Among limit examples features are common functions with roots, which is not always clear how to disclose. It's easier when there is an example of a border with a root function of the form

The solution of such limits is simple and clear to everyone.
Difficulties arise if there are the following examples of functions with roots.

Example 1 . Calculate function limit

With a direct substitution of the point x = 1, it is clear that both the numerator and denominator of the function

turn to zero, that is, we have an uncertainty of the form 0/0 .
To reveal the uncertainty, one should multiply the expression containing the root by its conjugate and apply the difference of squares rule. For given example transformations will be as follows



Limit of a function with roots is 6 . Without the above rule, it would be difficult to find it.
Consider similar examples boundary calculations with the given rule

Example 2 Find the limit of a function

We make sure that when substituting x = 3, we get an uncertainty of the form 0/0.
It is revealed by multiplying the numerator and denominator by the conjugate to the numerator.


Next, we decompose the numerator according to the rule of the difference of squares

That's just how we found the limit of a function with roots.

Example 3 Define function limit

We see that we have an uncertainty of the form 0/0.
Getting rid of irrationality in the denominator

The function limit is 8 .

Now consider another type of examples, when the variable in the redistribution tends to infinity.

Example 4 . Calculate function limit

Many of you don't know how to find the limit of a function. The calculation technique will be disclosed below.
We have a limit of type infinity minus infinity. Multiply and divide by the conjugate factor and use the difference of squares rule

The function bounds is -2.5 .

The calculation of such limits is actually reduced to the disclosure of irrationality, and then the substitution of a variable

Example 5 Find the limit of a function

The limit is equivalent - infinity minus infinity
.
Multiply and divide by the adjoint expression and simplify

This mathematical calculator online will help you if needed calculate function limit. Program limit solutions not only gives the answer to the problem, it leads detailed solution with explanations, i.e. displays the progress of the limit calculation.

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Enter a function expression

Calculate Limit

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A bit of theory.

The limit of the function at x-> x 0

Let the function f(x) be defined on some set X and let the point \(x_0 \in X \) or \(x_0 \notin X \)

Take from X a sequence of points other than x 0:
x 1 , x 2 , x 3 , ..., x n , ... (1)
converging to x*. The function values ​​at the points of this sequence also form a numerical sequence
f(x 1), f(x 2), f(x 3), ..., f(x n), ... (2)
and one can pose the question of the existence of its limit.

Definition. The number A is called the limit of the function f (x) at the point x \u003d x 0 (or at x -> x 0), if for any sequence (1) of values ​​\u200b\u200bof the argument x that converges to x 0, different from x 0, the corresponding sequence (2) of values function converges to the number A.

$$ \lim_(x\to x_0)( f(x)) = A $$

The function f(x) can have only one limit at the point x 0. This follows from the fact that the sequence
(f(x n)) has only one limit.

There is another definition of the limit of a function.

Definition The number A is called the limit of the function f(x) at the point x = x 0 if for any number \(\varepsilon > 0 \) there exists a number \(\delta > 0 \) such that for all \(x \in X, \; x \neq x_0 \) satisfying the inequality \(|x-x_0| Using logical symbols, this definition can be written as
\((\forall \varepsilon > 0) (\exists \delta > 0) (\forall x \in X, \; x \neq x_0, \; |x-x_0| Note that the inequalities \(x \neq x_0 , \; |x-x_0| The first definition is based on the notion of a limit number sequence, which is why it is often referred to as the "sequence language" definition. The second definition is called the "language \(\varepsilon - \delta \)" definition.
These two definitions of the limit of a function are equivalent, and you can use either of them, whichever is more convenient for solving a particular problem.

Note that the definition of the limit of a function "in the language of sequences" is also called the definition of the limit of a function according to Heine, and the definition of the limit of a function "in the language \(\varepsilon - \delta \)" is also called the definition of the limit of a function according to Cauchy.

Function limit at x->x 0 - and at x->x 0 +

In what follows, we will use the concepts of one-sided limits of a function, which are defined as follows.

Definition The number A is called the right (left) limit of the function f (x) at the point x 0 if for any sequence (1) converging to x 0, whose elements x n are greater (less) than x 0 , the corresponding sequence (2) converges to A.

Symbolically it is written like this:
$$ \lim_(x \to x_0+) f(x) = A \; \left(\lim_(x \to x_0-) f(x) = A \right) $$

One can give an equivalent definition of one-sided limits of a function "in the language \(\varepsilon - \delta \)":

Definition the number A is called the right (left) limit of the function f(x) at the point x 0 if for any \(\varepsilon > 0 \) there exists \(\delta > 0 \) such that for all x satisfying the inequalities \(x_0 Symbolic entries:

\((\forall \varepsilon > 0) (\exists \delta > 0) (\forall x, \; x_0

\begin(equation) a^4-b^4=(a-b)\cdot(a^3+a^2 b+ab^2+b^3)\end(equation)

Example #4

Find $\lim_(x\to 4)\frac(\sqrt(5x-12)-\sqrt(x+4))(16-x^2)$.

Since $\lim_(x\to 4)\left(\sqrt(5x-12)-\sqrt(x+4)\right)=0$ and $\lim_(x\to 4)(16-x^ 2)=0$, then we are dealing with an uncertainty of the form $\frac(0)(0)$. To get rid of the irrationality that caused this uncertainty, you need to multiply the numerator and denominator by the expression conjugate to the numerator. won't help here anymore, because multiplication by $\sqrt(5x-12)+\sqrt(x+4)$ will lead to the following result:

$$ \left(\sqrt(5x-12)-\sqrt(x+4)\right)\left(\sqrt(5x-12)+\sqrt(x+4)\right)=\sqrt((5x -12)^2)-\sqrt((x+4)^2) $$

As you can see, such a multiplication will not save us from the difference of the roots, which causes the indeterminacy of $\frac(0)(0)$. We need to multiply by another expression. This expression must be such that after multiplication by it the difference of cube roots disappears. And the cube root can only "remove" the third degree, so you need to use . Substituting in right side this formula $a=\sqrt(5x-12)$, $b=\sqrt(x+4)$, we get:

$$ \left(\sqrt(5x-12)- \sqrt(x+4)\right)\left(\sqrt((5x-12)^2)+\sqrt(5x-12)\cdot \sqrt( x+4)+\sqrt((x+4)^2) \right)=\\ =\sqrt((5x-12)^3)-\sqrt((x+4)^3)=5x-12 -(x+4)=4x-16. $$

So, after multiplying by $\sqrt((5x-12)^2)+\sqrt(5x-12)\cdot \sqrt(x+4)+\sqrt((x+4)^2)$, the difference of cube roots disappeared. It is the expression $\sqrt((5x-12)^2)+\sqrt(5x-12)\cdot \sqrt(x+4)+\sqrt((x+4)^2)$ that will be conjugate to the expression $\ sqrt(5x-12)-\sqrt(x+4)$. Let's return to our limit and multiply the numerator and denominator by the expression conjugate to the numerator $\sqrt(5x-12)-\sqrt(x+4)$:

$$ \lim_(x\to 4)\frac(\sqrt(5x-12)-\sqrt(x+4))(16-x^2)=\left|\frac(0)(0)\right |=\\ =\lim_(x\to 4)\frac(\left(\sqrt(5x-12)- \sqrt(x+4)\right)\left(\sqrt((5x-12)^2 )+\sqrt(5x-12)\cdot \sqrt(x+4)+\sqrt((x+4)^2) \right))((16-x^2)\left(\sqrt((5x -12)^2)+\sqrt(5x-12)\cdot \sqrt(x+4)+\sqrt((x+4)^2) \right))=\\ =\lim_(x\to 4 )\frac(4x-16)((16-x^2)\left(\sqrt((5x-12)^2)+\sqrt(5x-12)\cdot \sqrt(x+4)+\sqrt ((x+4)^2) \right)) $$

The task is practically solved. It remains only to take into account that $16-x^2=-(x^2-16)=-(x-4)(x+4)$ (see ). In addition, $4x-16=4(x-4)$, so we rewrite the last limit in this form:

$$ \lim_(x\to 4)\frac(4x-16)((16-x^2)\left(\sqrt((5x-12)^2)+\sqrt(5x-12)\cdot \ sqrt(x+4)+\sqrt((x+4)^2) \right))=\\ =\lim_(x\to 4)\frac(4(x-4))(-(x-4 )(x+4)\left(\sqrt((5x-12)^2)+\sqrt(5x-12)\cdot \sqrt(x+4)+\sqrt((x+4)^2) \ right))=\\ =-4\cdot\lim_(x\to 4)\frac(1)((x+4)\left(\sqrt((5x-12)^2)+\sqrt(5x- 12)\cdot \sqrt(x+4)+\sqrt((x+4)^2) \right))=\\ =-4\cdot\frac(1)((4+4)\left(\ sqrt((5\cdot4-12)^2)+\sqrt(5\cdot4-12)\cdot \sqrt(4+4)+\sqrt((4+4)^2) \right))=-\ frac(1)(24). $$

Answer: $\lim_(x\to 4)\frac(\sqrt(5x-12)-\sqrt(x+4))(16-x^2)=-\frac(1)(24)$.

Consider one more example (example No. 5) in this part, where we apply . Fundamentally, the solution scheme is no different from the previous examples, except that the conjugate expression will have a different structure. By the way, it is worth noting that in typical calculations and tests, there are often tasks when, for example, expressions with cube root, and in the denominator - with a square root. In this case, you have to multiply both the numerator and denominator by various conjugate expressions. For example, when calculating the limit $\lim_(x\to 8)\frac(\sqrt(x)-2)(\sqrt(x+1)-3)$ containing an uncertainty of the form $\frac(0)(0 )$, the multiplication will look like:

$$ \lim_(x\to 8)\frac(\sqrt(x)-2)(\sqrt(x+1)-3)=\left|\frac(0)(0)\right|= \lim_ (x\to 8)\frac(\left(\sqrt(x)-2\right)\cdot \left(\sqrt(x^2)+2\sqrt(x)+4\right)\cdot\left (\sqrt(x+1)+3\right))(\left(\sqrt(x+1)-3\right)\cdot\left(\sqrt(x+1)+3\right)\cdot\ left(\sqrt(x^2)+2\sqrt(x)+4\right))=\\= \lim_(x\to 8)\frac((x-8)\cdot\left(\sqrt( x+1)+3\right))(\left(x-8\right)\cdot\left(\sqrt(x^2)+2\sqrt(x)+4\right))= \lim_(x \to 8)\frac(\sqrt(x+1)+3)(\sqrt(x^2)+2\sqrt(x)+4)=\frac(3+3)(4+4+4) =\frac(1)(2). $$

All the transformations applied above have already been considered earlier, so I suppose there are no particular ambiguities here. However, if the solution of your similar example raises questions, please unsubscribe about it on the forum.

Example #5

Find $\lim_(x\to 2)\frac(\sqrt(5x+6)-2)(x^3-8)$.

Since $\lim_(x\to 2)(\sqrt(5x+6)-2)=0$ and $\lim_(x\to 2)(x^3-8)=0$, we have with uncertainty $\frac(0)(0)$. To reveal this uncertainty, we use . The conjugate expression for the numerator has the form

$$\sqrt((5x+6)^3)+\sqrt((5x+6)^2)\cdot 2+\sqrt(5x+6)\cdot 2^2+2^3=\sqrt(( 5x+6)^3)+2\cdot\sqrt((5x+6)^2)+4\cdot\sqrt(5x+6)+8.$$

Multiplying the numerator and denominator of the fraction $\frac(\sqrt(5x+6)-2)(x^3-8)$ by the above conjugate expression, we get:

$$\lim_(x\to 2)\frac(\sqrt(5x+6)-2)(x^3-8)=\left|\frac(0)(0)\right|=\\ =\ lim_(x\to 2)\frac(\left(\sqrt(5x+6)-2\right)\cdot \left(\sqrt((5x+6)^3)+2\cdot\sqrt((5x +6)^2)+4\cdot\sqrt(5x+6)+8\right))((x^3-8)\cdot\left(\sqrt((5x+6)^3)+2\ cdot\sqrt((5x+6)^2)+4\cdot\sqrt(5x+6)+8\right))=\\ =\lim_(x\to 2)\frac(5x+6-16) ((x^3-8)\cdot\left(\sqrt((5x+6)^3)+2\cdot\sqrt((5x+6)^2)+4\cdot\sqrt(5x+6) +8\right))=\\ =\lim_(x\to 2)\frac(5x-10)((x^3-8)\cdot\left(\sqrt((5x+6)^3)+ 2\cdot\sqrt((5x+6)^2)+4\cdot\sqrt(5x+6)+8\right)) $$

Since $5x-10=5\cdot(x-2)$ and $x^3-8=x^3-2^3=(x-2)(x^2+2x+4)$ (see ), then:

$$ \lim_(x\to 2)\frac(5x-10)((x^3-8)\cdot\left(\sqrt((5x+6)^3)+2\cdot\sqrt((5x +6)^2)+4\cdot\sqrt(5x+6)+8\right))=\\ =\lim_(x\to 2)\frac(5(x-2))((x-2 )(x^2+2x+4)\cdot\left(\sqrt((5x+6)^3)+2\cdot\sqrt((5x+6)^2)+4\cdot\sqrt(5x+ 6)+8\right))=\\ \lim_(x\to 2)\frac(5)((x^2+2x+4)\cdot\left(\sqrt((5x+6)^3) +2\cdot\sqrt((5x+6)^2)+4\cdot\sqrt(5x+6)+8\right))=\\ \frac(5)((2^2+2\cdot 2 +4)\cdot\left(\sqrt((5\cdot 2+6)^3)+2\cdot\sqrt((5\cdot 2+6)^2)+4\cdot\sqrt(5\cdot 2+6)+8\right))=\frac(5)(384). $$

Answer: $\lim_(x\to 2)\frac(\sqrt(5x+6)-2)(x^3-8)=\frac(5)(384)$.

Example #6

Find $\lim_(x\to 2)\frac(\sqrt(3x-5)-1)(\sqrt(3x-5)-1)$.

Since $\lim_(x\to 2)(\sqrt(3x-5)-1)=0$ and $\lim_(x\to 2)(\sqrt(3x-5)-1)=0$, then we are dealing with the uncertainty of $\frac(0)(0)$. In such situations, when the expressions under the roots are the same, you can use the replacement method. It is required to replace the expression under the root (i.e. $3x-5$) by introducing some new variable. However, simply using the new letter won't do anything. Imagine that we simply replaced the expression $3x-5$ with the letter $t$. Then the fraction under the limit becomes: $\frac(\sqrt(t)-1)(\sqrt(t)-1)$. Irrationality has not disappeared anywhere, it has only changed somewhat, which has not made the task any easier.

Here it is appropriate to recall that the root can only remove the degree. But which degree to use? The question is not trivial, because we have two roots. One root of the fifth, and the other - of the third order. The degree must be such that both roots are removed at the same time! We need natural number, which would be simultaneously divisible by $3$ and $5$. Such numbers infinite set, but the smallest of them is the number $15$. He's called least common multiple numbers $3$ and $5$. And the replacement should be like this: $t^(15)=3x-5$. See what such a replacement does to the roots.

The theory of limits is one of the sections mathematical analysis. The question of solving limits is quite extensive, since there are dozens of methods for solving limits various kinds. There are dozens of nuances and tricks that allow you to solve one or another limit. Nevertheless, we will still try to understand the main types of limits that are most often encountered in practice.

Let's start with the very concept of a limit. But first a short history reference. Once upon a time there was a Frenchman Augustin Louis Cauchy in the 19th century, who laid the foundations of mathematical analysis and gave strict definitions, the definition of the limit, in particular. It must be said that this same Cauchy dreamed, dreams and will dream in nightmares of all students of physics and mathematics faculties, as he proved great amount theorems of mathematical analysis, and one theorem is more disgusting than the other. In this regard, we will not consider a strict definition of the limit, but will try to do two things:

1. Understand what a limit is.
2. Learn to solve the main types of limits.

I apologize for some unscientific explanations, it is important that the material is understandable even to a teapot, which, in fact, is the task of the project.

So what is the limit?

And immediately an example of why to shag your grandmother ....

Any limit consists of three parts:

1) The well-known limit icon.
2) Entries under the limit icon, in this case. The entry reads "x tends to unity." Most often - exactly, although instead of "x" in practice there are other variables. AT practical tasks in place of the unit, absolutely any number can be, as well as infinity ().
3) Functions under the limit sign, in this case .

The record itself reads like this: "the limit of the function when x tends to unity."

Let's analyze the following important question What does the expression "X" mean? seeks to unity? And what is “strive” anyway?
The concept of a limit is a concept, so to speak, dynamic. Let's construct a sequence: first , then , , …, , ….
That is, the expression "x seeks to one" should be understood as follows - "x" consistently takes the values which are infinitely close to unity and practically coincide with it.

How to solve the above example? Based on the above, you just need to substitute the unit in the function under the limit sign:

So the first rule is: When given any limit, first just try to plug the number into the function.

We have reviewed simplest limit, but these are also found in practice, and not so rare!

Infinity example:

Understanding what is it? This is the case when it increases indefinitely, that is: first, then, then, then, and so on ad infinitum.

And what happens to the function at this time?
, , , …

So: if , then the function tends to minus infinity:

Roughly speaking, according to our first rule, we substitute infinity into the function instead of "x" and get the answer .

Another example with infinity:

Again, we start increasing to infinity, and look at the behavior of the function:

Conclusion: for , the function increases indefinitely:

And another series of examples:

Please try to mentally analyze the following for yourself and remember the simplest types of limits:

, , , , , , , , ,
If there is any doubt somewhere, you can pick up a calculator and practice a little.
In the event that , try to build the sequence , , . If , then , , .

Note: strictly speaking, this approach with building sequences of several numbers is incorrect, but it is quite suitable for understanding the simplest examples.

Also pay attention to the following thing. Even if a limit is given a large number at the top, even with a million: then it doesn’t matter , because sooner or later "x" will take on such gigantic values ​​that a million compared to them will be a real microbe.

What should be remembered and understood from the above?

1) When given any limit, first we simply try to substitute a number into the function.

2) You must understand and immediately solve the simplest limits, such as , , etc.

Now we will consider the group of limits, when , and the function is a fraction, in the numerator and denominator of which are polynomials

Example:

Calculate Limit

According to our rule, we will try to substitute infinity into a function. What do we get at the top? Infinity. And what happens below? Also infinity. Thus, we have the so-called indeterminacy of the form. One might think that , and the answer is ready, but in general case this is not the case at all, and some solution must be applied, which we will now consider.

How to tackle limits of this type?

First we look at the numerator and find the highest power:

The highest power in the numerator is two.

Now we look at the denominator and also find the highest degree:

The highest power of the denominator is two.

We then choose the highest power of the numerator and denominator: in this example they coincide and are equal to two.

So, the solution method is as follows: in order to reveal the uncertainty, it is necessary to divide the numerator and denominator by to the highest degree.

Here it is, the answer, and not infinity at all.

What is essential in making a decision?

First, we indicate the uncertainty, if any.

Secondly, it is desirable to interrupt the solution for intermediate explanations. I usually use the sign, it does not carry any mathematical meaning, but means that the solution is interrupted for an intermediate explanation.

Thirdly, in the limit it is desirable to mark what and where it tends. When the work is drawn up by hand, it is more convenient to do it like this:

For notes, it is better to use a simple pencil.

Of course, you can do nothing of this, but then, perhaps, the teacher will note the shortcomings in the solution or start asking additional questions on assignment. And do you need it?

Example 2

Find the limit
Again in the numerator and denominator we find in the highest degree:

Maximum degree in the numerator: 3
Maximum degree in the denominator: 4
Choose greatest value, in this case four.
According to our algorithm, to reveal the uncertainty, we divide the numerator and denominator by .
A complete assignment might look like this:

Divide the numerator and denominator by

Example 3

Find the limit
The maximum degree of "x" in the numerator: 2
The maximum power of "x" in the denominator: 1 (can be written as)
To reveal the uncertainty, it is necessary to divide the numerator and denominator by . A clean solution might look like this:

Divide the numerator and denominator by

The record does not mean division by zero (it is impossible to divide by zero), but division by an infinitely small number.

Thus, when disclosing the indeterminacy of the form, we can get finite number , zero or infinity.

Limits with type uncertainty and a method for their solution

Next group limits is somewhat similar to the limits just considered: there are polynomials in the numerator and denominator, but “x” no longer tends to infinity, but to final number.

Example 4

Solve the limit
First, let's try to substitute -1 in a fraction:

In this case, the so-called uncertainty is obtained.

General rule : if there are polynomials in the numerator and denominator, and there is an uncertainty of the form , then for its disclosure factorize the numerator and denominator.

To do this, most often you need to solve a quadratic equation and (or) use abbreviated multiplication formulas. If these things are forgotten, then visit the page Mathematical formulas and tables and check out methodological material Hot Formulas school course mathematics. By the way, it is best to print it out, it is required very often, and information from paper is absorbed better.

So let's solve our limit

Factoring the numerator and denominator

In order to factorize the numerator, you need to solve the quadratic equation:

First we find the discriminant:

And the square root of it: .

If the discriminant is large, for example 361, we use a calculator, the extraction function square root is on the simplest calculator.

! If the root is not completely extracted (it turns out fractional number with a semicolon), it is very likely that the discriminant is calculated incorrectly or there is a typo in the task.

Next, we find the roots:

Thus:

Everything. The numerator is factored.

Denominator. The denominator is already the simplest factor, and there is no way to simplify it.

Obviously, it can be shortened to:

Now we substitute -1 in the expression that remains under the limit sign:

Naturally, in control work, at the test, exam, the decision is never painted in such detail. In the final version, the design should look something like this:

Let's factorize the numerator.

Example 5

Calculate Limit

First, a "clean" solution

Let's factorize the numerator and denominator.

Numerator:
Denominator:



,

What is important in this example?
First, you must understand well how the numerator is revealed, first we bracketed 2, and then used the difference of squares formula. This is the formula you need to know and see.

The theory of limits is one of the branches of mathematical analysis. The question of solving limits is quite extensive, since there are dozens of methods for solving limits of various types. There are dozens of nuances and tricks that allow you to solve one or another limit. Nevertheless, we will still try to understand the main types of limits that are most often encountered in practice.

Let's start with the very concept of a limit. But first, a brief historical background. Once upon a time there was a Frenchman Augustin Louis Cauchy in the 19th century, who gave strict definitions to many concepts of matan and laid its foundations. I must say that this respected mathematician dreamed, dreams and will dream in nightmares of all students of physics and mathematics faculties, as he proved a huge number of theorems of mathematical analysis, and one theorem is more killer than the other. For this reason, we will not consider determination of the Cauchy limit, but let's try to do two things:

1. Understand what a limit is.
2. Learn to solve the main types of limits.

I apologize for some unscientific explanations, it is important that the material is understandable even to a teapot, which, in fact, is the task of the project.

So what is the limit?

And immediately an example of why to shag your grandmother ....

Any limit consists of three parts:

1) The well-known limit icon.
2) Entries under the limit icon, in this case . The entry reads "x tends to unity." Most often - exactly, although instead of "x" in practice there are other variables. In practical tasks, in place of a unit, there can be absolutely any number, as well as infinity ().
3) Functions under the limit sign, in this case .

The record itself reads like this: "the limit of the function when x tends to unity."

Let's analyze the next important question - what does the expression "x seeks to unity? And what is “strive” anyway?
The concept of a limit is a concept, so to speak, dynamic. Let's construct a sequence: first , then , , …, , ….
That is, the expression "x seeks to one" should be understood as follows - "x" consistently takes the values which are infinitely close to unity and practically coincide with it.

How to solve the above example? Based on the above, you just need to substitute the unit in the function under the limit sign:

So the first rule is: When given any limit, first just try to plug the number into the function.

We considered the simplest limit, but such ones are also found in practice, and not so rarely!

Infinity example:

Understanding what is it? This is the case when it increases indefinitely, that is: first, then, then, then, and so on ad infinitum.

And what happens to the function at this time?
, , , …

So: if , then the function tends to minus infinity:

Roughly speaking, according to our first rule, we substitute infinity into the function instead of "x" and get the answer .

Another example with infinity:

Again, we start increasing to infinity and look at the behavior of the function:

Conclusion: for , the function increases indefinitely:

And another series of examples:

Please try to mentally analyze the following for yourself and remember the simplest types of limits:

, , , , , , , , ,
If there is any doubt somewhere, you can pick up a calculator and practice a little.
In the event that , try to build the sequence , , . If , then , , .

! Note: strictly speaking, such an approach with the construction of sequences of several numbers is incorrect, but it is quite suitable for understanding the simplest examples.

Also pay attention to the following thing. Even if a limit is given with a large number at the top, or at least with a million: , then all the same , because sooner or later "x" will begin to take on such gigantic values ​​that a million compared to them will be a real microbe.

What should be remembered and understood from the above?

1) When given any limit, first we simply try to substitute a number into the function.

2) You must understand and immediately solve the simplest limits, such as , , etc.

Moreover, the limit has a very good geometric meaning. For better understanding topics I recommend to read the methodological material Graphs and properties of elementary functions. After reading this article, you will not only finally understand what a limit is, but also get acquainted with interesting cases when the limit of the function is at all does not exist!

In practice, unfortunately, there are few gifts. And so we turn to the consideration of more complex limits. By the way, on this topic there is intensive course in pdf format, which is especially useful if you have VERY little time to prepare. But the materials of the site, of course, are no worse:

Now we will consider the group of limits, when , and the function is a fraction, in the numerator and denominator of which are polynomials

Example:

Calculate Limit

According to our rule, we will try to substitute infinity into a function. What do we get at the top? Infinity. And what happens below? Also infinity. Thus, we have the so-called indeterminacy of the form. One might think that , and the answer is ready, but in the general case this is not the case at all, and some solution must be applied, which we will now consider.

How to solve the limits of this type?

First we look at the numerator and find the highest power:

The highest power in the numerator is two.

Now we look at the denominator and also find the highest degree:

The highest power of the denominator is two.

Then we choose the highest power of the numerator and denominator: in this example, they are the same and equal to two.

So, the solution method is as follows: in order to reveal the uncertainty, it is necessary to divide the numerator and denominator by to the highest degree.

Here it is, the answer, and not infinity at all.

What is essential in making a decision?

First, we indicate the uncertainty, if any.

Secondly, it is desirable to interrupt the solution for intermediate explanations. I usually use the sign, it does not carry any mathematical meaning, but means that the solution is interrupted for an intermediate explanation.

Thirdly, in the limit it is desirable to mark what and where it tends. When the work is drawn up by hand, it is more convenient to do it like this:

For notes, it is better to use a simple pencil.

Of course, you can do nothing of this, but then, perhaps, the teacher will note the shortcomings in the solution or start asking additional questions on the assignment. And do you need it?

Example 2

Find the limit
Again in the numerator and denominator we find in the highest degree:

Maximum degree in the numerator: 3
Maximum degree in the denominator: 4
Choose greatest value, in this case four.
According to our algorithm, to reveal the uncertainty, we divide the numerator and denominator by .
A complete assignment might look like this:

Divide the numerator and denominator by

Example 3

Find the limit
The maximum degree of "x" in the numerator: 2
The maximum power of "x" in the denominator: 1 (can be written as)
To reveal the uncertainty, it is necessary to divide the numerator and denominator by . A clean solution might look like this:

Divide the numerator and denominator by

The record does not mean division by zero (it is impossible to divide by zero), but division by an infinitely small number.

Thus, when disclosing the indeterminacy of the form, we can get finite number, zero or infinity.

Limits with type uncertainty and a method for their solution

The next group of limits is somewhat similar to the limits just considered: there are polynomials in the numerator and denominator, but “x” no longer tends to infinity, but to final number.

Example 4

Solve the limit
First, let's try to substitute -1 in a fraction:

In this case, the so-called uncertainty is obtained.

General rule: if there are polynomials in the numerator and denominator, and there is an uncertainty of the form , then for its disclosure factorize the numerator and denominator.

To do this, most often you need to solve a quadratic equation and (or) use abbreviated multiplication formulas. If these things are forgotten, then visit the page Mathematical formulas and tables and get acquainted with the methodological material Hot School Mathematics Formulas. By the way, it is best to print it out, it is required very often, and information from paper is absorbed better.

So let's solve our limit

Factoring the numerator and denominator

In order to factorize the numerator, you need to solve the quadratic equation:

First we find the discriminant:

And the square root of it: .

If the discriminant is large, for example 361, we use a calculator, the square root function is on the simplest calculator.

! If the root is not extracted completely (a fractional number with a comma is obtained), it is very likely that the discriminant was calculated incorrectly or there is a typo in the task.

Next, we find the roots:

Thus:

Everything. The numerator is factored.

Denominator. The denominator is already the simplest factor, and there is no way to simplify it.

Obviously, it can be shortened to:

Now we substitute -1 in the expression that remains under the limit sign:

Naturally, in a test, on a test, an exam, the solution is never painted in such detail. In the final version, the design should look something like this:

Let's factorize the numerator.

Example 5

Calculate Limit

First, a "clean" solution

Let's factorize the numerator and denominator.

Numerator:
Denominator:



,

What is important in this example?
First, you must understand well how the numerator is revealed, first we bracketed 2, and then used the difference of squares formula. This is the formula you need to know and see.

Recommendation: If in the limit (of almost any type) it is possible to take a number out of the bracket, then we always do this.
Moreover, it is advisable to take such numbers beyond the limit sign. What for? Just so they don't get in the way. The main thing is not to lose these numbers in the course of the decision.

Please note that on final stage I took out the solution for the limit icon deuce, and then minus.

! Important
In the course of the solution, a type fragment occurs very often. Reduce this fractionit is forbidden . First you need to change the sign of the numerator or the denominator (put -1 out of brackets).
, that is, a minus sign appears, which is taken into account when calculating the limit and there is no need to lose it at all.

In general, I noticed that most often in finding limits of this type, you have to solve two quadratic equations, that is, both the numerator and the denominator contain square trinomials.

The method of multiplying the numerator and denominator by the adjoint expression

We continue to consider the uncertainty of the form

The next type of limits is similar to the previous type. The only thing, in addition to polynomials, we will add roots.

Example 6

Find the limit

We start to decide.

First, we try to substitute 3 in the expression under the limit sign
Once again I repeat - this is the first thing to do for ANY limit. This action usually carried out mentally or on a draft.

An uncertainty of the form , which needs to be eliminated, is obtained.

As you probably noticed, we have the difference of the roots in the numerator. And it is customary to get rid of the roots in mathematics, if possible. What for? And life is easier without them.