第一章 函數的導數與微分
(甲) 函數極限的概念
由上圖看出來, f(x) 在 x = 2 時的定義為 f(2) = 6 ; 但當 x → 2 時 , f(x) 為 4
由題意, g(x) = |x|/x , x ≠ 0 , 分析如下:
若 x > 0 , 即 x 為正數, g(x) = x / x = 1 , f(x) 為連續函數.
若 x < 0 , 即 x 為負數, g(x) = -x/x = -1 , f(x) 亦為連續函數.
但是若我們在0點左右討論, 即 x → 0 時討論如如下 :
當x → 0 + , lim g(x) = 1 ,但當x → 0 - , lim g(x) = -1 ; 左右極限不相等, 所以g(x) 在 x = 0 之極值不存在 . 極限存在必唯一.
h(x) = 1/x2 , x ≠ 0 , 當 x → 0 + , h(x) → ∞ ; 同理, 當 x → 0 - , h(x) → ∞ . ∞ 不是一個定值. Infinity 故不會趨近一個定值.
證:
給定一個正數ε，可以找到一個正數δ=δ(ε)，使得當 x 滿足 0<|x−a|<δ，|f(x)−l|<ε .
f(x) = x2 , l = 4 則 |f(x) -l| = | x2 - 4 | = |(x+2)(x-2)|
取 0 < |x-2| < δ1 , δ1 = 1
-1 < x-2 < 1
-1+2 < x < 1+2
1 < x < 3
1+2 < x+2 < 3+2
3 < x+2 < 5
因此| x2 - 4 | = |(x+2)(x-2)| = |x+2| |x-2| < 5 |x-2| < δ
取 δ2 = ε/5
∀ε > 0 , 取 δ = min {ε/5 ,1} .
則當 0 < |x-2| < δ , | x2 - 4 | = |(x+2)(x-2)| = |x+2| |x-2| < 5 |x-2| < 5 ∙ ε/5 = ε 成立.
故
得証
証 :
給定一個正數ε，可以找到一個正數δ=δ(ε)，使得當 x 滿足 0<|x−a|<δ，|f(x)−l|<ε
f(x) = 1 / x2 +1 , |f(x) - 1/10 | = | 1 / x2 +1 - 1/10 |
0 < | x - 3| < δ1 = 1 (暫取 δ = 1)
0 < | x - 3| < 1
-1 < x - 3 < 1
-1 + 3 < x < 1+3
2 < x < 4
2 + 3 < x+3 < 4+3
5 < x+3 < 7
5 < x2+1 < 17
50 < 10(x2+1) < 170
1/170 < 1/10(x2+1) < 1/50
| 9 - x2| / | 10(x2+1) | = | (x-3)(x+3) | /10 |x2+1| = |x-3| (|x+3|/10|x2+1|) < 7/50 |x-3| < δ , 取δ2 = 50/7 ε
∀ε > 0 , 取 δ = min {50ε/7 ,1} .
則當 0 < |x-3| < δ , |1 / x2 +1 - 1/10 | = | 9 - x2| / | 10(x2+1) | = |x-3|∙|x+3/x2+1|/10 < 7/50 ∙ (50ε/7) = ε 成立.
故
得証
∀ε > 0 , 取 δ = min {50ε/7 ,1} .
則當 0 < |x-3| < δ , |1 / x2 +1 - 1/10 | = | 9 - x2| / | 10(x2+1) | = |x-3|∙|x+3/x2+1|/10 < 7/50 ∙ (50ε/7) = ε 成立.
故
得証
(1) x > 0 , f(x) = x + 1 ; 當x → 0 + , the limit of f(x) when x approaches 0 + is 1 , and x < 0 , f(x) = -2x + 1 ; 當x → 0 - , the limit of f(x) when x approaches 0 - is 1. So the limit of f(x) is 1 ,but f(0) is undefined .
(2) as above .
(乙) 函數極限的四則運算
由題意, lim (f(x)+g(x)) = 6 , lim (f(x)- g(x)) = 2 ;
根據性質 , lim f(x) = a , lim g(x) = b ; a,b 必存在.
所以 a + b = 6 , a - b = 2 , 2a = 8 , a = 4 , 代回得 b = 2 故 ab =8 即 lim f(x)g(x) = 8 .
(丙) 極限的求法
(1) 先對其有理化. √x+6−√x+1 = (x+6) - (x+1) / √x+6+√x+1= 5 / √x+6+√x+1
(2)直接帶入, -1/-1 = 1
(3) 3√8-8-1/3√8 = -1/2
(5) 常數函數 , 故為 5
(1) 16 - 12 - 4 / 16 -28 + 12 = 0/0 型式. 故先找可約分的因式, 首先做因式分解(x-4)(x+1)/(x-4)(x-3) = x+1/x-3 , 故其值為4+1/ 4-3 = 5
(2) 1-1/1-1 = 0/0 型式. x - 1 = (3√x - 1)(3√x2+ 3√x + 1) 分子分母約分, 得1/ (3√x2+ 3√x + 1) 故將x = 1代入得1/1+1+1 = 1/3
(3) 1 -1 / 0 = 0/0 型式 . 展開尋求消去1, 1+6x+9X^2+2x^2+6x^3 - 1 / x = x(6+9x+2x+6x^2)/x = 6+9x+2x+6x^2 , x = 0 代入得 6 .
(1) [x-[x]] 為高斯, x - 1 < [x] ≤ x , -x ≤ -[x] < 1 -x , 0 ≤ x - [x] < 1 , 故 [ x-[x] ] = 0 , 所以 x → a , a 是任意值 ; 其極限值皆為 0 .
(2) (2-1)(2-3) = -1
(3) 3-5/3-2 = -2/1 = -2
(1) 分子通分並化簡 -3 / (x -3) , x = 4 代入, -3/1 = -3
(2) 分子因式分解 (x-2)(x2 + 2x + 4) 並與分母消去相同因式得(x2 + 2x + 4) 再將 x = 2 代入故得12 .
(3) 1-1/-2+2 = 0/0 型式. 故想辦法分解分子因式,並與分母消去共同因式.
(x+1)10 = ∑ 10Ci xi, i =0,1,2,3,4,5,7,8,9,10 ;
∑ 10Ci xi = 10C10x10 + 10C9x9 + 10C8x8+ 10C7x7+ 10C6x6+ 10C5x5+ 10C4x4+ 10C3x3+ 10C2x2+ 10C1x1+ 10C0x0 = x10+ 1 + (10C9x9 + 10C8x8+ 10C7x7+ 10C6x6+ 10C5x5+ 10C4x4+ 10C3x3+ 10C2x2+ 10C1x1 )
但是此法代麻煩, 是否觀察能更簡化上式: 令 t = x +1 , 則 t→ -1 , 上式可以寫成如下:
lim (t → -1) t10 -1 / t +1 = lim (t → -1) [ t9 - t8+ t7- t6+ t5- t4+ t3- t2+ t1- 1 ] = -1-1-1-1-1-1-1-1-1-1= -10
Let f(x) = 1/x-1 , g(x) = x^2+4x-8/x^2+x-2 . lim f(x) , lim g(x) don't exist . But lim (f(x)+g(x)) doesn't exist ?
通分並化簡可求出.
(1) 有理化後為 (x-5) (x+4)^1/2 + 3 / (x+4)-9 = (x-5) (x+4)^1/2 + 3 / (x-5) = (x+4)^1/2 + 3 , 代入 x = 5 得 3+3 = 6
(2) 有理化後為 1-(1-x) / x(1+(1-x)^1/2) = x / x(1+(1-x)^1/2) = 1/(1+(1-x)^1/2) ,代入 x = 0 得 1/1+1 = 1/2
(3) 有理化後為 (x+2)-2 / x [ (x+2)^2/3 + (2(x+2))^1/3 + 2^2/3 ] = 1 / (x+2)^2/3 + (2(x+2))^1/3 + 2^2/3 ] 代入 x = 0 得 1/2^2/3 + 2^2/3 + 2^2/3 = 1/ 3*(2^2/3) = 2^1/3 / 6
(1)通分, 並化簡
(2)分子有理化,消去共同因子x
(3)分子分母同時有理化,可以消除 x -3 的因式.
(4)通分後,分子分母互約相同因式.
(1) Let f(x) = a-4x+x^2 / 1- x ; when x → 1 , 1 - x → 0 and a-4x+x^2 is also → 0 , so a-4+1 = 0 , a = 3 因為 lim f(x) = b , b 是常數且 x → 1 時 , lim f(x) 為 0/0 型式, 故分子分母必有一個因式為(x-1) 可被消去, 原式 = (x-1)(x-3)/ (1-x) = 3-x , 所以 lim f(x) = 3-1 =2 , b = 2 .
(2) when x → 2 , lim f(x) = 0/0 form . So x^2+ax+b = 0 , when x → 2 , 4+2a+b = 0 , b = -(4+2a) and x^2+ax+b can be factored to (x-2)(x-c) . (x-2)(x+1)/(x-2)(x-c)= x+1/x-c , c = -5 代回(x-2)(x+5) = x^2 + 3x - 10 , 故 a = 3 , b = -10
Notes : 這是剛好除了(x-2)之外的因式其x一次項的係數為1,故可以寫成(x-c) , 否則多數的情況是錯誤的. 必須寫成 (cx+d) 形式.
(1) x = 1 帶入分母, 分母為 0 .故分子亦為 0 , 所以a + 5 + b = 0 , b = -
(a+5) 且分子應能分解為(x-1)(x+c)型式.消去共同因式(x-1) 得(x+c)/(x+1) , (1+c)/2 =3 , 1+ c = 6 , c = 5 ; (x-1)(x+5) = x^2+4x-5
(a+5) 且分子應能分解為(x-1)(x+c)型式.消去共同因式(x-1) 得(x+c)/(x+1) , (1+c)/2 =3 , 1+ c = 6 , c = 5 ; (x-1)(x+5) = x^2+4x-5
(2) 有理化分子, a^2(x^2+3) - b^2 / (x-1)(a(x^2+3)^1/2 + b ) . 分子式中必有x-1的因式.
證 :
(1) 若 f(x) 在x = 3上連續 , 若且為若 f(x) 在 x= 3 的極限值等於 f(3) .
x 趨近 1- , f(x) = 0 故極限值不存在. f(x) 在 x = 1 上不連續.
x ≠ -2 , f(x)可以化簡為 (x^2-2x+4) ; 所以 x → -2 , f(x)極限值為4+4+4 = 12 且等於f(-2). 故連續.
由題意,f(x) 在 x = -3 連續 , 故f(-3) = lim f(x) 在 x = - 3.
首先,討論 x ≠ 0 , f(x) = xsin(1/x) . 因為 |sin(1/x)| ≤ 1, ∀x ;
⇒ 0 ≤ |xsin(1/x)| ≤ |x| , ∀x . x = 0 的極限值為 0, 故f(x)在 x= 0 處連續.
(1) 已知 lim tant/t = 1 , x/tan3x = (1/3)∙(3x/tan3x) , 故 x/tan3x 在x 上的極限值為1/3
(2) 想辦法配成 sin t/t 型式. sin3x/sin5x = (sin3x/3x) ∙ (5x/sin5x) ∙ (3x/5x) = (3/5)∙(sin3x/3x)∙(5x/sin5x) = 3/5
(3) sinx∙ tanx / x2 = sinx ∙ sinx/x2cosx = sin2x / x2cosx = (sinx/x)2 ∙ 1/cosx
(4) 1-cos2x = 1-(1-2sin²x) = 2sin²x , 1-cos2x / xsinx = 2sin²x/xsinx = 2sinx/x
證明如上的性質 :
(a) Since f(x) , g(x) at x = a are continuous ,b is a constant , so lim f(x) = f(a) when x approaches a and lim g(x) = g(a) when x approaches a .
(1) lim [f(x)+g(x)] = lim f(x) + lim g(x) = f(a) + g(a) ( because f(x),g(x) at x = a are continuous )
(2)(3) same above
(4) lim [bf(x)] = lim [ f(x)+f(x)+f(x)+ ... +f(x)] = lim f(x) + lim f(x) + lim f(x) + ... + lim f(x) = f(a) + f(a) + ...+ f(a) = bf(a)
\____ b times ______/
證 :
|sin(1/x)| ≤ 1 , ∀x
0 ≤ |xsin(1/x)| ≤ |x| , ∀x
lim 0 = 0 , lim |x| = 0 由夾擊原理, lim xsin(1/x) = 0
|sin(1/x2)| ≤ 1
|(x2+1)sin(1/x2)| ≤ |(x2+1)|
|(x2+1)sin(1/x2)| = |x2sin(1/x2) + sin(1/x2) | ≥ | x2sin(1/x2) | + | sin(1/x2) |
| x2sin(1/x2) | ≤ |x2| , therefore lim |x2| = 0 , According to sandwich theorem , lim | x2sin(1/x2) | = 0 , so lim (x2sin(1/x2)) = 0
(1) 若 f(x) 在x = 3上連續 , 若且為若 f(x) 在 x= 3 的極限值等於 f(3) .
(2) 若 f(x) = 2x + 1 , x ≠ 1 ; 所以 x = a , f(a) = 2a +1 . 若 a ≠ 1 , a 在所有點的極限趨近值為2a+1, 其值相同於f(a) , 故 f(x) 在 x = a , a ≠ 1 為連續函數.
若 x = 1 時極限趨近值為 2+1 = 3 ,但 x =1 時的定義f(1) = 1 , 1 ≠ 3 ; 所以 x = 1 點是不連續的.
(1) f(x) 在 x = 0 , 上定義為 f(0) = 0 . 當 x ≠ 0 , x 去趨近於 0 時, f(x) = |x|/x ;
當x → 0+ , f(x) = x/x = 1
當x → 0- , f(x) = -x/x = -1 ; f(x) 在 x = 0 的左右極限值並不相同, 故f(x) 在x = 0 上的
極限值不存在 . 所以f(x) 在 x = 0 上不連續.
極限值不存在 . 所以f(x) 在 x = 0 上不連續.
(2) f(x) = [x] ; 由高斯符號定義 , x-1 < [x] <= x .
x = 1/2 , f(x) = 0 , lim f(x) = f(1/2) . 所以 f(x) 在 x =1/2 處連續.
x = 1 , f(x) = 1 , x 趨近 1+ , f(x) = 1 ; x 趨近 1- , f(x) = 0 故極限值不存在. f(x) 在 x = 1 上不連續.
(3) ∵ |sin(1/x)| ≤ 1 , therefore x = 0 上的極值不存在也不連續 .
由題意,f(x) 在 x = -3 連續 , 故f(-3) = lim f(x) 在 x = - 3.
f(-3) = a ,
當x ≠ -3 , f(x) = x-3 (約去共同因式). lim f(x) = - 6. 所以a = -6 .
首先,討論 x ≠ 0 , f(x) = xsin(1/x) . 因為 |sin(1/x)| ≤ 1, ∀x ;
⇒ 0 ≤ |xsin(1/x)| ≤ |x| , ∀x . x = 0 的極限值為 0, 故f(x)在 x= 0 處連續.
則f(x) 在 x 為任何實數上都連續 .
(戊)基本初等函數的連續性
(1) tanx / x = sinx / cosx / x = sinx/xcosx = (sinx/x)∙(1/cosx). Since lim sinx / x =1 when x approaches 0 . And 1/cosx = 1 when x approaches 0.
lim tanx/x = lim [(sinx/x)∙(1/cosx)] = lim (sinx/x) ∙ lim(1/cosx) = 1 .
(2) 如果熟悉三角函數公式, sin2x + cos2x = 1 , sin2x = 1 - cos2x . 所以
1 - cosx / x2 = 1 - cos2x / x2 (1 + cosx) = sin2x / x2 (1 + cosx) = (sinx/x)2 ∙ 1/(1 + cosx)
(戊)基本初等函數的連續性
(1) tanx / x = sinx / cosx / x = sinx/xcosx = (sinx/x)∙(1/cosx). Since lim sinx / x =1 when x approaches 0 . And 1/cosx = 1 when x approaches 0.
lim tanx/x = lim [(sinx/x)∙(1/cosx)] = lim (sinx/x) ∙ lim(1/cosx) = 1 .
(2) 如果熟悉三角函數公式, sin2x + cos2x = 1 , sin2x = 1 - cos2x . 所以
1 - cosx / x2 = 1 - cos2x / x2 (1 + cosx) = sin2x / x2 (1 + cosx) = (sinx/x)2 ∙ 1/(1 + cosx)
因為 lim sinx/x = 1 , 因為 lim (sinx/x)2 = (lim (sinx/x))2= 1 又因為 lim (1/1+cosx) = 1/2 , 所以 lim (1 - cosx / x2) = lim(sin2x / x2 (1 + cosx)) = (lim (sinx/x))2 ∙ lim (1/1+cosx) = 1 × 1/2 = 1/2 .
(2) 想辦法配成 sin t/t 型式. sin3x/sin5x = (sin3x/3x) ∙ (5x/sin5x) ∙ (3x/5x) = (3/5)∙(sin3x/3x)∙(5x/sin5x) = 3/5
(3) sinx∙ tanx / x2 = sinx ∙ sinx/x2cosx = sin2x / x2cosx = (sinx/x)2 ∙ 1/cosx
(4) 1-cos2x = 1-(1-2sin²x) = 2sin²x , 1-cos2x / xsinx = 2sin²x/xsinx = 2sinx/x
證明如上的性質 :
(a) Since f(x) , g(x) at x = a are continuous ,b is a constant , so lim f(x) = f(a) when x approaches a and lim g(x) = g(a) when x approaches a .
(1) lim [f(x)+g(x)] = lim f(x) + lim g(x) = f(a) + g(a) ( because f(x),g(x) at x = a are continuous )
(2)(3) same above
(4) lim [bf(x)] = lim [ f(x)+f(x)+f(x)+ ... +f(x)] = lim f(x) + lim f(x) + lim f(x) + ... + lim f(x) = f(a) + f(a) + ...+ f(a) = bf(a)
\____ b times ______/
Since g(x) at x = a is continuous , so lim g(x) = g(a) . And f(x) at x = g(a) be continuous , so limf(x) = b , b is a constant .
Now lim f(g(x)) = f(lim g(x)) = f (g(a))
(1)
證 :
According to the definition of limit , we can write it as below :
For any positive number ε , there exists another positive number δ , such that |f(x) - L| ≤ ε , when |x-a| < δ .
Let δ1 = 1
|x-a| < 1
-1 + a < x < 1+a
√a-1 < √x < √1+a
√a-1 + √a < √x + √a < √1+a + √a
1/(√1+a + √a ) < 1/(√x + √a ) < 1/(√1+a + √a ) < 1/(√a + √a ) < 1/(√a-1 + √a )
|√x - √a| = |(x -a)|/|√x + √a| ≤ 1/|(2√a ) | |(x -a)| ≤ ε ,
δ2 = (2√a )ε , δ = min { 1 , (2√a )ε }
For any positive number ε , there exists another positive number δ , δ = min { 1, (2√a )ε } such that |√x - √a| = |(x -a)|/|√x + √a| < 1/|(2√a ) | |(x -a)| ≤ (1/2√a )(2√a )ε = ε .
(2)
f(x) = √1+x2
Since x ∈ R , x2 ≥ 0 then x2 + 1 ≥ 1 , so the range of f(x) is [1,∞) and its domain is (-∞,∞) , R . So f(x) is defined on R . Therefore , f(x) is a continuous function on R .
f(x) at x = a is continuous , then lim f(x) = f(a) .
f(x) = ∑ gi(x) , i = 0,1,2, ... , n ; gi(x) = aixi
lim f(x) = lim ∑ gi(x) = ∑ lim gi(x) = lim g0(x) + lim g1(x) + ... + lim gn(x) = g0(a) + g1(a) + ... + gn(a) = f(a) ( because gi(x) is continuous function , i =0,1,2,...,n for any number a on R)
tan x = sin x / cos x , cos x ≠ 0 is its domain .
let f(x) = sin x , and g(x) = cos x ; then f(x)/g(x) = tan x = h(x) .
Since g(x) ≠ 0 , so cos x ≠ 0 . And the domain of sin x is R and the domain of cos x is [0,nπ/2] , n on Z . when n = 2k+1 , k ∈ Z , then cos x = 0 that causes h(x) = sin x / 0 is undefined . the domain of sin x is R , that is (-∞,∞) . So h(x)'s domain is {x ∈ R | x ≠ nπ/2 , n = 2k+1, k ∈ Z }.
(1) 依照函數的定義 , 對於任何一個定義域中的數 x , 其必然有一個數 b , 使得 f(x) = b .
當 x = -1 ,1 則 f(x) 是無定義的.(任何數除0是無意義) .
x > 1 , 1-x^2 < 0 , 根號內值不能為負 (因為目前討論的範圍在實數系上) ; 同理, x < -1 1-x^2 < 0 也一樣為負.換句話來說, 1-x^2 > 0 才能使 f(x) 有定義, 即求解 1-x^2 > 0 之解集合 .
(1+x)(1-x) > 0
1+x > 0 and 1-x > 0 implies x > -1 and x < 1
1+x < 0 and 1-x < 0 implies x < -1 and x > 1 ( >< )
the domain of f(x) is (-1,1)
Given any real number x and x ∈ (-1,1) , lim f(x) = f(x) then f(x) is continuous in (-1,1) .
(2) g(x) = |x^2 -1|
|x^2 -1| = |(x+1)(x-1)|
if (x+1)(x-1) < 0 implies x+1 > 0 and x-1 < 0 or x+1 < 0 and x-1 > 0 .
x > -1 and x < 1 (-∞ , 1) or
x < -1 and x > 1 (-∞,-1)∪(1,∞)
So (-∞,1)∪(-∞,-1)∪(1,∞) = R .
Since f(1) > 0 and f(3) > 0 , f(c) = 2.5 , but f(x) is floor function . For any number x , f(x) is the biggest integral number less than x .
So f(c) = 2.5 is impossible.
f(0) = -1 , f(1) = 1+2-1 =2 . f(0)f(1) < 0 , 故由勘根定理得知, [0,1]之間幣存在一個實數 c , 使得 f(c) = 0 , 即意為 c 點為交於 x 軸上的實數點,故為其實數解 .
綜合練習
證:
Given a positive number s , there exists another positive number p , such that 0 < |x- a| < p , |f(x) - L| < s .
Since | |f(x)| - |L| | < |f(x) - L| , so | |f(x)| - |L| | < s Q.E.D.
(1)
證 :
(1+x/x^2+1) - 2/5 = 5(1+x) - 2(x^2+1) / 5(x^2+1) = 5 + 5x - 2x^2 -2 / (5x^2 + 5) = 3 + 5x - 2x^2 / (5x^2 + 5) = (x-3)(2x-1) / 5(x^2+1) = (x-3) [2x-1/5(x^2+1)]
|x- 3| < 1 ( s1 = 1 )
-1 +3 < x < 1+3
2 < x < 4
4 < 2x < 8
3 < 2x - 1 < 7
4 < x^2< 16
5 < x^2 + 1 < 17
25 < 5(x^2 + 1) < 85
1/85 < 1/5(x^2 + 1) < 1/25
3/85 < (2x - 1)/5(x^2 + 1) < 7/25
|(1+x/x^2+1) - 2/5| = | (x-3) [2x-1/5(x^2+1)] | = | x-3 | | 2x-1/5(x^2+1)| < 7/25 | x-3 | , s2 = 25/7 p
S = min { 1, 25/7 p }
Therefore , given a positive number p then there exists another positive number s, s = min { 1 , 25/7 p } , such that |(1+x/x^2+1) - 2/5| = | (x-3) [2x-1/5(x^2+1)] | = | x-3 | | 2x-1/5(x^2+1)| < 7/25 | x-3 | = 7/25 (25/7 p) = p
(2)
證:
| x - 27 | = | x^1/3 - 3 | | x^2/3 - 3x^1/3 + 9 | =
| x^1/3 - 3 | = | x- 27 | / | x^2/3 - 3x^1/3 + 9 |
|x - 27 | < 1
-1 < x - 27 < 1
26 < x < 28
26^1/3 < x^1/3 < 28^1/3
3*26^1/3 < 3*x^1/3 < 3*28^1/3
- 3*28^1/3 < - 3*x^1/3 < - 3*26^1/3
-28 < - x < -26
26^2 < x^2 < 28^2
26^2/3 < x^2/3 < 28^2/3
26^2/3 - 3*28^1/3 + 9 < x^2/3 - 3*x^1/3 + 9 < 28^2/3 - 3*26^1/3 + 9
1/ 28^2/3 - 3*26^1/3 + 9 < 1 / ( x^2/3 - 3*x^1/3 + 9) < 1 / 26^2/3 - 3*28^1/3 + 9
|1 / ( x^2/3 - 3*x^1/3 + 9) | < 1 / (26^2/3 - 3*28^1/3 + 9)
(1) 此圖為連續函數,所以無處不連續, lim f(x) = f(a)
(2) 觀察f(x) 在x = a 處的極值為b , 但f(x)在x= a 定義為f(a)
(3) x= a , f(x) 出現斷點
(1) 見到絕對值必討論其中的式子, |x+1| .
case 1 : x+1 > 0 , x > -1
case 2 : x+1 < 0 , x < -1
x > -1 , |x+1| = x+ 1
x < -1 , |x+1|= -(x+1) 故x = -1 , f(x) 極值不存在 .
(2) 使用長除法 , 1-x^12 / 1- x = (1+x+x^2+...+x^11)
(3) 有理化, (x-6)^1/2+ (x-2)^1/2
(1) 消去 (x-1) , x-2/x-1 不存在
(2) 通分看看
(3) 有理化
(4) 分解因式消去使分子母同為0的因式
(1) 配成 sin t / t 形式
(2) sin3x/tan6x = (sin3x/3x)(6x/tan6x)(3x/6x) =1/2 (sin3x/3x)(6x/tan6x)
(3) sin(sinx)/x = (sin(sinx)/sinx)(sinx/x)
(4) 可用半角公式
觀察一下, 將x = 2代入分母為0 故分子必為0
8 + 2a + b = 0
2a + b = -8
同時分子可分解為(x-2)(cx+d) 且可以跟分母消去(x-2) 成為 (cx+d)/(x+1) 將 x = 2 代入其中,
2c+d/3 = 5/3
2c+d = 5
將(x-2)(cx+d) = cx^2 + (d-2c)x -2d , c =2 , d = 1 , b = - 2 , 2a - 2 = - 8 , 2a = - 6 , a = - 3
x = -1/2 代入分子, 分子為 0 . 所以分母必為0
2(1/4) - a/2 + b = 0
1/2 -a/2 + b = 0
1- a + 2b = 0
分子母皆有一個(x-1/2)的因式可消去. (2x+1)(x-1)/(x-1/2)(cx+d) = x-1/cx+d = -1/2 -1 / -c/2 +d = -3 ; -c/2 + d = 1/2 . 1/2 = d - c/2 , 1 = 2d - c
cx^2 - (c/2 +d)x -d/2 , c = 2 , d = 3/2
(1) [7/2] = 3
(2) x-1 < [x] <=x by sandwich theorem , lim (1-1/x) = 1 , lim 1 = 1 , lim [x]/x = 1
(3) 見絕對值討論其中式子(x^2-1) = (x+1)(x-1)
(x+1)(x-1) > 0 , x+1 > 0 and x -1 > 0 implies x > -1 and x > 1
(x+1)(x-1) < 0 , x+1 > 0 and x-1 < 0 or x+1 < 0 and x-1 > 0
x > -1 and x < 1 , (-1,1)
x < -1 and x > 1 , (-∞,-1)∪(1,∞)
x -> 1+ , (x-1)^2/(x^2-1) = (x-1)/(x+1) = 0
x -> 1- , (x-1)^2/-(x^2-1) = (x-1)/-(x+1) = 0
x^3+8 /(x+2)= (x+2)(x^2-2x+4)/(x+2) = (x^2-2x+4) , lim (x^2-2x+4) = 12 , x = -2 .
f(-2) <> lim f(x) . So f(x) is not continuous .
討論 x = 1 左右的極限值討論.
x (1-) 3 -1 = 2
x (1+) 1+1 = 2
lim f(x) = 2 , f(1) = 2
x = 0 上的討論
x → 0+ , f(0+) = 1
x → 0- , f(0-) = -1 ,
f(x) 在 x = 0 的左右極限不同. 故f(x) 在 x =0 上極值不存在.
y = x -[x] ,
x = 4 , then x → 4+ , [x] = 4 , x → 4- , [x] = 3
y = x - [x] = 0 , x → 4+
y = x - [x] = 1 , x → 4-
x = 3 , then x → 3+ , [x] = 3 , x → 3- , [x] = 2
y = x - [x] = 1 , x → 3-
x = 2 , then x → 2+ , [x] = 2 , x → 2- , [x] = 1
y = x - [x] = 1 , x → 2-
x = 1 , then x → 1+ , [x] = 1 , x → 1- , [x] = 0
y = x - [x] = 1 , x → 1-
x = 0 , then x → 0+ , [x] = 0 , x → 0- , [x] = -1
y = x - [x] = 1 , x → 0-
x = -1 , then x → -1+ , [x] = -1 , x → -1- , [x] = -2
y = x - [x] = 1 , x → -1-
同理, 若 x ∈ Z , 且 -4 ≤ x ≤ 4 , 每個整數點都不連續 , 即 lim f(x) 不存在.
lim f(x)/x = 10 , that is f(x) = xg1(x)
lim f(x)/x-1 = -4 , that is f(x) = (x-1)g2(x)
lim f(x)/x-2 = 26 , that is f(x) = (x-2)g3(x)
g(x) = g1(x)g2(x)g3(x)
所以f(x) 中必有x,(x-1),(x-2) 的三個因式 , 則 f(x) = x(x-1)(x-2)g(x) .
但是要如何決定g(x) ?
由題意欲求最低次數的f(x) , 問題是用假設的推論
g(x) = cx +d 則
x → 0 , f(x)/x = (x-1)(x-2)(cx+d) = 2d = 10 , d = 5
x → 1 , f(x)/x-1 = x(x-2)(cx+d) = -(c+d) = -4 , -(c+5) = - 4 , c = -1
既然 c, d 值接求出 將其帶入第三式, lim x(x-1)(x-2)(-x+5)/x-2 = lim x(x-1)(-x+5) = 6 錯的 !
所以 , g(x) 為一次式是錯的. 再假設 g(x) 為二次式 , g(x) = ax^2+bx+c
x → 0 , f(x)/x = (x-1)(x-2)(ax^2+bx+c) = 2c = 10 , c = 5
x → 1 , f(x)/x = x(x-2)(ax^2+bx+c) = - (a+b+c) = - 4 , a+b+c = 4 , a+b = -1
x → 2 , f(x)/x = x(x-1)(ax^2+bx+c) = 2(4a+2b+5) = 26 , 4a+2b+5 = 13 , 4a+2b = 8 , 2a+b = 4
2a+2b = -2 , 2a+b = 4 , b = -6 , a = 5
所以 f(x) = x(x-1)(x-2)(5x^2-6x+5) .
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