Second midline trapezoid properties. Thales's theorem. The middle line of the triangle. What we learned

The process of studying a function for continuity is inextricably linked with the skill of finding one-sided limits of a function. Therefore, in order to begin studying the material in this article, it is advisable to first examine the topic of the limit of a function.

Definition 1

Function f(x) is continuous at point x 0, if the limit on the left is equal to the limit on the right and coincides with the value of the function at point x 0, i.e.: lim x → x 0 - 0 f (x) = lim x → x 0 + 0 f (x) = f(x0)

This definition allows us to derive a corollary: the value of the limit of a function at points of continuity coincides with the value of the function at these points.

Example 1

The function f (x) = 1 6 (x - 8) 2 - 8 is given. It is necessary to prove its continuity at the point x 0 = 2.

Solution

First of all, we determine the existence of a limit on the left. To do this, we use a sequence of arguments x n, which reduces to x 0 = 2 · (x n< 2) . Например, такой последовательностью может быть:

2 , 0 , 1 , 1 1 2 , 1 3 4 , 1 7 8 , 1 15 16 , . . . , 1 1023 1024 , . . . → 2

The corresponding sequence of function values ​​looks like this:

f(-2); f (0) ; f (1) ; f 1 1 2 ; f 1 3 4 ; f 1 7 8 ; f 1 15 16 ; . . . ; f 1 1023 1024 ; . . . = = 8 . 667; 2. 667; 0 . 167; - 0 . 958; - 1 . 489; - 1 . 747; - 1 . 874; . . . ; - 1 . 998; . . . → - 2

in the drawing they are indicated in green.

It is quite obvious that such a sequence reduces to - 2, which means lim x → 2 - 0 1 6 (x - 8) 2 - 8 = - 2.

Let us determine the existence of a limit on the right: we use a sequence of arguments x n, which reduces to x 0 = 2 (x n > 2). For example, this sequence could be:

6 , 4 , 3 , 2 1 2 , 2 1 4 , 2 1 8 , 2 1 16 , . . . , 2 1 1024 , . . . → 2

The corresponding sequence of functions:

f (6) ; f (4) ; f (3) ; f 2 1 2 ; f 2 1 4 ; f 2 1 8 ; f 2 1 16 ; . . . ; f 2 1 1024 ; . . . = = - 7 . 333; - 5 . 333; - 3. 833; - 2. 958; - 2. 489; - 2. 247; - 2. 247; - 2. 124; . . . ; - 2. 001 ; . . . → - 2

indicated in blue in the figure.

And this sequence reduces to - 2, then lim x → 2 + 0 1 6 (x - 8) 2 - 8 = - 2.

The actions above showed that the limits on the right and left are equal, which means there is a limit of the function f (x) = 1 6 x - 8 2 - 8 at the point x 0 = 2, while lim x → 2 1 6 (x - 8 ) 2 - 8 = - 2 .

After calculating the value of the function in given point the equality is obvious:

lim x → 2 - 0 f (x) = lim x → 2 + 0 f (x) = f (2) = 1 6 (2 - 8) 2 - 8 = - 2 which indicates the continuity of the given function at a given point.

Let's show it graphically:

Answer: The continuity of the function f (x) = 1 6 (x - 8) 2 - 8 in the given part has been proven.

Removable rupture of the first kind

Definition 2

The function has removable rupture of the first kind at point x 0, when the limits on the right and left are equal, but not equal to the value of the function at the point, i.e.:

lim x → x 0 - 0 f (x) = lim x → x 0 + 0 f (x) ≠ f (x 0)

Example 2

The function f (x) = x 2 - 25 x - 5 is given. It is necessary to determine the points of its break and determine their type.

Solution

First, let's denote the domain of definition of the function: D (f (x)) ⇔ D x 2 - 25 x - 5 ⇔ x - 5 ≠ 0 ⇔ x ∈ (- ∞ ; 5) ∪ (5 ; + ∞)

In a given function, only the boundary point of the domain of definition can serve as a break point, i.e. x 0 = 5. Let us examine the function for continuity at this point.

Let's simplify the expression x 2 - 25 x - 5: x 2 - 25 x - 5 = (x - 5) (x + 5) x - 5 = x + 5.

Let's define the limits on the right and left. Since the function g(x) = x + 5 is continuous for any real x, then:

lim x → 5 - 0 (x + 5) = 5 + 5 = 10 lim x → 5 + 0 (x + 5) = 5 + 5 = 10

Answer: the limits on the right and left are equal, and the given function at the point x 0 = 5 is not defined, i.e. at this point the function has a removable discontinuity of the first kind.

An irremovable discontinuity of the first kind is also determined by the jump point of the function.

Definition 3 Example 3

Given a piecewise continuous function f (x) = x + 4 , x< - 1 , x 2 + 2 , - 1 ≤ x < 1 2 x , x ≥ 1 . Необходимо изучить given function for continuity, indicate the type of break points, draw up a drawing.

Solution

Discontinuities of this function can only be at the point x 0 = - 1 or at the point x 0 = 1.

Let us determine the limits to the right and left of these points and the value of the given function at these points:

• to the left of the point x 0 = - 1 the given function is f (x) = x + 4, then by continuity linear function: lim x → - 1 - 0 f (x) = lim x → - 1 - 0 (x + 4) = - 1 + 4 = 3 ;
• directly at the point x 0 = - 1 the function takes the form: f (x) = x 2 + 2, then: f (- 1) = (- 1) 2 + 2 = 3;
• on the interval (- 1 ; 1) the given function is: f (x) = x 2 + 2. Based on the property of continuity quadratic function, we have: lim x → - 1 + 0 f (x) = lim x → - 1 + 0 (x 2 + 2) = (- 1) 2 + 2 = 3 lim x → 1 - 0 f (x) = lim x → 1 - 0 (x 2 + 2) = (1) 2 + 2 = 3
• at point x 0 = - 1 the function has the form: f (x) = 2 x and f (1) = 2 1 = 2.
• to the right of the point x 0 the given function is f (x) = 2 x. Due to the continuity of the linear function: lim x → 1 + 0 f (x) = lim x → 1 + 0 (2 x) = 2 1 = 2

Answer: ultimately we got:

• lim x → - 1 - 0 f (x) = lim x → - 1 + 0 f (x) = f (- 1) = 3 - this means that at the point x 0 = - 1 the given piecewise function is continuous;
• lim x → - 1 - 0 f (x) = 3, lim x → 1 + 0 f (x) = 2 - thus, at the point x 0 = 1 an irremovable discontinuity of the first kind (jump) is defined.

All we have to do is prepare a drawing for this task.

Definition 4

The function has second kind discontinuity at the point x 0, when any of the limits on the left lim x → x 0 - 0 f (x) or on the right lim x → x 0 + 0 f (x) does not exist or is infinite.

Example 4

The function f (x) = 1 x is given. It is necessary to examine the given function for continuity, determine the type of break points, and prepare a drawing.

Solution

Let us write down the domain of definition of the function: x ∈ (- ∞ ; 0) ∪ (0 ; + ∞) .

Let's find the limits to the right and left of the point x 0 = 0.

Let us specify an arbitrary sequence of argument values ​​converging to x 0 on the left. Eg:

8 ; - 4 ; - 2 ; - 1 ; - 1 2 ; - 1 4 ; . . . ; - 1 1024 ; . . .

It corresponds to the sequence of function values:

f (- 8) ; f (- 4) ; f(-2); f (- 1) ; f - 1 2 ; f - 1 4 ; . . . ; f - 1 1024 ; . . . = = - 1 8 ; - 14 ; - 12 ; - 1 ; - 2; - 4 ; . . . ; - 1024; . . .

Obviously, this sequence is infinitely large negative, then lim x → 0 - 0 f (x) = lim x → 0 - 0 1 x = - ∞ .

Now let's specify an arbitrary sequence of argument values ​​converging to x 0 from the right. For example: 8 ; 4 ; 2 ; 1 ; 12 ; 14 ; . . . ; 1 1024 ; . . . , and it corresponds to the sequence of function values:

f (8) ; f (4) ; f (2) ; f (1) ; f 1 2 ; f 1 4 ; . . . ; f 1 1024 ; . . . = = 1 8 ; 14 ; 12 ; 1 ; 2 ; 4 ; . . . ; 1024 ; . . .

This sequence is infinitely large positive, which means lim x → 0 + 0 f (x) = lim x → 0 + 0 1 x = + ∞ .

Answer: point x 0 = 0 is the discontinuity point of a function of the second kind.

Let's illustrate:

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Definition of continuity according to Heine

The function of a real variable \(f\left(x \right)\) is said to be continuous at the point \(a \in \mathbb(R)\) (\(\mathbb(R)-\)set real numbers), if for any sequence \(\left\( ((x_n)) \right\)\), such that \[\lim\limits_(n \to \infty ) (x_n) = a,\] the relation holds \[\lim\limits_(n \to \infty ) f\left(((x_n)) \right) = f\left(a \right).\] In practice, it is convenient to use the following \(3\) conditions for the continuity of a function \(f\left(x \right)\) at the point \(x = a\) (which must be executed simultaneously):

1. The function \(f\left(x \right)\) is defined at the point \(x = a\);
2. The limit \(\lim\limits_(x \to a) f\left(x \right)\) exists;
3. The equality \(\lim\limits_(x \to a) f\left(x \right) = f\left(a \right)\) holds.

Definition of Cauchy continuity (notation \(\varepsilon - \delta\))

Consider a function \(f\left(x \right)\) that maps the set of real numbers \(\mathbb(R)\) to another subset \(B\) of the real numbers. The function \(f\left(x \right)\) is said to be continuous at the point \(a \in \mathbb(R)\), if for any number \(\varepsilon > 0\) there is a number \(\delta > 0\) such that for all \(x \in \mathbb (R)\), satisfying the relation \[\left| (x - a) \right| Definition of continuity in terms of increments of argument and function

The definition of continuity can also be formulated using increments of argument and function. The function is continuous at the point \(x = a\) if the equality \[\lim\limits_(\Delta x \to 0) \Delta y = \lim\limits_(\Delta x \to 0) \left[ ( f\left((a + \Delta x) \right) - f\left(a \right)) \right] = 0,\] where \(\Delta x = x - a\).

The above definitions of continuity of a function are equivalent on the set of real numbers.

The function is continuous on a given interval , if it is continuous at every point of this interval.

Continuity theorems

Theorem 1.
Let the function \(f\left(x \right)\) be continuous at the point \(x = a\) and \(C\) be a constant. Then the function \(Cf\left(x \right)\) is also continuous for \(x = a\).

Theorem 2.
Given two functions \((f\left(x \right))\) and \((g\left(x \right))\), continuous at the point \(x = a\). Then the sum of these functions \((f\left(x \right)) + (g\left(x \right))\) is also continuous at the point \(x = a\).

Theorem 3.
Suppose that two functions \((f\left(x \right))\) and \((g\left(x \right))\) are continuous at the point \(x = a\). Then the product of these functions \((f\left(x \right)) (g\left(x \right))\) is also continuous at the point \(x = a\).

Theorem 4.
Given two functions \((f\left(x \right))\) and \((g\left(x \right))\), continuous for \(x = a\). Then the ratio of these functions \(\large\frac((f\left(x \right)))((g\left(x \right)))\normalsize\) is also continuous for \(x = a\) subject to , that \((g\left(a \right)) \ne 0\).

Theorem 5.
Suppose that the function \((f\left(x \right))\) is differentiable at the point \(x = a\). Then the function \((f\left(x \right))\) is continuous at this point (i.e., differentiability implies continuity of the function at the point; the converse is not true).

Theorem 6 (Limit value theorem).
If a function \((f\left(x \right))\) is continuous on a closed and bounded interval \(\left[ (a,b) \right]\), then it is bounded above and below on this interval. In other words, there are numbers \(m\) and \(M\) such that \ for all \(x\) in the interval \(\left[ (a,b) \right]\) (Figure 1).

Fig.1		Fig.2

Theorem 7 (Intermediate value theorem).
Let the function \((f\left(x \right))\) be continuous on a closed and bounded interval \(\left[ (a,b) \right]\). Then, if \(c\) is some number greater than \((f\left(a \right))\) and less than \((f\left(b \right))\), then there exists a number \(( x_0)\), such that \ This theorem is illustrated in Figure 2.

Continuity elementary functions

All elementary functions are continuous at any point in their domain of definition.

The function is called elementary , if it is built from a finite number of compositions and combinations
(using \(4\) operations - addition, subtraction, multiplication and division) . A bunch of basic elementary functions includes:

This article is about continuous numerical function. For continuous mappings in various branches of mathematics, see continuous mapping.

Continuous function- a function without “jumps”, that is, one in which small changes in the argument lead to small changes in the value of the function.

A continuous function, generally speaking, is synonymous with the concept of continuous mapping, however, most often this term is used in a narrower sense - for mappings between number spaces, for example, on the real line. This article is devoted specifically to continuous functions defined on a subset of real numbers and taking real values.

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Subtitles

Definition

If you “correct” the function f (\displaystyle f) at the point of removable rupture and put f (a) = lim x → a f (x) (\displaystyle f(a)=\lim \limits _(x\to a)f(x)), then we get a function that is continuous at a given point. This operation on a function is called extending the function to continuous or redefinition of the function by continuity, which justifies the name of the point as a point removable rupture.

Break point "jump"

A “jump” discontinuity occurs if

lim x → a − 0 f (x) ≠ lim x → a + 0 f (x) (\displaystyle \lim \limits _(x\to a-0)f(x)\neq \lim \limits _(x \to a+0)f(x)).

Break point "pole"

A pole gap occurs if one of the one-sided limits is infinite.

lim x → a − 0 f (x) = ± ∞ (\displaystyle \lim \limits _(x\to a-0)f(x)=\pm \infty ) or lim x → a + 0 f (x) = ± ∞ (\displaystyle \lim \limits _(x\to a+0)f(x)=\pm \infty ). [ ]

Significant break point

At the point of significant discontinuity, one of the one-sided limits is completely absent.

Classification of isolated singular points in Rn, n>1

For functions f: R n → R n (\displaystyle f:\mathbb (R) ^(n)\to \mathbb (R) ^(n)) And f: C → C (\displaystyle f:\mathbb (C) \to \mathbb (C) ) there is no need to work with break points, but you often have to work with special points(points where the function is not defined). The classification is similar.

The concept of “leap” is missing. What's in R (\displaystyle \mathbb (R) ) is considered a jump; in spaces of higher dimensions it is an essential singular point.

Properties

Local

• Function continuous at a point a (\displaystyle a), is bounded in some neighborhood of this point.
• If the function f (\displaystyle f) continuous at a point a (\displaystyle a) And f (a) > 0 (\displaystyle f(a)>0)(or f(a)< 0 {\displaystyle f(a)<0} ), That f (x) > 0 (\displaystyle f(x)>0)(or f(x)< 0 {\displaystyle f(x)<0} ) for all x (\displaystyle x), quite close to a (\displaystyle a).
• If the functions f (\displaystyle f) And g (\displaystyle g) continuous at a point a (\displaystyle a), then the functions f + g (\displaystyle f+g) And f ⋅ g (\displaystyle f\cdot g) are also continuous at a point a (\displaystyle a).
• If the functions f (\displaystyle f) And g (\displaystyle g) continuous at a point a (\displaystyle a) and wherein g (a) ≠ 0 (\displaystyle g(a)\neq 0), then the function f / g (\displaystyle f/g) is also continuous at a point a (\displaystyle a).
• If the function f (\displaystyle f) continuous at a point a (\displaystyle a) and function g (\displaystyle g) continuous at a point b = f (a) (\displaystyle b=f(a)), then their composition h = g ∘ f (\displaystyle h=g\circ f) continuous at a point a (\displaystyle a).

Global

• compact set) is uniformly continuous on it.
• A function that is continuous on a segment (or any other compact set) is bounded and reaches its maximum and minimum values ​​on it.
• Function range f (\displaystyle f), continuous on the segment , is the segment [ min f , max f ] , (\displaystyle [\min f,\ \max f],) where the minimum and maximum are taken along the segment [ a , b ] (\displaystyle ).
• If the function f (\displaystyle f) continuous on the segment [ a , b ] (\displaystyle ) And f (a) ⋅ f (b)< 0 , {\displaystyle f(a)\cdot f(b)<0,} then there is a point at which f (ξ) = 0 (\displaystyle f(\xi)=0).
• If the function f (\displaystyle f) continuous on the segment [ a , b ] (\displaystyle ) and number φ (\displaystyle \varphi ) satisfies the inequality f(a)< φ < f (b) {\displaystyle f(a)<\varphi or inequality f (a) > φ > f (b) , (\displaystyle f(a)>\varphi >f(b),) then there is a point ξ ∈ (a , b) , (\displaystyle \xi \in (a,b),) wherein f (ξ) = φ (\displaystyle f(\xi)=\varphi ).
• A continuous mapping of a segment to the real line is injective if and only if the given function on the segment is strictly monotonic.
• Monotonic function on a segment [ a , b ] (\displaystyle ) is continuous if and only if its range of values ​​is a segment with ends f (a) (\displaystyle f(a)) And f (b) (\displaystyle f(b)).
• If the functions f (\displaystyle f) And g (\displaystyle g) continuous on the segment [ a , b ] (\displaystyle ), and f(a)< g (a) {\displaystyle f(a) And f (b) > g (b) , (\displaystyle f(b)>g(b),) then there is a point ξ ∈ (a , b) , (\displaystyle \xi \in (a,b),) wherein f (ξ) = g (ξ) .(\displaystyle f(\xi)=g(\xi).)

From here, in particular, it follows that any continuous mapping of a segment into itself has at least one fixed point.

Examples

Elementary functions This function is continuous at every point.

x ≠ 0 (\displaystyle x\neq 0) The point is the break point first kind

, and,

lim x → 0 − f (x) = − 1 ≠ 1 = lim x → 0 + f (x) (\displaystyle \lim \limits _(x\to 0-)f(x)=-1\neq 1= \lim \limits _(x\to 0+)f(x))

while at the point itself the function vanishes.

Step function

Step function defined as< 0 , x ∈ R {\displaystyle f(x)={\begin{cases}1,&x\geqslant 0\\0,&x<0\end{cases}},\quad x\in \mathbb {R} }

f (x) = ( 1 , x ⩾ 0 0 , x is continuous everywhere except the point x = 0 (\displaystyle x=0) is continuous everywhere except the point, where the function suffers a discontinuity of the first kind. However, at the point there is a right-hand limit that coincides with the value of the function at a given point. So this function is an example continuous on the right functions.

throughout the entire definition area

Similarly, the step function defined as

f (x) = ( 1 , x > 0 0 , x ⩽ 0 , x ∈ R (\displaystyle f(x)=(\begin(cases)1,&x>0\\0,&x\leqslant 0\end( cases)),\quad x\in \mathbb (R) ) is an example continuous on the right functions.

continuous on the left

Dirichlet function

f (x) = ( 1 , x ∈ Q 0 , x ∈ R ∖ Q (\displaystyle f(x)=(\begin(cases)1,&x\in \mathbb (Q) \\0,&x\in \ mathbb (R) \setminus \mathbb (Q) \end(cases))) Definition. Let the function y = f(x) be defined at the point x0 and some of its neighborhood. The function y = f(x) is called continuous at point x0

, If:
1. exists 2. this limit equal to the value

functions at point x0:

f (x) = ( 1 , x ∈ Q 0 , x ∈ R ∖ Q (\displaystyle f(x)=(\begin(cases)1,&x\in \mathbb (Q) \\0,&x\in \ mathbb (R) \setminus \mathbb (Q) \end(cases))) When defining the limit, it was emphasized that f(x) may not be defined at the point x0, and if it is defined at this point, then the value of f(x0) does not participate in any way in determining the limit. When determining continuity, it is fundamental that f(x0) exists, and this value must be equal to lim f(x). Let the function y = f(x) be defined at the point x0 and some of its neighborhood. A function f(x) is called continuous at the point x0 if for all ε>0 there existsδ such that for all x from the δ-neighborhood of the point x0 (i.e. |x-x0|
Here it is taken into account that the value of the limit must be equal to f(x0), therefore, in comparison with the definition of the limit, the condition of puncture of the δ-neighborhood 0 is removed
Let us give one more (equivalent to the previous) definition in terms of increments. Let's denote Δх = x - x0; we will call this value the increment of the argument. Since x->x0, then Δx->0, i.e. Δx - b.m. (infinitesimal) quantity. Let us denote Δу = f(x)-f(x0), we will call this value the increment of the function, since |Δу| should be (for sufficiently small |Δх|) less than an arbitrary number ε>0, then Δу- is also b.m. value, therefore

f (x) = ( 1 , x ∈ Q 0 , x ∈ R ∖ Q (\displaystyle f(x)=(\begin(cases)1,&x\in \mathbb (Q) \\0,&x\in \ mathbb (R) \setminus \mathbb (Q) \end(cases))) Let the function y = f(x) be defined at the point x0 and some of its neighborhood. The function f(x) is called Let the function y = f(x) be defined at the point x0 and some of its neighborhood. The function y = f(x) is called, if an infinitesimal increment in the argument corresponds to an infinitesimal increment in the function.

f (x) = ( 1 , x ∈ Q 0 , x ∈ R ∖ Q (\displaystyle f(x)=(\begin(cases)1,&x\in \mathbb (Q) \\0,&x\in \ mathbb (R) \setminus \mathbb (Q) \end(cases))) The function f(x), which is not continuous at the point x0, called discontinuous at this point.

f (x) = ( 1 , x ∈ Q 0 , x ∈ R ∖ Q (\displaystyle f(x)=(\begin(cases)1,&x\in \mathbb (Q) \\0,&x\in \ mathbb (R) \setminus \mathbb (Q) \end(cases))) A function f(x) is called continuous on a set X if it is continuous at every point of this set.

Theorem on the continuity of a sum, product, quotient

Theorem on passing to the limit under the sign continuous function

Theorem on the continuity of superposition of continuous functions

Let the function f(x) be defined on an interval and be monotonic on this interval. Then f(x) can have only discontinuity points of the first kind on this segment.

Intermediate value theorem. If the function f(x) is continuous on a segment and at two points a and b (a is less than b) takes unequal values ​​A = f(a) ≠ B = f(b), then for any number C lying between A and B, there is a point c ∈ at which the value of the function is equal to C: f(c) = C.

Theorem on the boundedness of a continuous function on an interval. If a function f(x) is continuous on an interval, then it is bounded on this interval.

Theorem on achieving minimum and maximum values. If the function f(x) is continuous on an interval, then it reaches its lower and upper bounds on this interval.

Theorem on the continuity of the inverse function. Let the function y=f(x) be continuous and strictly increasing (decreasing) on ​​the interval [a,b]. Then on the interval there exists inverse function x = g(y), also monotonically increasing (decreasing) on ​​and continuous.