Chapter 4 Applications of Derivatives (導數的應用)

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Chapter 4 Applications of Derivatives (導數的應用)

Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

Fall 2022

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本章預定授課範圍

4.1 Extreme Values of Functions on Closed Intervals 4.2 The Mean Value Theorem

4.3 Monotonic Functions and the First Derivative Test 4.4 Concavity and Curve Sketching

4.5 Indeterminate Forms and L’Hôpital’s Rule

4.8 Antiderivatives

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Section 4.1

Extreme Values of Functions on Closed Intervals

(函數在閉區間上的極值)

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Def (局部極值)

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

(1) f has a local maximum value (局部極大值) at c if ∃ open interval I containing c s.t. f(x)≤ f(c) ∀ x ∈ D ∩ I.

(2) f has a local minimum value (局部極小值) at c if ∃ open interval I containing c s.t. f(x)≥ f(c) ∀ x ∈ D ∩ I.

(3) Local maximum and minimum values are called the local extrema (局部極值) off.

Note: Local extrema of f are also called the relative extrema (相對

極值) of f.

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Some Questions

Let f be a real-valued function defined on D⊆ R.

Does f always have a local extreme value on D?

How to find the local extrema of f?

What is f^′(c) if f(c) is a local extreme value?

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Example (極值發生處的導數)

(a) The rational function

f(x) = 9(x²− 3)

x³ with f^′(x) = 9(9− x²) x⁴

has a rel. max. at the point (3, 2), and f^′(3) = 0 in this case.

(b) The function f(x) =|x| has a rel. min. value f(0) = 0 at the origin (0, 0), but f ^′(0)@. (Why?)

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示意圖 (承上頁)

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Def (臨界點的定義)

Iff^′(c) = 0 orf ^′(c)@ for some interior point c of D = dom(f), then the value c is called a critical point (臨界點) of f.

Thm 2 (發生局部極值的必要條件)

If f has a local extremum at an interior point c of D = dom(f), then f^′(c) = 0 or f^′(c) @,

i.e. x = c must be a critical point of f.

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Example (Thm 2 的反例)

For f(x) = x³, x = 0 is the only critical point of f, since f^′(x) = 3x² = 0⇐⇒ x = 0. But, f(0) = 0 is NOT a local extremum of f.

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Proof of Thm 2

Suppose that f(c) is a local extremum with f^′(c)̸= 0 ∃.

Without loss of generality, we may assume that f^′(c) > 0.

For ε = f ^′(c)

2 > 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ, then f(x)− f(c)

x− c − f ^′(c) < f^′(c)

2 or f(x)− f(c)

x− c > f ^′(c) 2 > 0.

Thus, we know that

f(x) > f(c) ∀ x ∈ (c, c+δ) and f(x) < f(c) ∀ x ∈ (c−δ, c).

This contradicts to the assumption and hence completes the proof.

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Def of Absolute Extrema (絕對極值)

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

(1) f(c) is an absolute maximum value (絕對極大值) of f on D if f(x)≤ f(c) ∀ x ∈ D.

(2) f(c) is an absolute minimum value (絕對極小值) of f on D if f(x)≥ f(c) ∀ x ∈ D.

(3) Absolute maximum and minimum values are called the absolute extrema (絕對極值) off on D.

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Thm 1 (Extreme Value Theorem; E.V.T. 極值定理)

If f isconti. on I = [a, b], then ∃ c1, c2∈ I s.t.

f(c₁)≤ f(x) ≤ f(c2) ∀ x ∈ I,

i.e., f(c₁) is the absolute minimum value of f and f(c₂) is the absolute maximum value of f on I, respectively.

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How to find the points c

₁

and c

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in Thm 1?

Step 1 find all critical numbers c₁, c₂, . . . , ck of f in the open interval (a, b), where k∈ N.

Step 2 evaluate f(a), f(b) and f(c_i) for i = 1, 2, . . . , k.

Step 3 compare the function values obtained in Step 2.

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Example 3 (Thm 1 的例子)

Find the absolute max. and min. values of f(x) = 10x(2− ln x) on the closed interval [1, e²].

Sol: Applying the Product Rule, we see that

f^′(x) = 10(2− ln x) + 10x−1 x

= 10(1− ln x), x > 0, and hence x = e is the only critical point of f in (1, e²).

Since f(1) = 10(2− 0) = 20, f(e) = 10e ≈ 27.2 and f(e²) = 0, we know that f(e²) = 0 is the absolute min. vale and f(e) = 10e is the absolute max. value, respectively.

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Example 3 的示意圖

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Example 4 (Thm 1 的例子)

Find the absolute max. and min. values of f(x) = x^2/3 on the closed interval [−2, 3].

Sol: Applying the Power Rule, we see that

f^′(x) = 2

3x^(2/3)⁻¹ = 2

3x^−1/3= 2 3√³

x, x̸= 0,

and f^′(0) @. Hence, x = 0 is the only critical point of f in (−2, 3).

Since f(−2) = (−2)^2/3= √³

4, f(0) = 0 and f(3) = √³

9≈ 2.08, we know that f(0) = 0 is the absolute min. vale and f(3) =√³

9 is the absolute max. value, respectively.

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Exammple 4 的示意圖

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Section 4.2

The Mean Value Theorem

(均值定理)

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Thm 3 (Rolle’s Theorem; 羅爾定里)

Suppose that f isconti. on [a, b]anddiff. on (a, b). If f(a) = f(b), then∃ c ∈ (a, b) s.t. f^′(c) = 0.

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Proof of Thm 3

Since f is conti. on [a, b] and f(a) = f(b), it follows from E.V.T. that ∃ c ∈ (a, b) s.t.f(c) is a relative extremum.

Otherwise, f must be a constant function on [a, b] and hence f ^′(x) = 0 ∀ x ∈ (a, b).

Next, we shall claim that f^′(c) = 0. If not, say f^′(c) > 0, then it fowws from the ε-δ Def of a limit that ∃ δ > 0 s.t.

f(x) < f(c) for x∈ (c − δ, c) and f(c) < f(x) for x ∈ (c, c + δ). Thus, f(c) is NOT a relative extremum and this gives a contradiction!

Similarly, we can deduce that f^′(c) < 0 =⇒ f(c) is NOT a relative extremum. Consequently, we must have f ^′(c) = 0 for some c∈ (a, b).

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Example 1 (Rolle’s Thm 的例子)

Show that the equation x³+ 3x + 1 = 0 has exactly one real solution.

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Solution of Example 1

Sice f(x) = x³+ 3x + 1 is conti. on the interval [−1, 0] and f(−1) = −3 < 0 < 1 = f(0), it follows from I.V.T. that

∃ c ∈ (−1, 0) s.t. f(c) = 0.

Suppose that∃ a, b ∈ R with a < b s.t.f(a) = f(b) = 0. Since f is conti. on [a, b] and diff. on (a, b), it follows from Rolle’s Thm that

∃ d ∈ (a, b) s.t. f^′(d) = 0, but we see that f ^′(x) = 3x²+ 3 > 0 ∀ x ∈ R,

which gives a contradiction. So, the equation f(x) = 0 has exactly one zero in (−∞, ∞) = R.

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Thm 4 (Mean Value Theorem; M.V.T. 均值定理)

If f is conti. on [a, b] and diff. on (a, b), then∃ c ∈ (a, b) s.t.

f ^′(c) = f(b)− f(a)

b− a or f(b)− f(a) = f^′(c)(b− a).

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Proof of Thm 4

Let g : [a, b]→ R be a function defined by g(x) = f(x)−f(b)− f(a)

b− a (x− a) − f(a) ∀ x ∈ [a, b].

Since f is conti. on [a, b] and diff. on (a, b), we know that g is conti. on [a, b], diff. on (a, b) and g(a) = 0 = g(b).

Since g^′(x) = f^′(x)−f(b)− f(a)

b− a ∀ x ∈ (a, b), it follows from Thm 3 (Rolle’s Thm) that ∃ c ∈ (a, b)s.t.

0 = g^′(c) = f ^′(c)−f(b)− f(a) b− a or, equivalently, we prove that ∃ c ∈ (a, b) s.t.

f ^′(c) = f(b)− f(a) b− a .

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Example 2 (均值定理的例子)

For f(x) = x², find a number c∈ (0, 2) s.t. f ^′(c) = f(2)− f(0) 2− 0 .

Sol: Since f is conti. on [0, 2] and diff. on (0, 2), it follows from

M.V.T. that∃ c ∈ (0, 2) s.t.

2c = f^′(c) = f(2)− f(0)

2− 0 = 4− 0

2− 0 = 2 or c = 1.

Note that c = 1∈ (0, 2) is the only number satisfying the conclusion of Mean Value Theorem.

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Example 2 的示意圖

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Example (M.V.T. 的補充題)

For any a, b∈ R, prove the following inequality

| sin a − sin b| ≤ |a − b|.

Proof: Let a, b

∈ R. Without loss of generality, we may assume that a < b. Since f(x) = sin x is conti. on [a, b] and diff. on (a, b), it follows from M.V.T. that∃ c ∈ (a, b) s.t.

sin b− sin a = f ^′(c)· (b − a) = (cos c) · (b − a).

So, we immediately see that

| sin a − sin b| = | cos c| · |a − b| ≤ |a − b|

because| cos c| ≤ 1, and hence this completes the proof.

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Cor 1 (均值定理的第一個推論)

If f^′(x) = 0 ∀ x ∈ (a, b), then f(x) = C ∀ x ∈ (a, b), where C is a constant.

pf: For any x

₁, x₂ ∈ (a, b) with x1< x₂, f is conti. on [x₁, x₂] and diff. on (x₁, x₂). Thus, from M.V.T., we see that

f(x₂)− f(x1) = f ^′(d)(x₂− x1) = 0 or f(x₁) = f(x₂) for some d∈ (x1, x2). We then conclude that

f(x) = C ∀ x ∈ (a, b), where C is a constant.

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Cor 2 (均值定理的第二個推論)

If f^′(x) = g ^′(x) ∀ x ∈ (a, b), then ∃ a cnstant C s.t.

f(x) = g(x) + C ∀ x ∈ (a, b).

pf: Let h(x) := f(x)

− g(x) ∀ x ∈ (a, b), then

h^′(x) = f ^′(x)− g^′(x) = 0 ∀ x ∈ (a, b). From Cor. 1, ∃ a cnstant C s.t. h(x) = C ∀ x ∈ (a, b) or f(x) = g(x) + C ∀ x ∈ (a, b).

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Example 4 (Cor 2 的例子)

Find a real-valued function f(x) satisfying f^′(x) = sin x and f(0) = 2.

Sol: Let

g(x) =− cos x. Then we see that

g ^′(x) = (− cos x)^′= sin x = f ^′(x) ∀ x ∈ R.

It follows from Cor. 2 that f(x) =− cos x + C for some constant C.

Moreover, since the graph of f passes through (0, 2), we have 2 = f(0) =−1 + C or C = 3. So, the desired function is f(x) =− cos x + 3.

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Section 4.3

Monotonic Functions and the First Derivative Test

(單調函數與一階導數測試)

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Strictly Monotonic Functions

Def (嚴格單調函數的定義)

Let f be a real-valued function defined on an interval I.

(1) f is (strictly) increasing (嚴格遞增;↗) on I if f(x₁) < f(x₂) whenever x₁, x₂ ∈ I withx₁< x₂.

(2) f is (strictly) decreasing (嚴格遞減; ↘) on I iff(x₁) > f(x₂) whenever x₁, x2 ∈ I withx₁< x2.

(3) The increasing or decreasing functions are called (strictly) monotonic functions (嚴格單調函數).

Note: Monotonic functions are one-to-one, but one-to-one

functions are NOT necessarily monotonic!

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Non-strictly Monotonic Functions

Def (非嚴格單調函數)

Let f be a real-valued function defined on an interval I.

(1) f isnon-strictly increasing or nondecreasing (非遞減)on I if f(x₁)≤ f(x2) whenever x₁, x₂ ∈ I withx₁< x₂.

(2) f isnon-strictly decreasing ornonincreasing (非遞增)on I if f(x₁)≥ f(x2) whenever x₁, x₂ ∈ I withx₁< x₂.

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單調函數的示意圖

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Cor 3 (單調函數的充分條件)

Suppose that f is conti. on [a, b] and diff. on (a, b).

(1) f ^′(x) > 0 ∀ x ∈ (a, b) =⇒ fis increasing (↗)on [a, b].

(2) f ^′(x) < 0 ∀ x ∈ (a, b) =⇒ fis decreasing (↘)on [a, b].

Example (Cor 3 的反例)

The function f(x) = x^1/3 isincreasing on R, but its first derivative satisfiesf ^′(x) = _3x¹_2/3 > 0 ∀ x ∈ R\{0}.

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Proof of Cor 3

For any x₁, x2 ∈ (a, b) withx₁ < x2, since f is conti. on [x₁, x2] and diff. on (x₁, x₂), it follows from M.V.T. that ∃ c ∈ (x1, x₂) s.t.

f(x₂)− f(x1) = f^′(c)(x2− x1).

(1) If f ^′(x) > 0 ∀ x ∈ (a, b), thenf^′(c) > 0 and hence f(x₂)− f(x1) > 0 orf(x₂) > f(x1). This implies that f is increasing (↗) on (a, b).

(2) If f ^′(x) < 0 ∀ x ∈ (a, b), thenf^′(c) < 0 and hence f(x₂)− f(x1) < 0 orf(x₂) < f(x₁). This implies that f is decreasing (↘) on (a, b).

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Example 1 (Cor 3 的例子)

Find the critical points of f(x) = x³− 12x − 5 and identify the open intervals on which f is increasing or decreasing.

Sol: Since the first derivative of f is given by

f^′(x) = 3x²− 12 = 3(x²− 4) = 3(x + 2)(x − 2), the critical points of f are x =−2 and x = 2, respectively.

Hence we know that f^′(x) > 0 for x∈ (−∞, −2) ∪ (2, ∞) and f^′(x) < 0 for x∈ (−2, 2). From Cor 3, f is increasing on the open intervals (−∞, −2) and (2, ∞), and decreasing on (−2, 2).

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Example 1 的示意圖 (1/2)

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Example 1 的示意圖 (2/2)

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Thm (First Derivative Test; 一階導數測試)

Let f be diff. on an open interval containing c except possibly at c.

If x = c is a critical proint of f, then

(1) sign of f ^′ changes from (+) to (−) at c =⇒ f(c) is a local max. value of f.

(2) sign of f ^′ changes from (−) to (+) at c =⇒ f(c) is a local min. value of f.

(3) sign of f^′ does notchange on both sides of c =⇒ f(c) isnota local extremum.

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示意圖 (承上頁)

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Example 3 (一階導數測試法的例子)

Let f(x) = (x²− 3)e^x for all x∈ R.

(a) Find the critical points of f.

(b) Identify the open intervals on which f is increasing or decreasing.

(c) Find the local and absolute extreme values of f.

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Solution of Example 3 (1/2)

(a) From the first derivative of f given by

f ^′(x) = 2xe^x+ (x²− 3)e^x= e^x(x²+ 2x− 3) = e^x(x + 3)(x− 1), we see that x =−3 and x = 1 are critical points of f.

(b) Since f ^′(x) > 0 for x∈ (−∞, −3) ∪ (1, ∞) and f^′(x) < 0 for x∈ (−3, 1), it follows from Cor 3 that f is increasing on (−∞, −3) and (1, ∞), and decreasing on (−3, 1), respectively.

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Example 3 的示意圖 (1/2)

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Solution of Example 3 (2/2)

(c) From the First Derivative Test, since the sign of f^′ changes from (+) to (−) at x = −3, f(−3) = 6e⁻³ ≈ 0.299 is a local max. value. Similarly, f(1) =−2e ≈ −5.437 is a local min.

value because the sign of f ^′ changes from (−) to (+) at x = 1.

In fact, sincef(x) > 0 for |x| >√

3, we see that f(1) =−2e is the absolute min. value and there is no absolute maxi. value because lim

x→∞f(x) = lim

x→∞(x²− 3)e^x=∞.

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Example 3 的示意圖 (2/2)

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Section 4.4

Concavity and Curve Sketching

(凹性與曲線描繪)

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Concavity (函數圖形的凹性)

Let f be diff. on an open interval I = (a, b).

(1) The graph of f is concave up (凹向上; C.U.) on I if its first derivative f^′ is ↗ on I.

(2) The graph of f is concave down (凹向下; C.D.) on I if its first derivative f^′ is ↘ on I.

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凹性的示意圖 (承上頁)

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Thm (Test for Concavity; 凹性測試法)

Suppose that f^′′(x) exists on an open interval I.

(1) f ^′′(x) > 0 ∀ x ∈ I =⇒ the graph of f is C.U. on I.

(2) f ^′′(x) < 0 ∀ x ∈ I =⇒ the graph of f is C.D. on I.

pf: It follows immediately from the Def of Concavity and

f^′′(x) =_dx^d[f^′(x)] that

(1) f ^′′(x) > 0 ∀ x ∈ I =⇒ f^′ is increasing on I =⇒ the graph of f is C.U. on I.

(2) f ^′′(x) < 0 ∀ x ∈ I =⇒ f^′ is decreasing on I =⇒ the graph of f is C.D. on I.

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Example 1 (判斷凹性的例子)

(a) The graph of f(x) = x³ is C.D. on the open interval (−∞, 0) because f^′′(x) = 6x < 0 for x < 0, and its graph is C.U. on (0,∞) because f^′′(x) = 6x > 0 for x > 0.

(b) The graph of f(x) = x² is C.U. onR = (−∞, ∞) because f ^′′(x) = 2 > 0 for all x∈ R.

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Example 1 的示意圖

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Points of Inflection (反曲點的定義; P.I.)

Let f be conti. on an open interval containing c. If the graph of f

1 has a (vertical) tangent line at (c, f(c)), and

2 its concavity changes on both sides of c,

then (c, f(c)) is called a point of inflection of the graph of f.

(函數圖形凹性改變的轉折點即為反曲點!)

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反曲點的示意圖 (1/2)

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反曲點的示意圖 (2/2)

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Thm P.I. (反曲點的必要條件)

Suppose thatf ^′′(x) exists on an open interval containing c. If (c, f(c)) is a point of inflection of the graph of f, then

f ^′′(c) = 0 or f^′′(c)@.

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Proof of Thm P.I.

Without loss of generality, we assume that f ^′′(c) > 0 ∃. Since f ^′′ exists at c, we know that,

xlim→c

f^′(x)− f^′(c)

x− c = f ^′′(c).

Thus, for ε = ^f^′′₂^(c) > 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ, then

f^′(x)− f^′(c)

x− c − f^′′(c) <f^′′(c)

2 or f^′(x)− f^′(c)

x− c >f^′′(c) 2 > 0.

Thenf ^′(x) < f ^′(c) for x∈ (c − δ, c) and f^′(c) < f ^′(x) for x∈ (c, c + δ). Thus, it follows from the Def of concavity that the graph of f is C.U. on I = (c− δ, c + δ). This contradicts to our assumption that (c, f(c)) is a point of inflection of f, and hence we complete the proof.

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Example (Thm P.I. 的反例)

Consider f(x) =√³

x²= x^2/3 ∀ x ∈ R. Then its first and second derivatives are given by

f^′(x) = 2

3x^−1/3 and f^′′(x) = −2

9 x^−4/3< 0

for all x̸= 0. In this case, we see that f ^′′(0) @ and the origin (0, 0) is NOT a point of inflection!

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示意圖 (承上例)

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. . . . . . . .

. . .

. .

. . . . .

Example 3 (反曲點的例子)

Determine the concavity and find the inflection points of f(x) = x³− 3x²+ 2.

Sol: Since the first and second derivatives of f are given by

f ^′(x) = 3x²− 6x, f^′′(x) = 6x− 6 = 6(x − 1), we see thatf^′′(x) < 0 for x < 1 and f^′′(x) > 0 for x > 1.

So, it follows from Thm P.I. that the graph of f is C.D. on (−∞, 1) and is C.U. on (1,∞), respectively. Since the concavity of f changes on both sides of x = 1, (1, f(1)) = (1, 0) is the only infection point of the graph of f.

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Example 3 的示意圖

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. . .

. .

. . . . .

Thm 5 (Second Derivative Test; 二階導數測試法)

Suppose thatf ^′(c) = 0and f ^′′ ∃ on an open interval containing c.

(1) f ^′′(c) > 0=⇒ f(c) is a local min. value.

(2) f ^′′(c) < 0=⇒ f(c) is a local max. value.

(3) f ^′′(c) = 0=⇒ the test is inconclusive.

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示意圖 (承上頁)

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Section 4.5

Indeterminate Forms and L’Hôpital’s Rule

(不定型與羅必達法則)

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Types of Indeterminate Forms

For the limit of a quotient f(x)/g(x) as x→ c, we may have

xlim→c

f(x) g(x) = 0

0, ∞

∞, 0· ∞, 1^∞, ∞⁰, 0⁰, ∞ − ∞, which are called the indeterminate forms (不定型).

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. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Thm 6 (L’Hôpital’s Rule; 羅必達法則)

Suppose that f and g are diff. on an open interval I containing c, and that g^′(x)̸= 0 ∀ x ∈ I \{c}. If the limit satisfies

xlim→c

f(x) g(x) = 0

0 or ±∞

±∞, then we have

xlim→c

f(x) g(x) = lim

x→c

f^′(x)

g ^′(x), (分子和分母各別微分喔!) assuming that the limit on the right exists.

Note: Thm 6 also holds for the one-sided limits. (羅必達法則也

適用於求解單邊極限值!)

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Remark (與 Sec. 2.4 的結果比較)

A statement in Thm of Section 2.4 can be derived immediately from the L’Hôpital’s Rule, since we see that the limit is an indeterminate form of type 0

0 and hence lim

θ→0

sin θ θ = lim

θ→0

cos θ

1 = cos(0) = 1.

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. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Example 2 (Type

⁰₀

的不定型)

Applying the L’Hôpital’s Rule, we see that

xlim→0

1− cos x

x + x² (Type 0 0)

= lim

x→0

sin x

1 + 2x = 0

1 + 0 = 0.

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Example 4 (Thpe

^∞_∞

的不定型)

(b) Applying the L’Hôpital’s Rule, we immediately obtain

xlim→∞

ln x 2√

x (Type ∞

∞)

= lim

x→∞

1/x 1/√

x = lim

x→∞

√1 x = 0.

(c) Applying the L’Hôpital’s Rule twice, we have

xlim→∞

e^x

x² = lim

x→∞

e^x

2x = lim

x→∞

e^x 2 =∞.

HW: read the part (a) by yourself.

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. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Indeterminate Forms of Type 0

· ∞ If lim

x→cf(x) = 0 and lim

x→cg(x) =±∞, then consider

xlim→c[f(x)g(x)] = lim

x→c

f(x)

1/g(x) (Type 0 0) or

limx→c[f(x)g(x)] =lim

x→c

g(x)

1/fx) (Type ∞

∞).

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Example 5 (Type 0 · ∞ 的不定型)

(b) Applying the L’Hôpital’s Rule, we obtain

xlim→0⁺

√x ln x (Type 0· ∞) = lim

x→0⁺

ln x 1/√

x (Type ∞

∞)

= lim

x→0⁺

1/x

(−1/2)x^−3/2 = lim

x→0⁺(−2)x^1/2 = 0.

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. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Indeterminate Forms of Type

∞ − ∞

The original limit will become an indeterminate form of type 00, if we apply the technique of reduction to common denominator.

(使用通分技巧將原極限問題變成標準不定型!)

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Example (補充題; Type ∞ − ∞ 的不定型)

Evaluate the following limit

xlim→1⁺

1

ln x− 1 x− 1

, which is an indeterminate form of type∞ − ∞.

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. . . . . .

. . . . . . . .

. . .

. .

. . . . .

Solution of Example

Applying the L’Hôpital’s Ruletwice, we see that lim

x→1⁺

1

ln x− 1 x− 1

= lim

x→1⁺

x− 1 − ln x

(x− 1) ln x (Type 0

0; 通分!)

= lim

x→1⁺

1− 1/x

ln x + (x− 1)(1/x) (使用 L’H Rule!)

= lim

x→1⁺

1− 1/x

ln x + (x− 1)(1/x)·x x

= lim

x→1⁺

x− 1

x ln x + x− 1 (Type 0 0)

= lim

x→1⁺

1

ln x + x(1/x) + 1 (再次使用 L’H Rule!)

= 1

0 + 1 + 1 = 1

2. (直接代入求極限喔!)

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Indeterminate Forms of Type 1

^∞

,

∞⁰

or 0

⁰ Assume that lim

x→c[f(x)]^g(x) = 1^∞,∞⁰, 0⁰. If we know that limx→cg(x) ln[f(x)] = L (Type 0

0 or ∞

∞) using the L’Hôpital’s Rule, then

limx→c[f(x)]^g(x) =lim

x→ceg(x) ln[f(x)] = e^L.

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. . . . . . . .

. . .

. .

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Example 7 (Type 1

^∞

的不定型)

Show that lim

x→0⁺(1 +x)^1/x= e.

Sol: Note that the given limit can be rewritten as

lim

x→0⁺(1 + x)^1/x= lim

x→0⁺e^ln(1+x)/x. From the L’Hp̂pital’s Rule, we see that

xlim→0⁺

ln(1 + x)

x = lim

x→0⁺

1/(1 + x)

1 = lim

x→0⁺

1

1 + x = 1 := L, and hence lim

x→0⁺(1 + x)^1/x= e^L= e.

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Example 8 (補充題; Type ∞

⁰

的不定型)

Evaluate lim

x→∞x^1/x. ( Type ∞⁰)

Sol: From the L’Hôpital’s Rule, we see that

xlim→∞

ln x x = lim

x→∞

1x

1 = lim

x→∞

1

x = 0 := L.

So, we immediately obtain

xlim→∞x^1/x= lim

x→∞e^{ln x}^x = e^L = e⁰ = 1.