Definition of a continuous function at a given point. Concept of continuity of function
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 of real numbers), if for any sequence \(\left\( ((x_n)) \right\)\ ), such that \[\lim\limits_(n \to \infty ) (x_n) = a,\] the relation \[\lim\limits_(n \to \infty ) f\left(((x_n)) \right) = f\left(a \right).\] In practice, it is convenient to use the following \(3\) conditions for the continuity of the function \(f\left(x \right)\) at the point \(x = a\) ( which must be executed simultaneously):
- The function \(f\left(x \right)\) is defined at the point \(x = a\);
- The limit \(\lim\limits_(x \to a) f\left(x \right)\) exists;
- 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).
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Fig.1 |
Fig.2 |
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 of 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:
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 is equal to the value of the function at point x0: 
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).
Definition. 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 a point x0 if for all ε>0 there is a positive number δ such that for all x in 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
Definition. Let the function y = f(x) be defined at the point x0 and some of its neighborhood. The function f(x) is called continuous at point x0, if an infinitesimal increment in the argument corresponds to an infinitesimal increment in the function.
Definition. The function f(x), which is not continuous at the point x0, called discontinuous at this point.
Definition. 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 the passage to the limit under the sign of a 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 reaching 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 segment there exists an inverse function x = g(y), also monotonically increasing (decreasing) on and continuous.
The study of a function for continuity at a point is carried out according to an already established routine scheme, which consists of checking three conditions of continuity:
Example 1
Examine the function for continuity. Determine the nature of the function discontinuities, if they exist. Execute the drawing.
Solution:
1) The only point within the scope is where the function is not defined.
One-sided limits are finite and equal.
Thus, at the point the function suffers a removable discontinuity.
What does the graph of this function look like?
I would like to simplify 
, and it seems like an ordinary parabola is obtained. BUT the original function is not defined at point , so the following clause is required:
Let's make the drawing: 
Answer: the function is continuous on the entire number line except the point at which it suffers a removable discontinuity.
The function can be further defined in a good or not so good way, but according to the condition this is not required.
You say this is a far-fetched example? Not at all. This has happened dozens of times in practice. Almost all of the site’s tasks come from real independent work and tests.
Let's get rid of our favorite modules:
Example 2
Explore function 
for continuity. Determine the nature of the function discontinuities, if they exist. Execute the drawing.
Solution: For some reason, students are afraid and don’t like functions with a module, although there is nothing complicated about them. We have already touched on such things a little in the lesson. Geometric transformations of graphs. Since the module is non-negative, it is expanded as follows: 
, where “alpha” is some expression. In this case, and our function should be written piecewise: 
But the fractions of both pieces must be reduced by . The reduction, as in the previous example, will not take place without consequences. The original function is not defined at the point since the denominator goes to zero. Therefore, the system should additionally specify the condition , and make the first inequality strict: 
Now about a VERY USEFUL decision technique: before finalizing the task on a draft, it is advantageous to make a drawing (regardless of whether it is required by the conditions or not). This will help, firstly, to immediately see points of continuity and points of discontinuity, and, secondly, it will 100% protect you from errors when finding one-sided limits.
Let's do the drawing. In accordance with our calculations, to the left of the point it is necessary to draw a fragment of a parabola (blue color), and to the right - a piece of a parabola (red color), while the function is not defined at the point itself: 
If in doubt, take a few x values and plug them into the function 
(remembering that the module destroys the possible minus sign) and check the graph.
Let us examine the function for continuity analytically:
1) The function is not defined at the point, so we can immediately say that it is not continuous at it.
2) Let’s establish the nature of the discontinuity; to do this, we calculate one-sided limits: 
The one-sided limits are finite and different, which means that the function suffers a discontinuity of the 1st kind with a jump at the point . Note that it doesn't matter whether the function at the break point is defined or not.
Now all that remains is to transfer the drawing from the draft (it was made as if with the help of research ;-)) and complete the task:
Answer: the function is continuous on the entire number line except for the point at which it suffers a discontinuity of the first kind with a jump.
Sometimes they require additional indication of the discontinuity jump. It is calculated simply - from the right limit you need to subtract the left limit: , that is, at the break point our function jumped 2 units down (as the minus sign tells us).
Example 3
Explore function 
for continuity. Determine the nature of the function discontinuities, if they exist. Make a drawing.
This is an example for you to solve on your own, a sample solution at the end of the lesson.
Let's move on to the most popular and widespread version of the task, when the function consists of three parts:
Example 4
Examine a function for continuity and plot a graph of the function
.
Solution: it is obvious that all three parts of the function are continuous on the corresponding intervals, so it remains to check only two points of “junction” between the pieces. First, let's make a draft drawing; I commented on the construction technique in sufficient detail in the first part of the article. The only thing is that we need to carefully follow our singular points: due to the inequality, the value belongs to the straight line (green dot), and due to the inequality, the value belongs to the parabola (red dot): 
Well, in principle, everything is clear =) All that remains is to formalize the decision. For each of the two “joining” points, we standardly check 3 continuity conditions:
I)
1) 
The one-sided limits are finite and different, which means that the function suffers a discontinuity of the 1st kind with a jump at the point .
Let us calculate the discontinuity jump as the difference between the right and left limits:
, that is, the graph jerked up one unit.
II) We examine the point for continuity
1) 
- the function is defined at a given point.
2) Find one-sided limits:
- one-sided limits are finite and equal, which means there is a general limit.
3) 
At the final stage, we transfer the drawing to the final version, after which we put the final chord:
Answer: the function is continuous on the entire number line, except for the point at which it suffers a discontinuity of the first kind with a jump.
Example 5
Examine a function for continuity and construct its graph 
.
This is an example for independent solution, a short solution and an approximate sample of the problem at the end of the lesson.
You may get the impression that at one point the function must be continuous, and at another there must be a discontinuity. In practice, this is not always the case. Try not to neglect the remaining examples - there will be several interesting and important features:
Example 6
Given a function 
. Investigate the function for continuity at points. Build a graph.
Solution: and again immediately execute the drawing on the draft: 
The peculiarity of this graph is that the piecewise function is given by the equation of the abscissa axis. Here this area is drawn in green, but in a notebook it is usually highlighted in bold with a simple pencil. And, of course, don’t forget about our rams: the value belongs to the tangent branch (red dot), and the value belongs to the straight line.
Everything is clear from the drawing - the function is continuous along the entire number line, all that remains is to formalize the solution, which is brought to full automation literally after 3-4 similar examples:
I) We examine the point for continuity
2) Let's calculate one-sided limits:
, which means there is a general limit.
A little funny thing happened here. The fact is that I created a lot of materials about the limits of a function, and several times I wanted to, but several times I forgot about one simple question. And so, with an incredible effort of will, I forced myself not to lose the thought =) Most likely, some “dummies” readers doubt: what is the limit of the constant? The limit of a constant is equal to the constant itself. In this case, the limit of zero is equal to zero itself (left-handed limit).
3) 
- the limit of a function at a point is equal to the value of this function at a given point.
Thus, a function is continuous at a point by the definition of continuity of a function at a point.
II) We examine the point for continuity
1) - the function is defined at a given point.
2) Find one-sided limits:
And here, in the right-hand limit, the limit of unity is equal to unity itself.
- there is a general limit.
3) 
- the limit of a function at a point is equal to the value of this function at a given point.
Thus, a function is continuous at a point by the definition of continuity of a function at a point.
As usual, after research we transfer our drawing to the final version.
Answer: the function is continuous at the points.
Please note that in the condition we were not asked anything about studying the entire function for continuity, and it is considered good mathematical form to formulate precise and clear the answer to the question posed. By the way, if the conditions do not require you to build a graph, then you have every right not to build it (although later the teacher can force you to do this).
A small mathematical “tongue twister” for solving it yourself:
Example 7
Given a function 
.
Investigate the function for continuity at points. Classify breakpoints, if any. Execute the drawing.
Try to “pronounce” all the “words” correctly =) And draw the graph more precisely, accuracy, it will not be superfluous everywhere;-)
As you remember, I recommended immediately completing the drawing as a draft, but from time to time you come across examples where you can’t immediately figure out what the graph looks like. Therefore, in some cases, it is advantageous to first find one-sided limits and only then, based on the study, depict the branches. In the final two examples we will also learn a technique for calculating some one-sided limits:
Example 8
Examine the function for continuity and construct its schematic graph.
Solution: the bad points are obvious: (reduces the denominator of the exponent to zero) and (reduces the denominator of the entire fraction to zero). It’s not clear what the graph of this function looks like, which means it’s better to do some research first:
I) We examine the point for continuity
2) Find one-sided limits:
pay attention to typical method for calculating a one-sided limit: instead of “x” we substitute . There is no crime in the denominator: the “addition” “minus zero” does not play a role, and the result is “four”. But in the numerator there is a little thriller going on: first we kill -1 and 1 in the denominator of the indicator, resulting in . Unit divided by , is equal to “minus infinity”, therefore: . And finally, the “two” in infinitely large negative degree equal to zero: . Or, to be even more specific: 
.
Let's calculate the right-hand limit:
And here - instead of “X” we substitute . In the denominator, the “additive” again does not play a role: . In the numerator, actions similar to the previous limit are carried out: we destroy opposite numbers and divide one by : 
The right-hand limit is infinite, which means that the function suffers a discontinuity of the 2nd kind at the point .
II) We examine the point for continuity
1) The function is not defined at this point.
2) Let's calculate the left-sided limit:
The method is the same: we substitute “X” into the function. There is nothing interesting in the numerator - it turns out to be a finite positive number. And in the denominator we open the brackets, remove the “threes”, and the “additive” plays a decisive role.
As a result, the final positive number divided by infinitesimal positive number, gives “plus infinity”: .
The right-hand limit is like a twin brother, with the only exception that it appears in the denominator infinitesimal negative number:
One-sided limits are infinite, which means that the function suffers a discontinuity of the 2nd kind at the point .
Thus, we have two break points, and, obviously, three branches of the graph. For each branch, it is advisable to carry out a point-by-point construction, i.e. take several “x” values and substitute them into . Please note that the condition allows for the construction of a schematic drawing, and such relaxation is natural for manual work. I build graphs using a program, so I don’t have such difficulties, here’s a fairly accurate picture:
Direct are vertical asymptotes for the graph of this function.
Answer: the function is continuous on the entire number line except for points at which it suffers discontinuities of the 2nd kind.
A simpler function to solve on your own:
Example 9
Examine the function for continuity and make a schematic drawing.
An approximate example of a solution at the end that crept up unnoticed.
See you soon!
Solutions and answers:
Example 3:Solution
: transform the function: 
. Considering the modulus disclosure rule 
and the fact that , we rewrite the function in piecewise form:
Let's examine the function for continuity.
1) The function is not defined at the point .
The one-sided limits are finite and different, which means that the function suffers a discontinuity of the 1st kind with a jump at the point . Let's make the drawing:
Answer: the function is continuous on the entire number line except the point , in which it suffers a discontinuity of the first kind with a jump. Jump Gap: (two units up).
Example 5:Solution
: Each of the three parts of the function is continuous on its own interval.
I)
1)
2) Let's calculate one-sided limits:
, which means there is a general limit.
3)
- the limit of a function at a point is equal to the value of this function at a given point.
So the function continuous at a point by defining the continuity of a function at a point.
II)
We examine the point for continuity
1) - the function is defined at a given point. the function suffers a discontinuity of the 2nd kind at the point
How to find the domain of a function?
Examples of solutions
If something is missing somewhere, it means there is something somewhere
We continue to study the “Functions and Graphs” section, and the next station on our journey is Function Domain. An active discussion of this concept began in the first lesson. about function graphs, where I looked at elementary functions, and, in particular, their domains of definition. Therefore, I recommend that dummies start with the basics of the topic, since I will not dwell on some basic points again.
It is assumed that the reader knows the domains of definition of the basic functions: linear, quadratic, cubic functions, polynomials, exponential, logarithm, sine, cosine. They are defined on . For tangents, arcsines, so be it, I forgive you =) Rarer graphs are not immediately remembered.
The scope of definition seems to be a simple thing, and a logical question arises: what will the article be about? In this lesson I will look at common problems of finding the domain of a function. Moreover, we will repeat inequalities with one variable, the solution skills of which will also be required in other problems of higher mathematics. The material, by the way, is all school material, so it will be useful not only for students, but also for students. The information, of course, does not pretend to be encyclopedic, but here are not far-fetched “dead” examples, but roasted chestnuts, which are taken from real practical works.
Let's start with a quick dive into the topic. Briefly about the main thing: we are talking about a function of one variable. Its domain of definition is many meanings of "x", for which exist meanings of "players". Let's look at a hypothetical example: 
The domain of definition of this function is a union of intervals:
(for those who forgot: - unification icon). In other words, if you take any value of “x” from the interval , or from , or from , then for each such “x” there will be a value “y”.
Roughly speaking, where the domain of definition is, there is a graph of the function. But the half-interval and the “tse” point are not included in the definition area, so there is no graph there.
Yes, by the way, if anything is not clear from the terminology and/or content of the first paragraphs, it is better to return to the article Graphs and properties of elementary functions.
This article is about a continuous number 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.
Encyclopedic YouTube
1 / 5
✪ Continuity of function and breakpoints of function
✪ 15 Continuous function
✪ Continuous features
✪ Mathematical analysis, lesson 5, Continuity of function
✪ Continuous random variable. Distribution function
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 often you have to work with singular 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), fairly 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
- 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)
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
f (x) = ( 1 , x ⩾ 0 0 , x< 0 , x ∈ R {\displaystyle f(x)={\begin{cases}1,&x\geqslant 0\\0,&x<0\end{cases}},\quad x\in \mathbb {R} }is continuous everywhere except the point x = 0 (\displaystyle x=0), where the function suffers a discontinuity of the first kind. However, at the point x = 0 (\displaystyle x=0) 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 left functions throughout the entire definition area.
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)))In this lesson we will learn how to establish the continuity of a function. We will do this using limits, one-sided ones at that - right and left, which are not at all scary, despite the fact that they are written as and .
But what is continuity of a function anyway? Until we get to a strict definition, it's easiest to imagine a line that can be drawn without lifting the pencil from the paper. If such a line is drawn, then it is continuous. This line is the graph of a continuous function.
Graphically, a function is continuous at a point if its graph does not “break” at this point. The graph of such a continuous function is 
shown in the figure below.
Determination of continuity of a function through a limit. A function is continuous at a point if three conditions are met:
1. The function is defined at point .
If at least one of the listed conditions is not met, the function is not continuous at the point. In this case, they say that the function suffers a discontinuity, and the points on the graph at which the graph is interrupted are called discontinuity points of the function. The graph of such a function that suffers a discontinuity at the point x=2 is in the figure below.
Example 1. Function f(x) is defined as follows:
Will this function be continuous at each of the boundary points of its branches, that is, at the points x = 0 , x = 1 , x = 3 ?
Solution. We check all three conditions for the continuity of a function at each boundary point. The first condition is met, since what function defined at each of the boundary points follows from the definition of the function. It remains to check the remaining two conditions.
Dot x= 0 . Let's find the left-hand limit at this point:
.
Let's find the right-hand limit:
x= 0 must be found for that branch of the function that includes this point, that is, the second branch. We find them:
As we can see, the limit of the function and the value of the function at the point x= 0 are equal. Therefore, the function is continuous at the point x = 0 .
Dot x= 1 . Let's find the left-hand limit at this point:
Let's find the right-hand limit:
Limit of a function and value of a function at a point x= 1 must be found for that branch of the function that includes this point, that is, the second branch. We find them:
.
Limit of a function and value of a function at a point x= 1 are equal. Therefore, the function is continuous at the point x = 1 .
Dot x= 3 . Let's find the left-hand limit at this point:
Let's find the right-hand limit:
Limit of a function and value of a function at a point x= 3 must be found for that branch of the function that includes this point, that is, the second branch. We find them:
.
Limit of a function and value of a function at a point x= 3 are equal. Therefore, the function is continuous at the point x = 3 .
The main conclusion: this function is continuous at each boundary point.
What is continuous change of function?
A continuous change in a function can be defined as a gradual change, without jumps, in which a small change in the argument entails a small change in the function.
Let us illustrate this continuous change in function with an example.
Let a weight hang on a thread above the table. Under the influence of this load, the thread stretches, so the distance l load from the point of suspension of the thread is a function of the mass of the load m, that is l = f(m) , m≥0 .
If you slightly change the mass of the load, then the distance l will change little: small changes m small changes correspond l. However, if the mass of the load is close to the tensile strength of the thread, then a slight increase in the mass of the load can cause the thread to break: distance l will increase abruptly and become equal to the distance from the suspension point to the table surface. Graph of a function l = f(m) shown in the figure. At a section, this graph is a continuous (solid) line, and at a point it is interrupted. The result is a graph consisting of two branches. At all points except , the function l = f(m) is continuous, but at a point it has a discontinuity.
Studying a function for continuity can be either an independent task or one of the stages of a complete study of the function and constructing its graph.
Continuity of a function on an interval
Let the function y = f(x) defined in the interval ] a, b[ and is continuous at every point of this interval. Then it is called continuous in the interval ] a, b[ . The concept of continuity of a function on intervals of the form ]- ∞ is defined similarly, b[ , ]a, + ∞[ , ]- ∞, + ∞[ . Let now the function y = f(x) defined on the interval [ a, b] . The difference between an interval and a segment: the boundary points of an interval are not included in the interval, but the boundary points of a segment are included in the segment. Here we should mention the so-called one-sided continuity: at the point a, remaining on the segment [ a, b] , we can only approach from the right, and to the point b- only on the left. The function is said to be continuous on the interval [ a, b] , if it is continuous at all interior points of this segment, continuous on the right at the point a and is left continuous at the point b.
An example of a continuous function can be any of the elementary functions. Each elementary function is continuous on any interval on which it is defined. For example, the functions and are continuous on any interval [ a, b], the function is continuous on the interval [ 0 , b] , the function is continuous on any segment not containing a point a = 2 .
Example 4. Examine the function for continuity.
Solution. Let's check the first condition. The function is not defined at points - 3 and 3. At least one of the conditions for the continuity of the function along the entire number line is not satisfied. Therefore, this function is continuous on the intervals
.Example 5. Determine at what value of the parameter a continuous throughout domain of definition function 
Solution.
Let's find the right-hand limit at:
.
Obviously, the value at the point x= 2 should be equal ax :
a = 1,5 .
Example 6. Determine at what parameter values a And b continuous throughout domain of definition function 
Solution.
Let's find the left-sided limit of the function at the point:
.
Therefore, the value at the point must be 1:
Let's find the left-hand function at the point:
Obviously, the value of the function at a point should be equal to:
Answer: the function is continuous over the entire domain of definition when a = 1; b = -3 .
Basic properties of continuous functions
Mathematics came to the concept of a continuous function by studying, first of all, various laws of motion. Space and time are infinite, and dependence, for example, paths s from time t, expressed by law s = f(t) , gives an example of a continuous functions f(t) . The temperature of the heated water also changes continuously; it is also a continuous function of time: T = f(t) .
In mathematical analysis, some properties that continuous functions have are proven. Let us present the most important of these properties.
1. If a function continuous on an interval takes values of different signs at the ends of the interval, then at some point of this interval it takes a value equal to zero. In a more formal statement, this property is given in a theorem known as the first Bolzano-Cauchy theorem.
2. Function f(x) , continuous on the interval [ a, b] , takes all intermediate values between the values at the end points, that is, between f(a) And f(b) . In a more formal statement, this property is given in a theorem known as the second Bolzano-Cauchy theorem.
