The definition of the limit of the function proves itself, and how to prove the limit of the functio

For example, it is proved that when n, the limit of lim 1 n is 0, and the limit definition is; For any given positive number, there is always a positive integer n, such that when n > n, all x, inequalities [xn-a]< are true, and the lateral constant a is its limit. My questions are:1.

When proving by definition, "for any given positive number, there is always a positive integer n", this is known, or 2 needs to be proved. The proof is mainly to prove what is to show that the limit is a constant a.

Answer: For any given positive number, find an n such that n > n, [xn-a]< Of course, the selection of this n is related to , which can be understood as a function of . For example, you can show that for any given positive number, there is n=[1 ]+1, and when n > n, there is |xn-a|=|1/n|<1 n< (1 n<1 n because n > n).

Function. There are two variables, one is the independent variable and the other is the dependent variable.

When the independent variable takes a value, the dependent variable has a unique value corresponding to it.

There are many, column one, y is equal to the square of x, when x takes 2, y can only be equal to 4, when x takes minus 2, y can only be equal to 4, understand that when the independent variable takes a value, the dependent variable has a unique value corresponding to it, just rely on two to judge, generally as if a is not equal to zero.

Proving limits with definitions is all written in the format:

Limited |x-1/2|<1 4, there is |x-1| >1/2-|x-1/2| >1/2-1/4 = 1/4。Arbitrarily given >0, to make.

x/(x-1)-(1)| 2|(x-1/2)/(x-1)|

2|x-1/2|/|x-1| <2|x-1/2|/(1/4)

8|x-1/2|"Just |x-2|Take δ (min > 0, then 0< |x-1/2|"When there is|x/(x-1)-(1) <8|x-1/2|According to the definition of the limit, it is proven.

NEED NOTICE:

In the seventeenth century, Galileo, in his book The Two New Sciences, almost entirely included the concept of functions or variable relations, expressing the relations of functions in the language of words and proportions.

Around 1637, Descartes had noticed the dependence of one variable on another in his analytic geometry, but because he was not aware of the need to refine the concept of functions at that time, no one had clarified the general meaning of functions until Newton and Leibniz established calculus in the late 17th century, and most functions were studied as curves.

The prenumerator and denominator are physical and chemical. Multiply the numerator and denominator by a positive number: (x +1) +x to get:

lim [(x²+1) -x²]/x²+1) +x]=lim 1/[√x²+1) +x]

At this point, you can prove it with the limit definition of the function.

lim(x tends to positive infinity) (x +1)-x

lim(xtends to +)1 (x+1)+x).

lim(xtends to +)1 x) (1+1x)+1

When x +, the function ( x +1) is the same as the function x =|x|The graphs are more and more coincidental, and their values are both positive infinity, so ( x +1)-x can be regarded as positive infinity minus positive infinity, which is equal to 0 in the current mathematical cognition.

Verification When x approaches x0, the limit of the function f(x) is equal to a Prov: Just prove: For an arbitrarily small e>0, there is d>0, when |x-x0|0, there is d=e 6>0, when |x-x0|At

lim(x->+x^2+1) -x]

There are physicochemical molecules.

lim(x->+x^2+1) -x].[x^2+1) +x]/ x^2+1)+x]

Utilize (a-b)(a+b)=a 2-b 2=lim(x->+x 2+1) -x 2] x 2+1)+x]=lim(x->+1 [x 2+1)+x] numerator=1, denominator->+

Limited |62616964757a686964616fe58685e5aeb931333431373836x-1/2|<1 4, there is |x-1| >1/2-|x-1/2| >1/2-1/4 = 1/4。Arbitrarily given >0, to make.

x/(x-1)-(1)| = 2|(x-1/2)/(x-1)

2|x-1/2|/|x-1| <2|x-1/2|/(1/4)

8|x-1/2|"Just |x-2|Take δ (min > 0, then 0< |x-1/2|"When there is|x/(x-1)-(1) <= 8|x-1/2|According to the definition of the limit, it is proven.

Functions are related to inequalities and equations (elementary functions). Let the value of the function be equal to zero, and from a geometric point of view, the value of the corresponding independent variable is the abscissa of the intersection point of the image and the x-axis; From an algebraic point of view, the corresponding independent variable is the solution of the equation. In addition, if you replace the "=" in the expression of the function (except for functions without expression) with "<" or ">" and replace the "y" with another algebraic formula, the function becomes an inequality and the range of independent variables can be found.

The image of the function f is a set of points on a plane, where x is taken to define all the members of the domain. Function image icons to help understand and prove some theorems.

If x and y are both continuous lines, then the image of the function has a very visual representation of the binary relationship between the two sets x and y that there are two definitions: one is a triplet (x, y, g), where g is the graph of the relationship; The second is to simply define it in terms of the diagram of the relationship. With the second definition, the function f is equal to its image.

|bai uses definitions to prove that the extreme limit is the writing of the format, and the drawing of the gourd is:

DAO limit |Edition X-1 2|<1 4, there is |x-1| >1/2-|x-1/2| >1/2-1/4 = 1/4。Given >0, it is necessary to make.

x/(x-1)-(1)| = 2|(x-1/2)/(x-1)|= 2|x-1/2|/|x-1| <2|x-1/2|/(1/4)= 8|x-1/2|"Just |x-2|0, then 0< |x-1/2|"When there is|

x/(x-1)-(1) <= 8|x-1/2|According to the definition of the limit, it is proven.

(1) Let f(x) = (2x+3) 3x, due to |f(x)-a|=|f(x)-2/3|=|1/x|, any >0, to prove the existence of m>0, when |x|At >m, inequality |(1/x)-0|< was founded.

Because this inequality is equivalent to 1 |x|1/ε.From this, it can be seen that if m=1 is taken, then when |x|>m=1, the inequality |1/x-0|, limf(x)=2 3

3) The little brother is not talented, this question will not be...

Other netizens' answers:

x-2]<δ0

1 (x-1)-1]=[2-x] [x-1]<δ (1-δ)= , can be set to δ= (1+).

Let's use -δ to prove that when x approaches 2, the limit of 1 (x-1) is 1.

For any small 0< <1, take a= (1+).

When [x-2](1+), >x-2](1+)=[x-2]+[x-2],x-2]< 1-[x-2]),1 (x-1)-1]=[x-2] [x-2+1]<[x-2] (1-[x-2])<

So, when x approaches 2, the limit of 1 (x-1) is 1.

4) If the limit of this problem is 2, it can be proved as follows:

The function is undefined at the point x=1, but the existence or absence of the limit of the function at x->1 has nothing to do with it. In fact, any >0 will be an inequality |f(x)-2|< After removing the non-zero factor x-1, it becomes |x-1|< Therefore, as long as δ= is taken, then it should be 0<|x-1|"When there is|f(x)-2|< So, the original limit is established.

Proving the limit with a definition is actually a way to write a format, and drawing a gourd in the same way is:

Prove that any >0 is to be made.

2x+1)-5| = 2|x-2|Just |x-2|< 2, take = 2, then 0<|x-2|< times, there is.

2x+1)-5| = 2|x-2|<2 = , proven.

Use definitions to prove that the limit is the way to write the format, and draw a gourd in the same way, help you write a :

1 (2) Arbitrarily given >0, to make.

x²-1)/(x-1)-2| = |x-1|"Only 0 < x-1|Take δ (= 0, then 0< |x-1|"When there is|

x²-1)/(x-1)-2| = |x-1|According to the definition of the limit, it is proven.

Function Limit Definition:

Let the function f(x) be defined in a decentered neighborhood at x0, and if there is a constant a, for any >0, there is always a positive δ, such that dang.

x-xo|<δ, |f(x)-a|< is true, then a is the limit of the function f(x) at x0.

e.g. limx 3=27

The limit when x approaches 3:

Since x is close to 3, we only consider the x value near x 3x-3|<10, there is always a positive number δ min(1, 37) takes the minimum value such that when |x-3|<δ, |f(x)-27|< holds, therefore, 27 is the limit of the function f(x)=x 3 at x=3.

Proving the limit with a definition is a way to write the format, and drawing a gourd in the same way is:

Limited |x-1/2|<1 4, there is |

x-1|1/2-|x-1/2|

1/4。Arbitrarily given >0, to make.

x/(x-1)-(1)|

2|(x-1/2)/(x-1)|

2|x-1/2|/|x-1|

2|x-1/2|/(1/4)

8|x-1/2|"Just |

x-2|min, take.

min0, then.

x-1/2|

"When there is|x/(x-1)-(1)

8|x-1/2|According to the definition of the limit, it is proven.

Proving the limit with a definition is a way to write the format, and drawing a gourd in the same way is:

Proof of arbitrarily given >0, to make.

sinx/√(x+1)-0|< = 1 x < only x > 1 takes x = x( )= 1 0, then when x > x, there is.

sinx/√(x+1)-0|< = 1 x < 1 x = , according to the definition of the limit, proven.

Any e>0, left and right limits are discussed.

Right limit: When x approaches 1 from the right side of 1, it can be seen from the monotonicity of the logarithmic function that there is a constant lnx>0 takes δ1=e e-1>0, then when 10, then when 1-δ2lnx-0|=-lnx<-ln(1-δ2)=-ln(e^(-e))=e

lim(x→1-)lnx=0

The left and right limits exist and are equal, lim(x 1)lnx=0