Usually the second remarkable limit is written in this form:

\begin(equation) \lim_(x\to\infty)\left(1+\frac(1)(x)\right)^x=e\end(equation)

The number $e$ indicated on the right side of equality (1) is irrational. The approximate value of this number is: $e\approx(2(,)718281828459045)$. If we make the replacement $t=\frac(1)(x)$, then formula (1) can be rewritten as follows:

\begin(equation) \lim_(t\to(0))\biggl(1+t\biggr)^(\frac(1)(t))=e\end(equation)

As for the first remarkable limit, it does not matter which expression stands in place of the variable $x$ in formula (1) or instead of the variable $t$ in formula (2). The main thing is to fulfill two conditions:

1. The base of the degree (i.e., the expression in brackets of formulas (1) and (2)) should tend to unity;
2. The exponent (i.e. $x$ in formula (1) or $\frac(1)(t)$ in formula (2)) must tend to infinity.

The second remarkable limit is said to reveal the uncertainty of $1^\infty$. Please note that in formula (1) we do not specify which infinity ($+\infty$ or $-\infty$) we are talking about. In any of these cases, formula (1) is correct. In formula (2), the variable $t$ can tend to zero both on the left and on the right.

I note that there are also several useful consequences from the second remarkable limit. Examples of the use of the second remarkable limit, as well as its consequences, are very popular among compilers of standard standard calculations and tests.

Example No. 1

Calculate the limit $\lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7)$.

Let us immediately note that the base of the degree (i.e. $\frac(3x+1)(3x-5)$) tends to unity:

$$ \lim_(x\to\infty)\frac(3x+1)(3x-5)=\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(3+\frac(1)(x))(3-\frac(5)(x)) =\frac(3+0)(3-0) = 1. $$

In this case, the exponent (expression $4x+7$) tends to infinity, i.e. $\lim_(x\to\infty)(4x+7)=\infty$.

The base of the degree tends to unity, the exponent tends to infinity, i.e. we are dealing with uncertainty $1^\infty$. Let's apply a formula to reveal this uncertainty. At the base of the power of the formula is the expression $1+\frac(1)(x)$, and in the example we are considering, the base of the power is: $\frac(3x+1)(3x-5)$. Therefore, the first action will be a formal adjustment of the expression $\frac(3x+1)(3x-5)$ to the form $1+\frac(1)(x)$. First, add and subtract one:

$$ \lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(3x+1)(3x-5)-1\right)^(4x+7) $$

Please note that you cannot simply add a unit. If we are forced to add one, then we also need to subtract it so as not to change the value of the entire expression. To continue the solution, we take into account that

$$ \frac(3x+1)(3x-5)-1 =\frac(3x+1)(3x-5)-\frac(3x-5)(3x-5) =\frac(3x+1- 3x+5)(3x-5) =\frac(6)(3x-5). $$

Since $\frac(3x+1)(3x-5)-1=\frac(6)(3x-5)$, then:

$$ \lim_(x\to\infty)\left(1+ \frac(3x+1)(3x-5)-1\right)^(4x+7) =\lim_(x\to\infty)\ left(1+\frac(6)(3x-5)\right)^(4x+7) $$

Let's continue the adjustment. In the expression $1+\frac(1)(x)$ of the formula, the numerator of the fraction is 1, and in our expression $1+\frac(6)(3x-5)$ the numerator is $6$. To get $1$ in the numerator, drop $6$ into the denominator using the following conversion:

$$ 1+\frac(6)(3x-5) =1+\frac(1)(\frac(3x-5)(6)) $$

Thus,

$$ \lim_(x\to\infty)\left(1+\frac(6)(3x-5)\right)^(4x+7) =\lim_(x\to\infty)\left(1+ \frac(1)(\frac(3x-5)(6))\right)^(4x+7) $$

So, the basis of the degree, i.e. $1+\frac(1)(\frac(3x-5)(6))$, adjusted to the form $1+\frac(1)(x)$ required in the formula. Now let's start working with the exponent. Note that in the formula the expressions in the exponents and in the denominator are the same:

This means that in our example, the exponent and the denominator must be brought to the same form. To get the expression $\frac(3x-5)(6)$ in the exponent, we simply multiply the exponent by this fraction. Naturally, to compensate for such a multiplication, you will have to immediately multiply by the reciprocal fraction, i.e. by $\frac(6)(3x-5)$. So we have:

$$ \lim_(x\to\infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(4x+7) =\lim_(x\to\ infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\frac(3x-5)(6)\cdot\frac(6)(3x-5 )\cdot(4x+7)) =\lim_(x\to\infty)\left(\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\ frac(3x-5)(6))\right)^(\frac(6\cdot(4x+7))(3x-5)) $$

Let us separately consider the limit of the fraction $\frac(6\cdot(4x+7))(3x-5)$ located in the power:

$$ \lim_(x\to\infty)\frac(6\cdot(4x+7))(3x-5) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(6\cdot\left(4+\frac(7)(x)\right))(3-\frac(5)(x)) =6\cdot\ frac(4)(3) =8. $$

Answer: $\lim_(x\to(0))\biggl(\cos(2x)\biggr)^(\frac(1)(\sin^2(3x)))=e^(-\frac(2) (9))$.

Example No. 4

Find the limit $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)$.

Since for $x>0$ we have $\ln(x+1)-\ln(x)=\ln\left(\frac(x+1)(x)\right)$, then:

$$ \lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right) =\lim_(x\to+\infty)\left(x\cdot\ln\ left(\frac(x+1)(x)\right)\right) $$

Expanding the fraction $\frac(x+1)(x)$ into the sum of fractions $\frac(x+1)(x)=1+\frac(1)(x)$ we get:

$$ \lim_(x\to+\infty)\left(x\cdot\ln\left(\frac(x+1)(x)\right)\right) =\lim_(x\to+\infty)\left (x\cdot\ln\left(1+\frac(1)(x)\right)\right) =\lim_(x\to+\infty)\left(\ln\left(\frac(x+1) (x)\right)^x\right) =\ln(e) =1. $$

Answer: $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)=1$.

Example No. 5

Find the limit $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))$.

Since $\lim_(x\to(2))(3x-5)=6-5=1$ and $\lim_(x\to(2))\frac(2x)(x^2-4)= \infty$, then we are dealing with uncertainty of the form $1^\infty$. Detailed explanations are given in example No. 2, but here we will limit ourselves to a brief solution. Making the replacement $t=x-2$, we get:

$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=x-2 ;\;x=t+2\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\biggl(1+3t\biggr)^(\frac(2t+4)(t^2+4t))=\\ =\lim_(t\to(0) )\biggl(1+3t\biggr)^(\frac(1)(3t)\cdot 3t\cdot\frac(2t+4)(t^2+4t)) =\lim_(t\to(0) )\left(\biggl(1+3t\biggr)^(\frac(1)(3t))\right)^(\frac(6\cdot(t+2))(t+4)) =e^ 3. $$

You can solve this example in a different way, using the replacement: $t=\frac(1)(x-2)$. Of course, the answer will be the same:

$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=\frac( 1)(x-2);\;x=\frac(2t+1)(t)\\&t\to\infty\end(aligned)\right| =\lim_(t\to\infty)\left(1+\frac(3)(t)\right)^(t\cdot\frac(4t+2)(4t+1))=\\ =\lim_ (t\to\infty)\left(1+\frac(1)(\frac(t)(3))\right)^(\frac(t)(3)\cdot\frac(3)(t) \cdot\frac(t\cdot(4t+2))(4t+1)) =\lim_(t\to\infty)\left(\left(1+\frac(1)(\frac(t)( 3))\right)^(\frac(t)(3))\right)^(\frac(6\cdot(2t+1))(4t+1)) =e^3. $$

Answer: $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))=e^3$.

Example No. 6

Find the limit $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) $.

Let's find out what the expression $\frac(2x^2+3)(2x^2-4)$ tends to under the condition $x\to\infty$:

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

Thus, in the given limit we are dealing with an uncertainty of the form $1^\infty$, which we will reveal using the second remarkable limit:

$$ \lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(2x^2+3)(2x^2-4)-1\right)^(3x)=\\ =\lim_(x\to \infty)\left(1+\frac(7)(2x^2-4)\right)^(3x) =\lim_(x\to\infty)\left(1+\frac(1)(\frac (2x^2-4)(7))\right)^(3x)=\\ =\lim_(x\to\infty)\left(1+\frac(1)(\frac(2x^2-4 )(7))\right)^(\frac(2x^2-4)(7)\cdot\frac(7)(2x^2-4)\cdot 3x) =\lim_(x\to\infty) \left(\left(1+\frac(1)(\frac(2x^2-4)(7))\right)^(\frac(2x^2-4)(7))\right)^( \frac(21x)(2x^2-4)) =e^0 =1. $$

Answer: $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x)=1$.

LESSON 20

20.1 DISCLOSURE OF SPECIES UNCERTAINTY

Example 1

Solve limit First, let's try to substitute -1 into the fraction: In this case, the so-called uncertainty is obtained.

General rule: if the numerator and denominator contain polynomials, and there is uncertainty of the form, then to reveal it you need to factor the numerator and denominator.

To do this, most often you need to solve a quadratic equation and/or use abbreviated multiplication formulas.

Let's factorize the numerator.

Example 2

Calculate limit

Let's factor the numerator and denominator.

Numerator denominator: ,

Method of multiplying the numerator and denominator by the conjugate 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 3

Find the limit

Multiply the numerator and denominator by the conjugate expression.

20.2 DISCLOSURE OF SPECIES UNCERTAINTY

Now we will consider the group of limits when , and the function is a fraction whose numerator and denominator contain polynomials

Example 4

Calculate limit

According to our rule, we will try to substitute infinity into the function. What do we get at the top? Infinity. And what happens below? Also infinity. Thus we have what is called species uncertainty. One might think that the answer is ready, but in the general case this is not at all the case, and it is necessary to apply some solution technique, which we will now consider.

How to solve limits of this type?

First we look at the numerator and find the highest power: The leading power in the numerator is two.

Now we look at the denominator and also find it to the highest power: The highest degree 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: to reveal uncertaintyyou need to divide the numerator and denominator byin the senior degree.

Divide the numerator and denominator by

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

What is fundamentally important in the design of a decision?

First, we indicate uncertainty, if any.

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

Thirdly, in the limit it is advisable to mark what is going where. When the work is drawn up by hand, it is more convenient to do it this way: It is better to use a simple pencil for notes.

Of course, you don’t have to do any of this, but then, perhaps, the teacher will point out shortcomings in the solution or start asking additional questions about the assignment. Do you need it?

Example 5

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 Select greatest value, in this case four. According to our algorithm, to reveal uncertainty, we divide the numerator and denominator by. The complete assignment might look like this:

Example 6

Find the limit Maximum degree of “X” in the numerator: 2 Maximum degree of “X” in the denominator: 1 (can be written as) To reveal uncertainty, it is necessary to divide the numerator and denominator by. The final solution might look like this:

Divide the numerator and denominator by

Notation does not mean division by zero (you cannot divide by zero), but division by an infinitesimal number.

Thus, by uncovering species uncertainty, we may be able to final number, zero or infinity.

PRACTICUM 20

TASK N 1

Solution: If instead of the variable we put the value 7 to which it tends, then we get an uncertainty of the form

TASK N 2Topic: Disclosure of uncertainty of the “zero to zero” type

Solution: If instead of a variable we put the value 0 to which it tends, then we get an uncertainty of the form

TASK N 3Topic: Disclosure of uncertainty of the “zero to zero” type

Solution: If instead of the variable we put the value 6 to which it tends, then we get uncertainty of the form

TASK N 4

Solution: Because And

TASK N 5Topic: Disclosure of uncertainty of the form "infinity to infinity"

Solution: Because And then there is uncertainty of the form. To reveal it, you need to divide each term of the numerator and denominator by. Then, knowing what we get:

INDEPENDENT WORK 20

TASK N 1Topic: Disclosure of uncertainty of the “zero to zero” type

TASK N 2Topic: Disclosure of uncertainty of the “zero to zero” type

TASK N 3Topic: Disclosure of uncertainty of the “zero to zero” type

TASK N 4Topic: Disclosure of uncertainty of the form "infinity to infinity"

TASK N 5Topic: Disclosure of uncertainty of the form "infinity to infinity" Function limit equal...

TASK N 6Topic: Disclosure of uncertainty of the form "infinity to infinity"

Limits give all mathematics students a lot of trouble. To solve a limit, sometimes you have to use a lot of tricks and choose from a variety of solution methods exactly the one that is suitable for a particular example.

In this article we will not help you understand the limits of your capabilities or comprehend the limits of control, but we will try to answer the question: how to understand limits in higher mathematics? Understanding comes with experience, so at the same time we will give several detailed examples of solving limits with explanations.

The concept of limit in mathematics

The first question is: what is this limit and the limit of what? We can talk about the limits of numerical sequences and functions. We are interested in the concept of the limit of a function, since this is what students most often encounter. But first, the most general definition of a limit:

Let's say there is some variable value. If this value in the process of change unlimitedly approaches a certain number a , That a – the limit of this value.

For a function defined in a certain interval f(x)=y such a number is called a limit A , which the function tends to when X , tending to a certain point A . Dot A belongs to the interval on which the function is defined.

It sounds cumbersome, but it is written very simply:

Lim- from English limit- limit.

There is also a geometric explanation for determining the limit, but here we will not delve into the theory, since we are more interested in the practical rather than the theoretical side of the issue. When we say that X tends to some value, this means that the variable does not take on the value of a number, but approaches it infinitely close.

Let's give a specific example. The task is to find the limit.

To solve this example, we substitute the value x=3 into a function. We get:

By the way, if you are interested in basic operations on matrices, read a separate article on this topic.

In examples X can tend to any value. It can be any number or infinity. Here's an example when X tends to infinity:

Intuitively, the larger the number in the denominator, the smaller the value the function will take. So, with unlimited growth X meaning 1/x will decrease and approach zero.

As you can see, to solve the limit, you just need to substitute the value to strive for into the function X . However, this is the simplest case. Often finding the limit is not so obvious. Within the limits there are uncertainties of the type 0/0 or infinity/infinity . What to do in such cases? Resort to tricks!

Uncertainties within

Uncertainty of the form infinity/infinity

Let there be a limit:

If we try to substitute infinity into the function, we will get infinity in both the numerator and the denominator. In general, it is worth saying that there is a certain element of art in resolving such uncertainties: you need to notice how you can transform the function in such a way that the uncertainty goes away. In our case, we divide the numerator and denominator by X in the senior degree. What will happen?

From the example already discussed above, we know that terms containing x in the denominator will tend to zero. Then the solution to the limit is:

To resolve type uncertainties infinity/infinity divide the numerator and denominator by X to the highest degree.

By the way! For our readers there is now a 10% discount on any type of work

Another type of uncertainty: 0/0

As always, substituting values ​​into the function x=-1 gives 0 in the numerator and denominator. Look a little more closely and you will notice that we have a quadratic equation in the numerator. Let's find the roots and write:

Let's reduce and get:

So, if you are faced with type uncertainty 0/0 – factor the numerator and denominator.

To make it easier for you to solve examples, we present a table with the limits of some functions:

L'Hopital's rule within

Another powerful way to eliminate both types of uncertainty. What is the essence of the method?

If there is uncertainty in the limit, take the derivative of the numerator and denominator until the uncertainty disappears.

L'Hopital's rule looks like this:

Important point : the limit in which the derivatives of the numerator and denominator stand instead of the numerator and denominator must exist.

And now - a real example:

There is typical uncertainty 0/0 . Let's take the derivatives of the numerator and denominator:

Voila, uncertainty is resolved quickly and elegantly.

We hope that you will be able to usefully apply this information in practice and find the answer to the question “how to solve limits in higher mathematics.” If you need to calculate the limit of a sequence or the limit of a function at a point, and there is absolutely no time for this work, contact a professional student service for a quick and detailed solution.

In the previous article we talked about how to correctly calculate the limits of elementary functions. If we take more complex functions, then we will have expressions with an undefined value in our calculations. They are called uncertainties.

The following main types of uncertainties are distinguished:

1. Divide 0 by 0 0 0 ;
2. Dividing one infinity by another ∞ ∞;

0 raised to the zero power 0 0 ;

3. infinity raised to the zero power ∞ 0 .

We have listed all the main uncertainties. Other expressions may take on finite or infinite values ​​under different conditions and therefore cannot be considered uncertainties.

Uncovering Uncertainties

Uncertainty can be resolved by:

1. By simplifying the type of function (using abbreviated multiplication formulas, trigonometric formulas, additional multiplication by conjugate expressions and subsequent reduction, etc.);

With the help of wonderful limits;

Using L'Hopital's rule;

By replacing one infinitesimal expression with its equivalent expression (as a rule, this action is performed using a table of infinitesimal expressions).

All the information presented above can be clearly presented in the form of a table. On the left side it shows the type of uncertainty, on the right - a suitable method for revealing it (finding the limit). This table is very convenient to use in calculations related to finding limits.

Uncertainty	Uncertainty Disclosure Method
1. Divide 0 by 0	Transformation and subsequent simplification of an expression. If the expression is sin (k x) k x or k x sin (k x) then you need to use the first remarkable limit. If this solution is not suitable, we use L'Hopital's rule or a table of equivalent infinitesimal expressions
2. Dividing infinity by infinity	Transform and simplify an expression or use L'Hopital's rule
3. Multiplying zero by infinity or finding the difference between two infinities	Conversion to 0 0 or ∞ ∞ followed by application of L'Hopital's rule
4. Unit to the power of infinity	Using the Second Great Limit
5. Raising zero or infinity to the zero power	Taking the logarithm of an expression using the equality lim x → x 0 ln (f (x)) = ln lim x → x 0 f (x)

Let's look at a couple of problems. These examples are quite simple: in them the answer is obtained immediately after substituting the values ​​and there is no uncertainty.

Example 1

Calculate the limit lim x → 1 x 3 + 3 x - 1 x 5 + 3 .

Solution

We perform value substitution and get the answer.

lim x → 1 x 3 + 3 x - 1 x 5 + 3 = 1 3 + 3 1 - 1 1 5 + 3 = 3 4 = 3 2

Answer: lim x → 1 x 3 + 3 x - 1 x 5 + 3 = 3 2 .

Example 2

Calculate the limit lim x → 0 (x 2 + 2 , 5) 1 x 2 .

Solution

We have an exponential power function, into the base of which we need to substitute x = 0.

(x 2 + 2, 5) x = 0 = 0 2 + 2, 5 = 2, 5

This means we can transform the limit into the following expression:

lim x → 0 (x 2 + 2, 5) 1 x 2 = lim x → 0 2, 5 1 x 2

Now let's look at the indicator - the power function 1 x 2 = x - 2. Let's look at the table of limits for power functions with an exponent less than zero and get the following: lim x → 0 + 0 1 x 2 = lim x → 0 + 0 x - 2 = + ∞ and lim x → 0 + 0 1 x 2 = lim x → 0 + 0 x - 2 = + ∞

Thus, we can write that lim x → 0 (x 2 + 2, 5) 1 x 2 = lim x → 0 2, 5 1 x 2 = 2, 5 + ∞.

Now we take the table of limits of exponential functions with bases greater than 0, and we get:

lim x → 0 (x 2 + 2, 5) 1 x 2 = lim x → 0 2, 5 1 x 2 = 2, 5 + ∞ = + ∞

Answer: lim x → 0 (x 2 + 2 , 5) 1 x 2 = + ∞ .

Example 3

Calculate the limit lim x → 1 x 2 - 1 x - 1 .

Solution

We perform value substitution.

lim x → 1 x 2 - 1 x - 1 = 1 2 - 1 1 - 1 = 0 0

As a result, we ended up with uncertainty. Use the table above to select a solution method. It indicates that you need to simplify the expression.

lim x → 1 x 2 - 1 x - 1 = 0 0 = lim x → 1 (x - 1) (x + 1) x - 1 = = lim x → 1 (x - 1) (x + 1) · (x + 1) x - 1 = lim x → 1 (x + 1) · x - 1 = = 1 + 1 · 1 - 1 = 2 · 0 = 0

As we can see, simplification has led to the revelation of uncertainty.

Answer: lim x → 1 x 2 - 1 x - 1 = 0

Example 4

Calculate the limit lim x → 3 x - 3 12 - x - 6 + x .

Solution

We substitute the value and get the following entry.

lim x → 3 x - 3 12 - x - 6 + x = 3 - 3 12 - 3 - 6 + 3 = 0 9 - 9 = 0 0

We have come to the need to divide zero by zero, which is uncertainty. Let's look at the required solution method in the table - this is simplification and transformation of the expression. Let us additionally multiply the numerator and denominator by the conjugate expression 12 - x + 6 + x:

lim x → 3 x - 3 12 - x - 6 + x = 0 0 = lim x → 3 x - 3 12 - x + 6 + x 12 - x - 6 + x 12 - x + 6 + x

The denominator is multiplied so that you can then use the abbreviated multiplication formula (difference of squares) to perform the reduction.

lim x → 3 x - 3 12 - x + 6 + x 12 - x - 6 + x 12 - x + 6 + x = lim x → 3 x - 3 12 - x + 6 + x 12 - x 2 - 6 + x 2 = lim x → 3 (x - 3) 12 - x + 6 + x 12 - x - (6 + x) = = lim x → 3 (x - 3) 12 - x + 6 + x 6 - 2 x = lim x → 3 (x - 3) 12 - x + 6 + x - 2 (x - 3) = = lim x → 3 12 - x + 6 + x - 2 = 12 - 3 + 6 + 3 - 2 = 9 + 9 - 2 = - 9 = - 3

As we can see, as a result of these actions we were able to get rid of uncertainty.

Answer: lim x → 3 x - 3 12 - x - 6 + x = - 3 .

It is important to note that the multiplication approach is used very often when solving problems like this, so we advise you to remember exactly how this is done.

Example 5

Calculate the limit lim x → 1 x 2 + 2 x - 3 3 x 2 - 5 x + 2 .

Solution

We perform the substitution.

lim x → 1 x 2 + 2 x - 3 3 x 2 - 5 x + 2 = 1 2 + 2 1 - 3 3 1 2 - 5 1 + 2 = 0 0

As a result, we ended up with uncertainty. The recommended way to solve the problem in this case is to simplify the expression. Since when x is equal to one, the numerator and denominator turn to 0, we can factor them and then reduce them by x - 1, and then the uncertainty will disappear.

We factorize the numerator:

x 2 + 2 x - 3 = 0 D = 2 2 - 4 1 (- 3) = 16 ⇒ x 1 = - 2 - 16 2 = - 3 x 2 = - 2 + 16 2 = 1 ⇒ x 2 + 2 x - 3 = x + 3 x - 1

Now we do the same with the denominator:

3 x 2 - 5 x + 2 = 0 D = - 5 2 - 4 3 2 = 1 ⇒ x 1 = 5 - 1 2 3 = 2 3 x 2 = 5 + 1 2 3 = 1 ⇒ 3 x 2 - 5 x + 3 = 3 x - 2 3 x - 1

We got a limit of the following form:

lim x → 1 x 2 + 2 x - 3 3 x 2 - 5 x + 2 = 0 0 = lim x → 1 x + 3 x - 1 3 x - 2 3 x - 1 = = lim x → 1 x + 3 3 x - 2 3 = 1 + 3 3 1 - 2 3 = 4

As we can see, during the transformation we managed to get rid of uncertainty.

Answer: lim x → 1 x 2 + 2 x - 3 3 x 2 - 5 x + 2 = 4 .

Next we need to consider the cases of limits at infinity from power expressions. If the exponents of these expressions are greater than 0, then the limit at infinity will also be infinite. In this case, the largest degree is of primary importance, and the rest can be ignored.

For example, lim x → ∞ (x 4 + 2 x 3 - 6) = lim x → ∞ x 4 = ∞ or lim x → ∞ x 4 + 4 x 3 + 21 x 2 - 11 5 = lim x → ∞ x 4 5 = ∞.

If under the limit sign we have a fraction with power expressions in the numerator and denominator, then as x → ∞ we have an uncertainty of the form ∞ ∞. To get rid of this uncertainty, we need to divide the numerator and denominator of the fraction by x m a x (m, n). Let us give an example of solving such a problem.

Example 6

Calculate the limit lim x → ∞ x 7 + 2 x 5 - 4 3 x 7 + 12 .

Solution

lim x → ∞ x 7 + 2 x 5 - 4 3 x 7 + 12 = ∞ ∞

The powers of the numerator and denominator are equal to 7. Divide them by x 7 and get:

lim x → ∞ x 7 + 2 x 5 - 4 3 x 7 + 12 = lim x → ∞ x 7 + 2 x 5 - 4 x 7 3 x 7 + 12 x 7 = = lim x → ∞ 1 + 2 x 2 - 4 x 7 3 + 12 x 7 = 1 + 2 ∞ 2 - 4 ∞ 7 3 + 12 ∞ 7 = 1 + 0 - 0 3 + 0 = 1 3

Answer: lim x → ∞ x 7 + 2 x 5 - 4 3 x 7 + 12 = 1 3 .

Example 7

Calculate the limit lim x → ∞ x 8 + 11 3 x 2 + x + 1 .

Solution

lim x → ∞ x 8 + 11 3 x 2 + x + 1 = ∞ ∞

The numerator has a power of 8 3 and the denominator has a power of 2. Let's divide the numerator and denominator by x 8 3:

lim x → ∞ x 8 + 11 3 x 2 + x + 1 = ∞ ∞ = lim x → ∞ x 8 + 11 3 x 8 3 x 2 + x + 1 x 8 3 = = lim x → ∞ 1 + 11 x 8 3 1 x 2 3 + 1 x 5 3 + 1 x 8 3 = 1 + 11 ∞ 3 1 ∞ + 1 ∞ + 1 ∞ = 1 + 0 3 0 + 0 + 0 = 1 0 = ∞

Answer: lim x → ∞ x 8 + 11 3 x 2 + x + 1 = ∞ .

Example 8

Calculate the limit lim x → ∞ x 3 + 2 x 2 - 1 x 10 + 56 x 7 + 12 3 .

Solution

lim x → ∞ x 3 + 2 x 2 - 1 x 10 + 56 x 7 + 12 3 = ∞ ∞

We have a numerator to the power of 3 and a denominator to the power of 10 3 . This means we need to divide the numerator and denominator by x 10 3:

lim x → ∞ x 3 + 2 x 2 - 1 x 10 + 56 x 7 + 12 3 = ∞ ∞ = lim x → ∞ x 3 + 2 x 2 - 1 x 10 3 x 10 + 56 x 7 + 12 3 x 10 3 = = lim x → ∞ 1 x 1 3 + 2 x 4 3 - 1 x 10 3 1 + 56 x 3 + 12 x 10 3 = 1 ∞ + 2 ∞ - 1 ∞ 1 + 56 ∞ + 12 ∞ 3 = 0 + 0 - 0 1 + 0 + 0 3 = 0

Answer: lim x → ∞ x 3 + 2 x 2 - 1 x 10 + 56 x 7 + 12 3 = 0 .

conclusions

In the case of a ratio limit, there are three main options:

If the degree of the numerator is equal to the degree of the denominator, then the limit will be equal to the ratio of the coefficients of the higher powers.

If the degree of the numerator is greater than the degree of the denominator, then the limit will be equal to infinity.

If the degree of the numerator is less than the degree of the denominator, then the limit will be zero.

We will discuss other methods for disclosing uncertainties in separate articles.

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Type and species uncertainty are the most common uncertainties that need to be disclosed when solving limits.

Most of the limit problems encountered by students contain just such uncertainties. To reveal them or, more precisely, to avoid uncertainties, there are several artificial techniques for transforming the type of expression under the limit sign. These techniques are as follows: term-by-term division of the numerator and denominator by the highest power of the variable, multiplication by the conjugate expression and factorization for subsequent reduction using solutions to quadratic equations and abbreviated multiplication formulas.

Species uncertainty

Example 1.

n is equal to 2. Therefore, we divide the numerator and denominator term by term by:

.

Comment on the right side of the expression. Arrows and numbers indicate what fractions tend to after substitution n meaning infinity. Here, as in example 2, the degree n There is more in the denominator than in the numerator, as a result of which the entire fraction tends to be infinitesimal or “super-small.”

We get the answer: the limit of this function with a variable tending to infinity is equal to .

Example 2. .

Solution. Here the highest power of the variable x is equal to 1. Therefore, we divide the numerator and denominator term by term by x:

.

Commentary on the progress of the decision. In the numerator we drive “x” under the root of the third degree, and so that its original degree (1) remains unchanged, we assign it the same degree as the root, that is, 3. There are no arrows or additional numbers in this entry, so try it mentally, but by analogy with the previous example, determine what the expressions in the numerator and denominator tend to after substituting infinity instead of “x”.

We received the answer: the limit of this function with a variable tending to infinity is equal to zero.

Species uncertainty

Example 3. Uncover uncertainty and find the limit.

Solution. The numerator is the difference of cubes. Let’s factorize it using the abbreviated multiplication formula from the school mathematics course:

The denominator contains a quadratic trinomial, which we will factorize by solving a quadratic equation (once again a link to solving quadratic equations):

Let's write down the expression obtained as a result of the transformations and find the limit of the function:

Example 4. Unlock uncertainty and find the limit

Solution. The quotient limit theorem does not apply here, since

Therefore, we transform the fraction identically: multiplying the numerator and denominator by the binomial conjugate to the denominator, and reduce by x+1. According to the corollary of Theorem 1, we obtain an expression, solving which we find the desired limit:

Example 5. Unlock uncertainty and find the limit

Solution. Direct value substitution x= 0 into a given function leads to uncertainty of the form 0/0. To reveal it, we perform identical transformations and ultimately obtain the desired limit:

Example 6. Calculate

Solution: Let's use the theorems on limits

Answer: 11

Example 7. Calculate

Solution: in this example the limits of the numerator and denominator at are equal to 0:

; . We have received, therefore, the theorem on the limit of the quotient cannot be applied.

Let us factorize the numerator and denominator in order to reduce the fraction by a common factor tending to zero, and, therefore, make it possible to apply Theorem 3.

Let's expand the square trinomial in the numerator using the formula , where x 1 and x 2 are the roots of the trinomial. Having factorized and denominator, reduce the fraction by (x-2), then apply Theorem 3.

Answer:

Example 8. Calculate

Solution: When the numerator and denominator tend to infinity, therefore, when directly applying Theorem 3, we obtain the expression , which represents uncertainty. To get rid of uncertainty of this type, you should divide the numerator and denominator by the highest power of the argument. In this example, you need to divide by X:

Answer:

Example 9. Calculate

Solution: x 3:

Answer: 2

Example 10. Calculate

Solution: When the numerator and denominator tend to infinity. Let's divide the numerator and denominator by the highest power of the argument, i.e. x 5:

=

The numerator of the fraction tends to 1, the denominator tends to 0, so the fraction tends to infinity.

Answer:

Example 11. Calculate

Solution: When the numerator and denominator tend to infinity. Let's divide the numerator and denominator by the highest power of the argument, i.e. x 7:

Answer: 0

Derivative.

Derivative of the function y = f(x) with respect to the argument x is called the limit of the ratio of its increment y to the increment x of the argument x, when the increment of the argument tends to zero: . If this limit is finite, then the function y = f(x) is said to be differentiable at x. If this limit exists, then they say that the function y = f(x) has an infinite derivative at point x.

Derivatives of basic elementary functions:

1. (const)=0 9.

3. 11.

4. 12.

5. 13.

6. 14.

Rules of differentiation:

a)

V)

Example 1. Find the derivative of a function

Solution: If the derivative of the second term is found by the rule of differentiation of fractions, then the first term is a complex function, the derivative of which is found by the formula:

, Where , Then

When solving the following formulas were used: 1,2,10,a,c,d.

Answer:

Example 21. Find the derivative of a function

Solution: both terms are complex functions, where for the first , , and for the second , , then

Answer:

Derivative applications.

1. Speed ​​and acceleration

Let the function s(t) describe position object in some coordinate system at time t. Then the first derivative of the function s(t) is instantaneous speed object:
v=s′=f′(t)
The second derivative of the function s(t) represents the instantaneous acceleration object:
w=v′=s′′=f′′(t)

2. Tangent equation
y−y0=f′(x0)(x−x0),
where (x0,y0) are the coordinates of the tangent point, f′(x0) is the value of the derivative of the function f(x) at the tangent point.

3. Normal equation
y−y0=−1f′(x0)(x−x0),

where (x0,y0) are the coordinates of the point at which the normal is drawn, f′(x0) is the value of the derivative of the function f(x) at this point.

4. Increasing and decreasing functions
If f′(x0)>0, then the function increases at the point x0. In the figure below the function is increasing as x x2.
If f′(x0)<0, то функция убывает в точке x0 (интервал x1If f′(x0)=0 or the derivative does not exist, then this criterion does not allow us to determine the nature of the monotonicity of the function at the point x0.

5. Local extrema of a function
The function f(x) has local maximum at the point x1, if there is a neighborhood of the point x1 such that for all x from this neighborhood the inequality f(x1)≥f(x) holds.
Similarly, the function f(x) has local minimum at the point x2, if there is a neighborhood of the point x2 such that for all x from this neighborhood the inequality f(x2)≤f(x) holds.

6. Critical points
Point x0 is critical point function f(x), if the derivative f′(x0) in it is equal to zero or does not exist.

7. The first sufficient sign of the existence of an extremum
If the function f(x) increases (f′(x)>0) for all x in some interval (a,x1] and decreases (f′(x)<0) для всех x в интервале и возрастает (f′(x)>0) for all x from the interval )