solution of Example 4

Let f(x) =⌊x⌋ be the greatest integer function (最大整數函 數) defined on R. For each n ∈ Z, we have

xlim→n⁻f(x) = n− 1 and lim

x→n⁺f(x) = n = f(n).

So, f is right-conti. and has a jump disconti. at each n∈ Z, but NOT left-conti. there! In addition, f is conti. at every point x∈ R\Z.

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Continuity on a Closed Interval

Def (在閉區間上的連續性)

We say that f is conti. on I = [a, b] if the following conditions hold:

1 f is conti. on the open interval (a, b).

2 f is right-conti. at a, i.e., lim

x→a⁺f(x) = f(a).

3 f is left-conti. at b, i.e., lim

x→b⁻f(x) = f(b).

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Example 2 (在閉區間上的連續性)

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Solution of Example 2

Note that the domain of f(x) =√

4− x² is D = [−2, 2].

Since f is conti. on the open interval (−2, 2), right-conti. at x =−2 and left-conti. at x = 2, we conclude that f is conti.

on the closed interval D.

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Properties of Continuity

Thm (連續函數的性質)

1 If f and g are conti. at c and b∈ R, then f ± g, bf, fg and f/g with g(c)̸= 0 are conti. at c, respectively.

2 If g is conti. atc and f is conti. atg(c), then (f◦ g)(x) = f(g(x)) is conti. at c. (see Theorem 10)

3 Elementary functions and inverse trigonometric functions are conti. on their domains.

Note: the above properties are also true for

one-sided continuity!

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Examples 6 and 7

(a) The polynomial P(x) is everywhere conti., since

xlim→cP(x) = P(c) for any c∈ R.

(b) The rational function P(x)/Q(x) is conti. at every point c∈ R with Q(c)̸= 0, since its limit at c satisfies

xlim→c

P(x)

Q(x) = P(c) Q(c) ∃.

(c) The function f(x) =|x| =

{ x, x≥ 0

−x, x < 0 is everywhere conti., since f is a polynomial for x̸= 0 and

xlim→0f(x) = 0 = f(0).

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Example 8 (判斷合成函數的連續性)

(a) The funtion f(x) =√

x²− 2x − 5 is conti. on its domain dom(f) = (−∞, 1 −√

6]∪ [1 +√ 6,∞).

(b) y = _1+x^x^2/3₄ is everywhere continuous, since 1 + x⁴> 0 ∀ x ∈ R.

(c) y =_x^x²⁻²₋₂ is conti. for all x ̸= ±√ 2.

(d) y =^{x sin x}_x²₊₂ is everywhere conti., since x²+ 2 > 0 ∀ x ∈ R.

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Example 9 (計算合成還數的極限值)

(a) Applying the continuity of the composite function, we have lim

x→π/2cos( 2x+sin

(3π 2 +x

))

= cos(π+sin(2π)) = cos π =−1.

(b) lim

x→1sin⁻¹(

1−x 1−x²

)

= sin⁻¹ (lim

x→1 1 1+x

)

= sin⁻¹(1/2) = π/6.

(c) lim

x→0

√x + 1e^{tan x}=√

0 + 1· e⁰ = 1.

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Intermediate Value Theorem

Thm 11 (I.V.T.; 中間值定理)

If f isconti. on [a, b],f(a)̸= f(b)andk is any number between f(a) and f(b), then∃ c ∈ [a, b] s.t.f(c) = k.

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Example 11 (中間值定理的應用)

Use the I.V.T. to show that the equation

√2x + 5 = 4− x²

has a solution in the closed interval [0, 2].

pf: Since f(x) =

√

2x + 5 + x²− 4 is conti. on dom(f) = [−5/2, ∞), it follows that f is also conti. on [0, 2]. Moreover, it is easily seen thatf(0) =√

5− 4 ≈ −1.76 < 0 < 3 =√

9 = f(2). Thus, for the value k = 0 in the I.V.T.,∃ c ∈ (0, 2) s.t. f(c) = k = 0, i.e., the nonlinear equation (非線性方程式) f(x) = 0 has at least one solution c∈ (o, 2).

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Example 11 的示意圖

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Section 2.6

Limits Involving Infinity; Asymptotes of Graphs

(無窮極限: 函數圖形的漸近線)

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Limits at Infinity

Def (無窮遠處的極限)

(1) lim

x→∞f(x) = L⇐⇒ ∀ ε > 0,∃ M > 0s.t. ifx > M, then

|f(x) − L| < ε.

(2) lim

x→−∞f(x) = L⇐⇒ ∀ ε > 0,∃ N < 0s.t. if x < N, then

|f(x) − L| < ε.

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示意圖 (承上頁)

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Thm (重要的極限法則)

(1) If r > 0 is arational number andc∈ R, then

xlim→∞

c

x^r = 0 = lim

x→−∞

c x^r. (2) lim

x→∞e^−x= 0 = lim

x→−∞e^x.

HW: read Example 1, Section 2.6 by yourself!

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. .

. . . . .

Proof of (1)

Let ε > 0 be given arbitrarily. Choose M, N∈ R with M >(|c|

ε )1/r

> 0 and N <−(|c|

ε )1/r

< 0.

Thus we have the following inequalities

|c|

M^r < ε and |c|

(−N)^r < ε. (Check!) If x > M(> 0) or x < N(< 0), then

c

x^r−0 = |c|

x^r < |c|

M^r < ε or c

x^r−0 = |c|

|x|^r = |c|

(−x)^r < |c|

(−N)^r < ε.

So, it follows from the Def. that

xlim→∞

c

x^r = 0 = lim

x→−∞

c x^r.

(66)

Example 3 (有理函數在無窮遠處的極限)

(a) Applying above Thm, we see that

xlim→∞

5x²+ 8x− 3 3x²+ 2 = lim

x→∞

5 + (8/x)− (3/x²)

3 + (2/x²) = 5 + 0− 0 3 + 0 = 5

3. (b) Applying above Thm, we see that

x→−∞lim

11x + 2

2x³− 1 = lim

x→−∞

(11/x²) + (2/x³)

2− (1/x³) = 0 + 0 2− 0 = 0.

(67)

. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Example 3 的示意圖

(68)

Horizontal Asymptotes

Def (水平漸近線的定義)

A line y = L is callled a horizontal asymptote (水平漸近線) of the graph of f if either

xlim→∞f(x) = L or lim

x→−∞f(x) = L.

(69)

. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Example 4 (水平漸近線的例子)

Find the horizontal asymptotes of the graph of the function f(x) = x³− 2

|x|³+ 1.

Sol: Since the limits of f as x

→ ±∞ are given by

xlim→∞f(x) = lim

x→∞

x³− 2 x³+ 1 = lim

x→∞

1− (2/x³)

1 + (1/x³) = 1 and

x→−∞lim f(x) = lim

x→−∞

x³− 2

(−x)³+ 1 = lim

x→−∞

x³− 2

−x³+ 1 =−1, the horizontal asymptotes of the graph of f arey = 1andy =−1, respectively.

(70)

Example 4 的示意圖

(71)

. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Infinite Limits (1/2)

Def (無窮極限值的定義; 1/2)

(1) lim

x→cf(x) =∞ ⇐⇒ ∀ M > 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ, then f(x) > M.

(2) lim

x→cf(x) =−∞ ⇐⇒ ∀ N < 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ, then f(x) < N.

(72)

示意圖 (承上頁)

(73)

. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Infinite Limits (2/2)

Def (無窮極限值的定義; 2/2)

(3) lim

x→c⁺f(x) =∞ ( or lim

x→c⁻f(x) =∞ )

⇐⇒ ∀ M > 0, ∃ δ > 0 s.t. if c < x < c + δ (orc− δ < x < c), then f(x) > M.

(4) lim

x→c⁺f(x) =−∞ ( or lim

x→c⁻f(x) =−∞)

⇐⇒ ∀ N < 0, ∃ δ > 0 s.t. if c < x < c + δ (orc− δ < x < c), then f(x) < N.

(74)

示意圖 (承上頁)

(75)

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. . . . . . . .

. . .

. .

. . . . .

重要口訣 (切記!)

若 +0 與−0 分別代表接近零的正數與負數, 則

1

+0 =∞, ₋₀¹ =−∞

∞1 = _−∞¹ = 0,

其中，+∞ = ∞ 和 −∞ 分別為正負無窮遠處的符號。

(76)

Example (無窮極限的例子)

(a) The one-sided limits of sec x at x =−π/2 are lim

x→(−π/2)⁻sec x = lim

x→(−π/2)⁻

1

cos x = 1

−0 =−∞, lim

x→(−π/2)⁺sec x = lim

x→(−π/2)⁺

1

cos x = 1

+0 =∞.

(b) The one-sided limits of tan x at x = π/2 are

x→(π/2)lim ⁻tan x = lim

x→(π/2)⁻

sin x cos x = 1

+0 =∞,

x→(π/2)lim ⁺tan x = lim

x→(π/2)⁺

sin x cos x = 1

−0 =−∞.