Factoring the numerator creates: (x -1)(x2 +x+1), thus the
(x – 1) cancels, leaving
limx2 x 1 3 x1
This indicates that the function values are getting very, very, very close to 1 as the x values approach 3.
时，
BD
1x
oC
A
△AOB 的面积＜ 圆扇形AOB的面积＜△AOD的面积
即
1 2
sin
x
1 2
tan
x
亦故即有 显然有
sin x x tan x
(0
x
2
)
cos x sin x 1 x
(0
x
2
)
注
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example.
calculate
解:
lim
x0
tan x
x
lim (1 1 )x
x
x
2.7
1.2
Rigorous Study of Limits
Definition of Limit: L is the limit of f(x) as x approaches c if for any value of epsilon (no matter how small) around f(x)=A, there’s a value of sigma around x=c, such that all function values fall into the “target area” of epsilon about L.
故
1 a 0
因此
a 1
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Squeeze Theorem
A) g(x) is bounded above and below by f(x) and h(x)
B) the limit of f(x) as x approaches a = L and the limit of h(x) as x approaches a=L
THEN C) the limit of g(x) as x approaches a = L. (WHY is this useful? Sometimes we can’t
calculate a limit of a specific function, but we CAN calculate the limit of other functions right above & below it!)
xx0
lim f (x) lim f (x) A
xx0
xx0
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Function
f
(
x)
x 0
1, ,
x 1 ,
x0 x0 x0
y
y x 1
1
o 1
x
y x 1
讨论 x 0 时 f (x) 的极限是否存在 .
lim f (x) lim (x 1) 1
x0
有界
因此 为无穷小, f (x) A
g(x) 1 B
由极限与无穷小关系定B 理 ,
得
1 g(x)
2 B
x (x0 )
(详见P44)
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若
lim f (x) A, lim g(x) B , 则有
说明: 可推广到有限个函数相乘的情形 .
推论 1 . lim[C f (x)] C lim f (x) ( C 为常数 )
x
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3. 左极限与右极限
左极限 :
f
(x0 )
lim
xx0
f
(x)
A
0, 0, 当 x ( x0 , x0 )
时, 有
右极限 :
f
(x0 )
lim
xx0
f
(x)
A
0, 0, 当 x ( x0 , x0 )
时, 有
定理 3 .
lim f (x) A
Find
lim x3 1
x1 x 1
The function is not defined at x = 1, but since we’re only interested in what happens as we get VERY close to that
point, it does not matter.
How do you prove a limit exists?
For any given epsilon, you must find sigma, in terms of that epsilon such that it would always be true.
Pr ove: lim 2x 1 7 x3
x1
x 1 u2 1 u 1 x 1 u 1
∴ 原式 lim(u 1) 2
u 1
方法 2
lim (x 1)( x 1) lim( x 1)
x1 x 1
x1
2
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试确定常数 a 使
解: 令t1,则 x
0
lim
t0
3
1
1 t3
a t
3
lim
t0
t3
1 t
a
lim 3 t3 1 a 0 t0
0 x 3 (2x 1) 7
2x 6
2x3
x 3 , 2
2
Prove
Proof: f (x) A
0, To any 0, 当
时,
总有 So
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Prove
Proof: f (x) A
So 0, 取 , 当 x2 1 2
x0
lim f (x) lim (x 1) 1
x0
x0
So f (0 ) f (0 ) ,
lim f (x) NOT Exist .
x0
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2.6 Limit Theorems
1.lim k k xc
2.lim x c xc
3.lim k f ( x) k lim f ( x)
(1 n11)n (1 1x)x (1 1n)n1
lim (1
n
n11)n
lim
n
(1 n11)n1 e
1
1 n1
lim (1
n
1 n
)n1
lim [(1
n
1n)n（1
1n）]

e
lim (1
x
1x) x
e
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Calculate
lim (1 x)1/ x
x0
Calculate
such that f ( x) A , provided that x M .
lim f ( x) A
x
Similarly,
lim f (x) A
x
lim f (x) A
x
0,M 0, s.t x M f ( x) A 0,M 0, s.t x M f ( x) A 0,M 0, s.tx M f ( x) A
lim x0
sin x
x
1 cos
x
lim sin x lim 1 1 x0 x x0 cos x
example.
解: 令 t arcsin x, 则 x sin t , 因此
原式 lim t t0 sin t
sin t 1
t
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2.
证: 当 x 0 时, 设 n x n 1, 则
推论 2 . lim[ f (x)]n [ lim f (x) ] n ( n 为正整数 )
设 n 次多项式
试证
lim
xx0
Pn
(
x)
Pn
(
x0
).
证:
lim
x x0
Pn
(x)
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定理. 若 lim f (x) A, lim g(x) B , 且 B≠0 , 则有
x = 3 时分母为 0 !
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example
解: x = 1 时分母 = 0 , 分子≠0 , 但因
lim x2 5x 4 12 51 4 0
x1 2x 3
21 3
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例. 求
解: 方法 1 令 u x , 则 limu 1,
证: 因 lim f (x) A, lim g(x) B , 则有
f (x) A , g(x) B (其中 , 为无穷小) 于是 f (x) g(x) (A ) (B )
(A B) ( ) 由定理 1 可知 也是无穷小, 再利用极限与无穷小
的关系定理 , 知定理结论成立 .
CHAPTER 2
LIMITS
2.1
Introduction to Limits (Calculus is ALL about limits!)
An Intuitive Understanding
A limit implies what happens to our function as we get closer and closer to a specific x value. We may never get to that point, and that does not matter, we are only interested in the behavior of the function (y-values) as we get VERY close to a specific x value.