1.4 ⌿⌿⌿333ᦲᦲᦲ‏‏‏ᬿᬿᬿ▲▲▲

1.4 連 連 連續 續 續函 函 函數 數 數與 與 與極 極 極 限 限 限

8 第 1 章 基本函數與極限

(7) 由三角函數的定義, 對任何角度 θ, cos θ = 1 sec θ, 所以 cos sec⁻¹x = 1

sec sec⁻¹x = 1 x (8) 由 於 csc⁻¹x 的 取 值 ∈ (−π

2,π

2), 因 此 cos csc⁻¹x 的 值 必 ≥ 0. 又 同 前 習 題, 有 sin csc⁻¹x = 1

x, 故

sin²x + cos²x = 1 ⇒ (cos csc⁻¹x)²= 1− (sin csc⁻¹x)²= 1− (1 x)²

⇒ (cos csc⁻¹x)²= x²− 1 x²

⇒ cos csc⁻¹x =

√x²− 1

|x|

1.4 連續函數與極限

1.4.1 連續函數 1.4.2 數列的極限







習題解答 1.4.1.



如果此數列有兩個相異的極限 L₁ 和 L₂, 令 ϵ = ^|L1−L₄ ^2|̸= 0. 依照課本極限的定義, 這數 列終究會落到 (L₁− ϵ, L1+ ϵ) 內, 同時也終究會落到 (L₂− ϵ, L2+ ϵ) 內. 但是這兩個範 圍根本不相交, 這違反了數列的「終究」定義, 因此數列不可能有兩個以上的極限.







習題解答 1.4.2.



(1) 由題意知：an 終究會等於 L + (很小的數)；b_n終究會等於 M + (很小的數). 所以 a_n· bn= (L + (很小的數))· (M + (很小的數)) = L · M + (很小的數) 因此 lim

n→∞(an· bn) = L· M.

(2) 由題意知：b_n終究會等於 M + (很小的數). 所以 1

bn = 1

M + (很小的數) = 1

M (1 + (很小的數))

= 1

M · (1 + (很小的數)) = 1

M + (很小的數) 因此 lim

n→∞

1 bn = 1

M.

1.4. 連續函數與極限 9







習題解答 1.4.3.



nlim→∞

an

bn = lim

n→∞(an· 1

bn)⁽²⁾= lim

n→∞an· lim_n

→∞

1 bn

(3)= L· 1

M = L

M







習題解答 1.4.4.



(1)

nlim→∞

n²+ 1

n²+ n = lim

n→∞

1 +_n¹2

1 +_n¹ =

n→∞lim (1 +_n¹2)

nlim→∞(1 +¹_n)

= ( lim

n→∞1) + ( lim

n→∞

1 n)² ( lim

n→∞1) + ( lim

n→∞

1

n) = 1 + 0 1 + 0 = 1

(2)

n→∞lim n³+ 1 n²+ n= lim

n→∞

n +_n¹2

1 +_n¹ =

n→∞lim (n +_n¹2)

n→∞lim (1 +_n¹) = ( lim

n→∞n) + 0 1 + 0 = ∞ (3)

n→∞lim n + 1

n²+ n= lim

n→∞

1 n+_n¹2

1 +_n¹ =

n→∞lim (_n¹+ _n¹2)

nlim→∞(1 +_n¹) = 0 + 0 1 + 0 = 0 (4) 因為 lim

n→∞

1

n = 0, 且 y = sin x 是連續函數. 由性質 1.4.1(4) 知

nlim→∞ sin(1

n) = sin( lim

n→∞

1

n) = sin 0 = 0







習題解答 1.4.5.



(1) l > k:

n→∞lim

akn^k+ ak−1n^k−1+· · · + a0

bln^l+ bl−1n^l−1+· · · + b0

=

nlim→∞(ak 1

n^l−k + ak−1 1

n^l−k+1 +· · · + a01 n^l)

n→∞lim (bl+ bl−11

n+· · · + b01 n^l)

= 0 + 0 +· · · + 0 bl+ 0 +· · · + 0= 0

1

10 第 1 章 基本函數與極限

(2) l = k:

n→∞lim

akn^k+ ak−1n^k−1+· · · + a0

bkn^k+ bk−1n^k−1+· · · + b0

=

nlim→∞(a_k+ a_k−1n¹+· · · + a01 n^k)

nlim→∞(bk+ bk−11

n+· · · + b0 1 n^k)

= ak+ 0 +· · · + 0 bk+ 0 +· · · + 0 = ak

bk

(3) k > l

nlim→∞

akn^k+ ak−1n^k−1+· · · + a0

bln^l+ bl−1n^l−1+· · · + b0

=

nlim→∞(a_kn^k−l+ a_k−1n^k−l−1+· · · + a01 n^l)

nlim→∞(b_l+ b_l−1n¹+· · · + b01 n^l)

= ( lim

n→∞akn^k−l) +· · · + al+ 0 +· · · + 0 bl+ 0 +· · · + 0 =±∞







習題解答 1.4.6.



(1) 因為 Sn=∑_n

k=1n = ⁿ²₂⁺ⁿ, 所以 ∑_∞

k=1n = lim

n→∞

n²+n 2 =∞.

(2) 因為 Sn=∑n

k=1n²= ²ⁿ³⁺³ⁿ₆ ²⁺ⁿ, 所以∑_∞

k=1n²= lim

n→∞

2n³+3n²+n 6 =∞.

(3) Sn=∑_n

k=1(−1)ⁿ, 計算可知

Sn=





−1, n 是奇數 0, n 是偶數 所以 ∑^∞_k=1(−1)ⁿ= lim

n→∞Sn無極限.

(4)

Sn=

∑n k=1

1 2^k = 1

2·1− (¹₂)ⁿ

1− ¹₂ = 1− 1 2ⁿ 所以 ∑^∞_k=12¹_k = lim

n→∞(1−₂¹n) = 1.

10 第 1 章 基本函數與極限

(2) l = k:

n→∞lim

akn^k+ ak−1n^k−1+· · · + a0

bkn^k+ bk−1n^k−1+· · · + b0

=

nlim→∞(a_k+ a_k−1n¹+· · · + a01 n^k)

nlim→∞(b_k+ b_k−1n¹+· · · + b0 1 n^k)

= a_k+ 0 +· · · + 0 bk+ 0 +· · · + 0 = a_k

bk

(3) k > l

nlim→∞

a_kn^k+ a_k−1n^k−1+· · · + a0

bln^l+ bl−1n^l−1+· · · + b0

=

nlim→∞(akn^k−l+ ak−1n^k−l−1+· · · + a01 n^l)

nlim→∞(b_l+ b_l−1n¹+· · · + b01 n^l)

= ( lim

n→∞akn^k−l) +· · · + al+ 0 +· · · + 0 bl+ 0 +· · · + 0 =±∞







習題解答 1.4.6.



(1) 因為 Sn=∑n

k=1n = ⁿ²₂⁺ⁿ, 所以 ∑_∞

k=1n = lim

n→∞

n²+n 2 =∞.

(2) 因為 Sn=∑_n

k=1n²= ²ⁿ³⁺³ⁿ₆ ²⁺ⁿ, 所以∑_∞

k=1n²= lim

n→∞

2n³+3n²+n 6 =∞.

(3) Sn=∑_n

k=1(−1)ⁿ, 計算可知

Sn=





−1, n 是奇數 0, n 是偶數 所以 ∑^∞_k=1(−1)ⁿ= lim

n→∞Sn無極限.

(4)

Sn=

∑n k=1

1 2^k = 1

2·1− (¹₂)ⁿ

1− ¹₂ = 1− 1 2ⁿ 所以 ∑^∞_k=12¹_k = lim

n→∞(1−₂¹ⁿ) = 1.

2

1.4. 連續函數與極限 11

1.4.3 函數的極限







習題解答 1.4.7.



假設 f(x) 和 g(x) 是連續函數.























xlim→a(f (x)± g(x)) = lim_x

→af (x)± lim_x

→ag(x) = f (a)± g(a)

xlim→a(f (x)· g(x)) = lim_x

→af (x)· lim_x

→ag(x) = f (a)· g(a)

x→alim f (x) g(x) =

x→alim f (x)

x→alim g(x) = f (a) g(a)

xlim→af (g(x)) = f ( lim

x→ag(x)) = f (g(a)) 所以 f(x) ± g(x), f(x) · g(x), f (x)

g(x), f (g(x)) 是連續函數.







習題解答 1.4.8.



(1)

x→alim

x³− a³ x− a = lim

x→a

(x− a)(x²+ ax + a²)

x− a = lim

x→a(x²+ ax + a²) = 3a² (2)

xlim→a

x⁻²− a⁻²

x− a = lim

x→a a²−x²

a²·x²

x− a = lim

x→a

(

− 1 a²· x²

(x− a)(x + a) x− a

)

= lim

x→a−x + a a²· x² =−2

a³ (3) n > 0.

xlim→a

xⁿ− aⁿ

x− a = lim

x→a

(x− a)(xⁿ⁻¹+ axⁿ⁻²+· · · + aⁿ⁻²x + aⁿ⁻¹) x− a

= lim

x→a(xⁿ⁻¹+ axⁿ⁻²+· · · + aⁿ⁻²x + aⁿ⁻¹) = naⁿ⁻¹ n =−m < 0, a ̸= 0.

xlim→a

x^−m− a^−m x− a = lim

x→a a^m−x^m

a^m·x^m

x− a

= lim

x→a

(

− 1

a^m· x^m·(x− a)(x^m−1+ ax^m−2+· · · + a^m−2x + a^m−1) x− a

)

= lim

x→a − x^m−1+ ax^m−2+· · · + a^m−2x + a^m−1

a^m· x^m =− m

a^m+1 = naⁿ⁻¹ 都是 naⁿ⁻¹

12 第 1 章 基本函數與極限







習題解答 1.4.9.



(1) lim

x→1x²− 3x + 1 = 1²− 3 × 1 + 1 = −1 (2) lim

x→1

x²− 3x + 1 2x− 1 =

xlim→1x²− 3x + 1

x→1lim 2x− 1 = −1 1 =−1 (3) lim

x→0

√5

x− 1 =√⁵

x→0lim (x− 1) =√⁵

−1 = −1 (4) lim

x→^π₆ sin²2x = ( lim

x→^π₆ sin 2x)²= (sin( lim

x→^π₆ 2x))²= (sinπ 3)²= (

√3 2 )²= 3

4 (5) lim

x→⁴_π log(tan1

x) = log( lim

x→_π⁴

tan1

x) = log(tan( lim

x→_π⁴

1

x)) = log(tanπ

4) = log 1 = 0 (6) 先計算

x→1lim sin(2tan⁻¹x) = sin( lim

x→12tan⁻¹x) = sin(2tan⁻¹1) = sin(2·π 4) = 1 所以

xlim→12^{sin(2tan−1x)}= 2^{xlim→1sin(2tan−1x)}= 2¹= 2 (7) 已知 lim

x→1 ln x

x−1= 1, 所以

xlim→1⁺

ln x

|x − 1|= lim

x→1⁺

ln x x− 1 = 1 (8) 同前因為 lim

x→1 ln x

x−1= 1, 所以

x→1lim⁻ ln x

|x − 1| = lim

x→1⁻ − ln x x− 1=−1 (9) lim

x→−1⁺ sin⁻¹x =−π 2







習題解答 1.4.10.



(1) lim

x→1

x³− 1 x⁴− 1 = lim

x→1

(x− 1)(x²+ x + 1)

(x− 1)(x³+ x²+ x + 1) = lim

x→1

x²+ x + 1 x³+ x²+ x + 1= 3

4 (2) lim

x→0

x⁴− 2x

x⁵− x³+ 7x= lim

x→0

x³− 2

x⁴− x²+ 7 =−2 7 (3) lim

x→4

√x− 2 x− 4 = lim

x→4

(√

x− 2)(√ x + 2) (x− 4)(√

x + 2) = lim

x→4

x− 4 (x− 4)(√

x + 2)= lim

x→4

√ 1

x + 2 = 1 4 (4)

xlim→−2

√x³+ x²+ 8 + x

x + 2 = lim

x→−2

(√

x³+ x²+ 8 + x)(√

x³+ x²+ 8− x) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x³+ 8 (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

(x + 2)(x²− 2x + 4) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x²− 2x + 4

√x³+ x²+ 8− x = 12 4 = 3

12 第 1 章 基本函數與極限







習題解答 1.4.9.



(1) lim

x→1x²− 3x + 1 = 1²− 3 × 1 + 1 = −1 (2) lim

x→1

x²− 3x + 1 2x− 1 =

x→1lim x²− 3x + 1

xlim→12x− 1 = −1 1 =−1 (3) lim

x→0

√5

x− 1 =√⁵

xlim→0(x− 1) =√⁵

−1 = −1 (4) lim

x→^π6

sin²2x = ( lim

x→^π6

sin 2x)²= (sin( lim

x→^π6

2x))²= (sinπ 3)²= (

√3 2 )²= 3

4 (5) lim

x→⁴_π log(tan1

x) = log( lim

x→_π⁴ tan1

x) = log(tan( lim

x→_π⁴

1

x)) = log(tanπ

4) = log 1 = 0 (6) 先計算

xlim→1 sin(2tan⁻¹x) = sin( lim

x→12tan⁻¹x) = sin(2tan⁻¹1) = sin(2·π 4) = 1 所以

x→1lim 2^{sin(2tan−1x)}= 2^{x→1lim sin(2tan−1x)}= 2¹= 2 (7) 已知 lim

x→1 ln x

x−1= 1, 所以

x→1lim⁺ ln x

|x − 1|= lim

x→1⁺

ln x x− 1 = 1 (8) 同前因為 lim

x→1 ln x

x−1= 1, 所以

xlim→1⁻

ln x

|x − 1| = lim

x→1⁻ − ln x x− 1=−1 (9) lim

x→−1⁺ sin⁻¹x =−π 2







習題解答 1.4.10.



(1) lim

x→1

x³− 1 x⁴− 1 = lim

x→1

(x− 1)(x²+ x + 1)

(x− 1)(x³+ x²+ x + 1) = lim

x→1

x²+ x + 1 x³+ x²+ x + 1= 3

4 (2) lim

x→0

x⁴− 2x

x⁵− x³+ 7x= lim

x→0

x³− 2

x⁴− x²+ 7 =−2 7 (3) lim

x→4

√x− 2 x− 4 = lim

x→4

(√

x− 2)(√ x + 2) (x− 4)(√

x + 2) = lim

x→4

x− 4 (x− 4)(√

x + 2)= lim

x→4

√ 1

x + 2 = 1 (4) 4

x→−2lim

√x³+ x²+ 8 + x

x + 2 = lim

x→−2

(√

x³+ x²+ 8 + x)(√

x³+ x²+ 8− x) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x³+ 8 (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

(x + 2)(x²− 2x + 4) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x²− 2x + 4

√x³+ x²+ 8− x = 12 4 = 3

12 第 1 章 基本函數與極限







習題解答 1.4.9.



(1) lim

x→1x²− 3x + 1 = 1²− 3 × 1 + 1 = −1 (2) lim

x→1

x²− 3x + 1 2x− 1 =

xlim→1x²− 3x + 1

xlim→12x− 1 = −1 1 =−1 (3) lim

x→0

√5

x− 1 =√⁵

x→0lim (x− 1) =√⁵

−1 = −1 (4) lim

x→^π₆ sin²2x = ( lim

x→^π₆ sin 2x)²= (sin( lim

x→^π₆ 2x))²= (sinπ 3)²= (

√3 2 )²= 3

4 (5) lim

x→⁴_π log(tan1

x) = log( lim

x→_π⁴

tan1

x) = log(tan( lim

x→_π⁴

1

x)) = log(tanπ

4) = log 1 = 0 (6) 先計算

x→1lim sin(2tan⁻¹x) = sin( lim

x→12tan⁻¹x) = sin(2tan⁻¹1) = sin(2·π 4) = 1 所以

xlim→12^{sin(2tan−1x)}= 2^{xlim→1sin(2tan−1x)}= 2¹= 2 (7) 已知 lim

x→1 ln x

x−1= 1, 所以

xlim→1⁺

ln x

|x − 1|= lim

x→1⁺

ln x x− 1 = 1 (8) 同前因為 lim

x→1 ln x

x−1= 1, 所以 lim

x→1⁻

ln x

|x − 1| = lim

x→1⁻ − ln x x− 1=−1 (9) lim

x→−1⁺ sin⁻¹x =−π 2







習題解答 1.4.10.



(1) lim

x→1

x³− 1 x⁴− 1 = lim

x→1

(x− 1)(x²+ x + 1)

(x− 1)(x³+ x²+ x + 1) = lim

x→1

x²+ x + 1 x³+ x²+ x + 1= 3

4 (2) lim

x→0

x⁴− 2x

x⁵− x³+ 7x= lim

x→0

x³− 2

x⁴− x²+ 7 =−2 7 (3) lim

x→4

√x− 2 x− 4 = lim

x→4

(√

x− 2)(√ x + 2) (x− 4)(√

x + 2) = lim

x→4

x− 4 (x− 4)(√

x + 2)= lim

x→4

√ 1

x + 2 = 1 4 (4)

xlim→−2

√x³+ x²+ 8 + x

x + 2 = lim

x→−2

(√

x³+ x²+ 8 + x)(√

x³+ x²+ 8− x) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x³+ 8 (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

(x + 2)(x²− 2x + 4) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x²− 2x + 4

√x³+ x²+ 8− x = 12 4 = 3

3

12 第 1 章 基本函數與極限







習題解答 1.4.9.



(1) lim

x→1x²− 3x + 1 = 1²− 3 × 1 + 1 = −1 (2) lim

x→1

x²− 3x + 1 2x− 1 =

xlim→1x²− 3x + 1

x→1lim 2x− 1 = −1 1 =−1 (3) lim

x→0

√5

x− 1 =√⁵

x→0lim (x− 1) =√⁵

−1 = −1 (4) lim

x→^π₆ sin²2x = ( lim

x→^π₆ sin 2x)²= (sin( lim

x→^π₆ 2x))²= (sinπ 3)²= (

√3 2 )²= 3

4 (5) lim

x→⁴_π log(tan1

x) = log( lim

x→_π⁴

tan1

x) = log(tan( lim

x→_π⁴

1

x)) = log(tanπ

4) = log 1 = 0 (6) 先計算

x→1lim sin(2tan⁻¹x) = sin( lim

x→12tan⁻¹x) = sin(2tan⁻¹1) = sin(2·π 4) = 1 所以

xlim→12^{sin(2tan−1x)}= 2^{xlim→1sin(2tan−1x)}= 2¹= 2 (7) 已知 lim

x→1 ln x

x−1= 1, 所以

xlim→1⁺

ln x

|x − 1|= lim

x→1⁺

ln x x− 1 = 1 (8) 同前因為 lim

x→1 ln x

x−1= 1, 所以

x→1lim⁻ ln x

|x − 1| = lim

x→1⁻ − ln x x− 1=−1 (9) lim

x→−1⁺ sin⁻¹x =−π 2







習題解答 1.4.10.



(1) lim

x→1

x³− 1 x⁴− 1 = lim

x→1

(x− 1)(x²+ x + 1)

(x− 1)(x³+ x²+ x + 1) = lim

x→1

x²+ x + 1 x³+ x²+ x + 1= 3

4 (2) lim

x→0

x⁴− 2x

x⁵− x³+ 7x= lim

x→0

x³− 2

x⁴− x²+ 7 =−2 7 (3) lim

x→4

√x− 2 x− 4 = lim

x→4

(√

x− 2)(√ x + 2) (x− 4)(√

x + 2) = lim

x→4

x− 4 (x− 4)(√

x + 2)= lim

x→4

√ 1

x + 2 = 1 4 (4)

xlim→−2

√x³+ x²+ 8 + x

x + 2 = lim

x→−2

(√

x³+ x²+ 8 + x)(√

x³+ x²+ 8− x) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x³+ 8 (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

(x + 2)(x²− 2x + 4) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x²− 2x + 4

√x³+ x²+ 8− x = 12 4 = 3

12 第 1 章 基本函數與極限







習題解答 1.4.9.



(1) lim

x→1x²− 3x + 1 = 1²− 3 × 1 + 1 = −1 (2) lim

x→1

x²− 3x + 1 2x− 1 =

xlim→1x²− 3x + 1

xlim→12x− 1 = −1 1 =−1 (3) lim

x→0

√5

x− 1 =√⁵

x→0lim (x− 1) =√⁵

−1 = −1 (4) lim

x→^π6

sin²2x = ( lim

x→^π6

sin 2x)²= (sin( lim

x→^π6

2x))²= (sinπ 3)²= (

√3 2 )²= 3

4 (5) lim

x→⁴_π log(tan1

x) = log( lim

x→_π⁴ tan1

x) = log(tan( lim

x→_π⁴

1

x)) = log(tanπ

4) = log 1 = 0 (6) 先計算

x→1lim sin(2tan⁻¹x) = sin( lim

x→12tan⁻¹x) = sin(2tan⁻¹1) = sin(2·π 4) = 1 所以

xlim→12^{sin(2tan−1x)}= 2^{xlim→1sin(2tan−1x)}= 2¹= 2 (7) 已知 lim

x→1 ln x

x−1= 1, 所以

xlim→1⁺

ln x

|x − 1|= lim

x→1⁺

ln x x− 1 = 1 (8) 同前因為 lim

x→1 ln x

x−1= 1, 所以 lim

x→1⁻

ln x

|x − 1| = lim

x→1⁻ − ln x x− 1=−1 (9) lim

x→−1⁺ sin⁻¹x =−π 2







習題解答 1.4.10.



(1) lim

x→1

x³− 1 x⁴− 1 = lim

x→1

(x− 1)(x²+ x + 1)

(x− 1)(x³+ x²+ x + 1) = lim

x→1

x²+ x + 1 x³+ x²+ x + 1= 3

4 (2) lim

x→0

x⁴− 2x

x⁵− x³+ 7x= lim

x→0

x³− 2

x⁴− x²+ 7 =−2 7 (3) lim

x→4

√x− 2 x− 4 = lim

x→4

(√

x− 2)(√ x + 2) (x− 4)(√

x + 2) = lim

x→4

x− 4 (x− 4)(√

x + 2)= lim

x→4

√ 1

x + 2 = 1 4 (4)

xlim→−2

√x³+ x²+ 8 + x

x + 2 = lim

x→−2

(√

x³+ x²+ 8 + x)(√

x³+ x²+ 8− x) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x³+ 8 (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

(x + 2)(x²− 2x + 4) (x + 2)(√

x³+ x²+ 8− x)

= lim

x→−2

x²− 2x + 4

√x³+ x²+ 8− x = 12 4 = 3

1.4. 連續函數與極限 13

(5) lim

x→0

tan x sin 2x = lim

x→0

sin x cos x

2 sin x cos x = lim

x→0

1 2 cos²x = 1

2 (6) lim

x→0 tan 2x · csc x = lim

x→0 sin 2x cos 2x

sin x = lim

x→0

2 sin x cos x cos 2x · sin x= lim

x→0

2 cos x cos 2x = 2







習題解答 1.4.11.



(1) lim

x→0

sin(λx) x = lim

x→0λsin(λx) λx

y=λx= lim

y→0λsin y y = λ (2) 由 cos x − 1 = cos(2 ·x

2)− 1 = 1 − 2 sin²x

2 − 1 = −2 sin²x 2. 故

xlim→0

cos x − 1

x = lim

x→0

(

(−2) ·sin^x₂ x · sinx

2 )

= (−2) · lim

x→0

sin^x₂ x · lim

x→0 sinx 2

= (−2) ·1 2· 0 = 0 (3) 由上題 cos x − 1 = −2 sin^{2 x}₂. 故

xlim→0

cos x − 1

x² = lim

x→0

(

(−2) ·(sin^x₂)² x²

)

= (−2) · (

xlim→0

sin^x₂ x

)2

= (−2) · (1

2)²=−1 2







習題解答 1.4.12.



(1) lim

x→∞

x³− 1 x³+ 1 = lim

x→∞

1−_x¹3

1 +_x¹3

= 1− 0 1 + 0= 1 (2) lim

x→∞

x³− 1 x²+ 1 = lim

x→∞

x−_x¹2

1 +_x¹2

=∞

(3) lim

x→∞

x³− 1 x⁴+ 1 = lim

x→∞

1 x− _x¹4

1 +_x¹4

= 0

(4) 因為當 x 很大時必是正數, 因此

−1

x ≤ sin x x ≤ 1

x 故

x→∞lim

−1

x ≤ lim_x→∞ sin x

x ≤ lim_x→∞ 1 x 由夾擊法得 lim

x→∞

sin x x = 0.

(5) lim

x→∞tan⁻¹x = π 2 (6) lim

x→∞ sin(tan⁻¹x) = sin( lim

x→∞tan⁻¹x) = sinπ 2 = 1

1.4. 連續函數與極限 13

(5) lim

x→0

tan x sin 2x = lim

x→0

sin x cos x

2 sin x cos x = lim

x→0

1 2 cos²x = 1

2 (6) lim

x→0 tan 2x · csc x = lim

x→0 sin 2x cos 2x

sin x = lim

x→0

2 sin x cos x cos 2x · sin x= lim

x→0

2 cos x cos 2x = 2







習題解答 1.4.11.



(1) lim

x→0

sin(λx) x = lim

x→0λsin(λx) λx

y=λx= lim

y→0λsin y y = λ (2) 由 cos x − 1 = cos(2 ·x

2)− 1 = 1 − 2 sin²x

2 − 1 = −2 sin²x 2. 故

xlim→0

cos x − 1

x = lim

x→0

(

(−2) ·sin^x₂ x · sinx

2 )

= (−2) · lim

x→0

sin^x₂ x · lim

x→0 sinx 2

= (−2) ·1 2· 0 = 0 (3) 由上題 cos x − 1 = −2 sin^{2 x}2. 故

xlim→0

cos x − 1

x² = lim

x→0

(

(−2) ·(sin^x₂)² x²

)

= (−2) · (

xlim→0

sin^x₂ x

)2

= (−2) · (1

2)²=−1 2







習題解答 1.4.12.



(1) lim

x→∞

x³− 1 x³+ 1 = lim

x→∞

1−_x¹3

1 +_x¹3

= 1− 0 1 + 0= 1 (2) lim

x→∞

x³− 1 x²+ 1 = lim

x→∞

x−_x¹2

1 +_x¹2

=∞

(3) lim

x→∞

x³− 1 x⁴+ 1 = lim

x→∞

1 x− _x¹⁴ 1 +_x¹4

= 0

(4) 因為當 x 很大時必是正數, 因此

−1

x ≤ sin x x ≤ 1

x 故

x→∞lim

−1 x ≤ lim

x→∞

sin x x ≤ lim

x→∞

1 x 由夾擊法得 lim

x→∞

sin x x = 0.

(5) lim

x→∞tan⁻¹x = π 2 (6) lim

x→∞ sin(tan⁻¹x) = sin( lim

x→∞tan⁻¹x) = sinπ 2 = 1

1.4. 連續函數與極限 13

(5) lim

x→0

tan x sin 2x = lim

x→0

sin x cos x

2 sin x cos x = lim

x→0

1 2 cos²x = 1

2 (6) lim

x→0 tan 2x · csc x = lim_x

→0 sin 2x cos 2x

sin x = lim

x→0

2 sin x cos x cos 2x · sin x= lim

x→0

2 cos x cos 2x = 2







習題解答 1.4.11.



(1) lim

x→0

sin(λx) x = lim

x→0λsin(λx) λx

y=λx= lim

y→0λsin y y = λ (2) 由 cos x − 1 = cos(2 ·x

2)− 1 = 1 − 2 sin²x

2 − 1 = −2 sin²x 2. 故

xlim→0

cos x − 1

x = lim

x→0

(

(−2) ·sin^x₂ x · sinx

2 )

= (−2) · lim

x→0

sin^x₂ x · lim

x→0 sinx 2

= (−2) ·1 2· 0 = 0 (3) 由上題 cos x − 1 = −2 sin^{2 x}₂. 故

x→0lim

cos x − 1

x² = lim

x→0

(

(−2) ·(sin^x₂)² x²

)

= (−2) · (

x→0lim sin^x₂

x )2

= (−2) · (1

2)²=−1 2







習題解答 1.4.12.



(1) lim

x→∞

x³− 1 x³+ 1 = lim

x→∞

1−_x¹3

1 +_x¹3

= 1− 0 1 + 0= 1 (2) lim

x→∞

x³− 1 x²+ 1 = lim

x→∞

x−_x¹2

1 +_x¹2

=∞

(3) lim

x→∞

x³− 1 x⁴+ 1 = lim

x→∞

1x− _x¹4

1 +_x¹4

= 0

(4) 因為當 x 很大時必是正數, 因此

−1

x ≤ sin x x ≤ 1

x 故

xlim→∞

−1 x ≤ lim_x

→∞

sin x x ≤ lim_x

→∞

1 x 由夾擊法得 lim

x→∞

sin x x = 0.

(5) lim

x→∞tan⁻¹x = π 2 (6) lim

x→∞ sin(tan⁻¹x) = sin( lim

x→∞tan⁻¹x) = sinπ 2 = 1

14 第 1 章 基本函數與極限







習題解答 1.4.13.



同第四節下述習題結果：

nlim→∞

a_kn^k+ a_k−1n^k−1+· · · + a0

bln^l+ bl−1n^l−1+· · · + b0







習題解答 1.4.14.



因為

x→±∞lim P (x)

Q(x)= lim

x→±∞(多項式 +R(x) Q(x)) 但因為 R(x) 的次數小於 Q(x) 的次數, lim

x→±∞

R(x)

Q(x) = 0. 所以當 x→ ±∞ 時, 函數圖形 漸近於多項式的部分.







習題解答 1.4.15.



如果當 f(x) 隨著 x 趨近 ∞, 終究會落在 L 的附近, 其中整數點 x = n 所對應的 f (n) = an 當然終究會落在 L 的附近.

1.5 e 與自然對數







習題解答 1.5.1.



(1) ln e = log_ee = 1 (2) e^{ln x}= e^logex= x (3) log_ax = log_ex

log_ea = ln x



ln a





習題解答 1.5.2.



(1) lim

x→¹λ

ln x + ln λ λx− 1 = lim

x→¹λ

ln λx λx− 1

y=λx= lim

y→1

ln y y− 1 = 1 (2) lim

x→0

ln sec x tan²x = lim

x→0

1

2 ·ln(sec x)² sec²x− 1

y=sec²x

= lim

y→1

1 2 · ln y

y− 1= 1 2







習題解答 1.5.3.



(1) lim

x→∞(1 + 1

x)^λx= ( lim

x→∞(1 + 1

x)^x)^λ= e^λ (2) lim

x→∞(1 +λ

x)^x= lim

x→∞((1 + 1

(^x_λ))^xλ)^{λ y=}

x

=λ lim

y→±∞((1 +1

y)^y)^λ= e^λ







習題解答 1.5.4.



可用計算機或 Wolfram Alpha 自行嘗試.

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