# Chapter 4 Applications of Derivatives (導數的應用)

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## Applications of Derivatives(導數的應用)

### Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

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## (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.

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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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1

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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(ci) 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 4 (Thm 1 的例子)

Find the absolute max. and min. values of f(x) = x2/3 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 33

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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## (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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## 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) = x1/3 isincreasing on R, but its first derivative satisfiesf (x) = 3x12/3 > 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 (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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## Example 3 (一階導數測試法的例子)

Let f(x) = (x2− 3)ex 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) = 2xex+ (x2− 3)ex= ex(x2+ 2x− 3) = ex(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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## 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)ex=∞.

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## (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) =dxd[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 (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= x2/3 ∀ 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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## 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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## (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.

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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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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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## 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→∞

ex

x2 = lim

x→∞

ex

2x = lim

x→∞

ex 2 =∞.

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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)x1/2 = 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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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)] = eL.

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## 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+eln(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= eL= e.

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0

## 的不定型)

Evaluate lim

x→∞x1/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→∞x1/x= lim

x→∞eln xx = eL = e0 = 1.

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## References

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