Calculation of the limits of a function with a detailed solution. Sequence and function limit
Solving problems on finding limits When solving problems on finding limits, you should remember some limits so as not to recalculate them each time. Combining these known limits, we will find new limits using the properties indicated in § 4. For convenience, we present the most frequently encountered limits: Limits 1 lim x - a x a 2 lim 1 = 0 3 lim x- ± co X ± 00 4 lim -L, = oo X->o\X\ 5 lim sin*- l X -о X 6 lim f(x) = f(a), if f (x) is continuous x a If it is known that the function is continuous, then instead of finding the limit, we calculate the value of the function. Example 1. Find lim (x*-6l:+ 8). Since the multi-term X->2 term function is continuous, then lim (x*-6x4- 8) = 2*-6-2 + 8 = 4. x-+2 x*_2x 4-1 Example 2. Find lim -G. . First, we find the limit of the denominator: lim [xr-\-bx)= 12 + 5-1 =6; it is not equal to X-Y1 zero, which means we can apply property 4 § 4, then x™i *" + &* ~~ lim (x2 bx) - 12 + 5-1 ""6 1. The limit of the denominator X X is equal to zero, therefore, property 4 of § 4 cannot be applied. Since the numerator is a constant number, and the denominator [x2x) -> -0 for x - - 1, then the entire fraction increases unlimitedly in absolute value, i.e. lim " 1 X - * - - 1 x* + x Example 4. Find lim\-ll*"!"" "The limit of the denominator is zero: lim (xr-6lg+ 8) = 2*-6-2 + 8 = 0, so X property 4 § 4 not applicable. But the limit of the numerator is also equal to zero: lim (x2 - 5d; + 6) = 22 - 5-2-f 6 = 0. So, the limits of the numerator and denominator are simultaneously equal to zero. However, the number 2 is the root of both the numerator and the denominator, so the fraction can be reduced by the difference x-2 (according to Bezout’s theorem). In fact, x*-5x + 6 (x-2) (x-3) x-3 x"-6x + 8~ (x-2) (x-4) ~~ x-4" therefore, xr- -f- 6 g x-3 -1 1 Example 5. Find lim xn (n integer, positive). X with We have xn = X* X . . X, n times Since each factor grows without limit, the product also grows without limit, i.e. lim xn = oo. x oo Example 6. Find lim xn(n integer, positive). X -> - CO We have xn = x x... x. Since each factor grows in absolute value while remaining negative, then in the case of an even degree the product will grow unlimitedly while remaining positive, i.e. lim *n = + oo (for even n). *-* -о In the case of an odd degree, the absolute value of the product increases, but it remains negative, i.e. lim xn = - oo (for n odd). p -- 00 Example 7. Find lim . x x-*- co * If m>pu then we can write: m = n + kt where k>0. Therefore xm b lim -=- = lim -=-= lim x. UP Yn x - x> A x yu We came to example 6. If ti uTL xm I lim lim lim t. X - O x -* yu A X -> co Here the numerator remains constant, and the denominator grows in absolute value, so lim -ь = 0. X - *oo X* It is recommended to remember the result of this example in the following form: The power function grows as faster, the larger the exponent. $хв_Зхг + 7 Example 8. Find lim g L -г-=. In this example x-*® «J* "Г bХ -ох-о and the numerator and denominator increase without limit. Let us divide both the numerator and the denominator by the highest power of x, i.e. on xb, then 3 7_ Example 9. Find lira... Performing transformations, we obtain lira... ^ = lim X CO + 3 7 3 Since lim -5 = 0, lim -, = 0 , then the limit of the denominator rad-*® X X-+-CD X is zero, while the limit of the numerator is 1. Consequently, the entire fraction increases without limit, i.e. t. 7x hm X-+ yu Example 10. Find lim Let's calculate the limit S of the denominator, remembering that the cos*-function is continuous: lira (2 + cos x) = 2 + cozy = 2. Then x->- S lim (l-fsin*) Example 15. Find lim *<*-e>2 and lim e "(X"a)\ Polo X-+ ± co X ± CO press (l: - a)2 = z; since (Λ;-a)2 always grows non-negatively and without limit with x, then for x - ±oo the new variable z-*oc. Therefore we obtain qt £<*-«)* = X ->± 00 s=lim ег = oo (see note to §5). g -*■ co Similarly lim e~(X-a)2 = lim e~z=Q, since x ± oo g m - (x- a)z decreases without limit as x ->±oo (see note to §
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, 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
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 this topic we will consider all three groups of limits with irrationality listed above. Let's start with limits containing uncertainty of the form $\frac(0)(0)$.
Uncertainty disclosure $\frac(0)(0)$.
The solution to standard examples of this type usually consists of two steps:
- We get rid of the irrationality that caused uncertainty by multiplying by the so-called “conjugate” expression;
- If necessary, factor the expression in the numerator or denominator (or both);
- We reduce the factors leading to uncertainty and calculate the desired value of the limit.
The term "conjugate expression" used above will be explained in detail in the examples. For now there is no reason to dwell on it in detail. In general, you can go the other way, without using the conjugate expression. Sometimes a well-chosen replacement can eliminate irrationality. Such examples are rare in standard tests, so we will consider only one example No. 6 for the use of replacement (see the second part of this topic).
We will need several formulas, which I will write down below:
\begin(equation) a^2-b^2=(a-b)\cdot(a+b) \end(equation) \begin(equation) a^3-b^3=(a-b)\cdot(a^2 +ab+b^2) \end(equation) \begin(equation) a^3+b^3=(a+b)\cdot(a^2-ab+b^2) \end(equation) \begin (equation) a^4-b^4=(a-b)\cdot(a^3+a^2 b+ab^2+b^3)\end(equation)
In addition, we assume that the reader knows the formulas for solving quadratic equations. If $x_1$ and $x_2$ are the roots of the quadratic trinomial $ax^2+bx+c$, then it can be factorized using the following formula:
\begin(equation) ax^2+bx+c=a\cdot(x-x_1)\cdot(x-x_2) \end(equation)
Formulas (1)-(5) are quite sufficient for solving standard problems, which we will now move on to.
Example No. 1
Find $\lim_(x\to 3)\frac(\sqrt(7-x)-2)(x-3)$.
Since $\lim_(x\to 3)(\sqrt(7-x)-2)=\sqrt(7-3)-2=\sqrt(4)-2=0$ and $\lim_(x\ to 3) (x-3)=3-3=0$, then in the given limit we have an uncertainty of the form $\frac(0)(0)$. The difference $\sqrt(7-x)-2$ prevents us from revealing this uncertainty. In order to get rid of such irrationalities, multiplication by the so-called “conjugate expression” is used. We will now look at how such multiplication works. Multiply $\sqrt(7-x)-2$ by $\sqrt(7-x)+2$:
$$(\sqrt(7-x)-2)(\sqrt(7-x)+2)$$
To open the brackets, apply , substituting $a=\sqrt(7-x)$, $b=2$ into the right side of the mentioned formula:
$$(\sqrt(7-x)-2)(\sqrt(7-x)+2)=(\sqrt(7-x))^2-2^2=7-x-4=3-x .$$
As you can see, if you multiply the numerator by $\sqrt(7-x)+2$, then the root (i.e., irrationality) in the numerator will disappear. This expression $\sqrt(7-x)+2$ will be conjugate to the expression $\sqrt(7-x)-2$. However, we cannot simply multiply the numerator by $\sqrt(7-x)+2$, because this will change the fraction $\frac(\sqrt(7-x)-2)(x-3)$, which is under the limit . You need to multiply both the numerator and denominator at the same time:
$$ \lim_(x\to 3)\frac(\sqrt(7-x)-2)(x-3)= \left|\frac(0)(0)\right|=\lim_(x\to 3)\frac((\sqrt(7-x)-2)\cdot(\sqrt(7-x)+2))((x-3)\cdot(\sqrt(7-x)+2)) $$
Now remember that $(\sqrt(7-x)-2)(\sqrt(7-x)+2)=3-x$ and open the brackets. And after opening the parentheses and a small transformation $3-x=-(x-3)$, we reduce the fraction by $x-3$:
$$ \lim_(x\to 3)\frac((\sqrt(7-x)-2)\cdot(\sqrt(7-x)+2))((x-3)\cdot(\sqrt( 7-x)+2))= \lim_(x\to 3)\frac(3-x)((x-3)\cdot(\sqrt(7-x)+2))=\\ =\lim_ (x\to 3)\frac(-(x-3))((x-3)\cdot(\sqrt(7-x)+2))= \lim_(x\to 3)\frac(-1 )(\sqrt(7-x)+2) $$
The uncertainty $\frac(0)(0)$ has disappeared. Now you can easily get the answer of this example:
$$ \lim_(x\to 3)\frac(-1)(\sqrt(7-x)+2)=\frac(-1)(\sqrt(7-3)+2)=-\frac( 1)(\sqrt(4)+2)=-\frac(1)(4).$$
I note that the conjugate expression can change its structure, depending on what kind of irrationality it should remove. In examples No. 4 and No. 5 (see the second part of this topic) a different type of conjugate expression will be used.
Answer: $\lim_(x\to 3)\frac(\sqrt(7-x)-2)(x-3)=-\frac(1)(4)$.
Example No. 2
Find $\lim_(x\to 2)\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))$.
Since $\lim_(x\to 2)(\sqrt(x^2+5)-\sqrt(7x^2-19))=\sqrt(2^2+5)-\sqrt(7\cdot 2 ^2-19)=3-3=0$ and $\lim_(x\to 2)(3x^2-5x-2)=3\cdot2^2-5\cdot 2-2=0$, then we we are dealing with uncertainty of the form $\frac(0)(0)$. Let's get rid of the irrationality in the denominator of this fraction. To do this, we add both the numerator and denominator of the fraction $\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))$ to the expression $\sqrt(x^ 2+5)+\sqrt(7x^2-19)$ conjugate to the denominator:
$$ \lim_(x\to 2)\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))=\left|\frac(0 )(0)\right|= \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19))) ((\sqrt(x^2+5)-\sqrt(7x^2-19))(\sqrt(x^2+5)+\sqrt(7x^2-19))) $$
Again, as in example No. 1, you need to use parentheses to expand. Substituting $a=\sqrt(x^2+5)$, $b=\sqrt(7x^2-19)$ into the right side of the mentioned formula, we obtain the following expression for the denominator:
$$ \left(\sqrt(x^2+5)-\sqrt(7x^2-19)\right)\left(\sqrt(x^2+5)+\sqrt(7x^2-19)\ right)=\\ =\left(\sqrt(x^2+5)\right)^2-\left(\sqrt(7x^2-19)\right)^2=x^2+5-(7x ^2-19)=-6x^2+24=-6\cdot(x^2-4) $$
Let's return to our limit:
$$ \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19)))((\sqrt(x ^2+5)-\sqrt(7x^2-19))(\sqrt(x^2+5)+\sqrt(7x^2-19)))= \lim_(x\to 2)\frac( (3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19)))(-6\cdot(x^2-4))=\\ =-\ frac(1)(6)\cdot \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19)) )(x^2-4) $$
In example No. 1, almost immediately after multiplication by the conjugate expression, the fraction was reduced. Here, before the reduction, you will have to factorize the expressions $3x^2-5x-2$ and $x^2-4$, and only then proceed to the reduction. To factor the expression $3x^2-5x-2$ you need to use . First, let's solve the quadratic equation $3x^2-5x-2=0$:
$$ 3x^2-5x-2=0\\ \begin(aligned) & D=(-5)^2-4\cdot3\cdot(-2)=25+24=49;\\ & x_1=\ frac(-(-5)-\sqrt(49))(2\cdot3)=\frac(5-7)(6)=-\frac(2)(6)=-\frac(1)(3) ;\\ & x_2=\frac(-(-5)+\sqrt(49))(2\cdot3)=\frac(5+7)(6)=\frac(12)(6)=2. \end(aligned) $$
Substituting $x_1=-\frac(1)(3)$, $x_2=2$ into , we will have:
$$ 3x^2-5x-2=3\cdot\left(x-\left(-\frac(1)(3)\right)\right)(x-2)=3\cdot\left(x+\ frac(1)(3)\right)(x-2)=\left(3\cdot x+3\cdot\frac(1)(3)\right)(x-2) =(3x+1)( x-2). $$
Now it’s time to factorize the expression $x^2-4$. Let's use , substituting $a=x$, $b=2$ into it:
$$ x^2-4=x^2-2^2=(x-2)(x+2) $$
Let's use the results obtained. Since $x^2-4=(x-2)(x+2)$ and $3x^2-5x-2=(3x+1)(x-2)$, then:
$$ -\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2 -19)))(x^2-4) =-\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(x-2)(\sqrt(x ^2+5)+\sqrt(7x^2-19)))((x-2)(x+2)) $$
Reducing by the bracket $x-2$ we get:
$$ -\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(x-2)(\sqrt(x^2+5)+\sqrt(7x^ 2-19)))((x-2)(x+2)) =-\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(\sqrt( x^2+5)+\sqrt(7x^2-19)))(x+2). $$
All! The uncertainty has disappeared. One more step and we come to the answer:
$$ -\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(\sqrt(x^2+5)+\sqrt(7x^2-19)) )(x+2)=\\ =-\frac(1)(6)\cdot\frac((3\cdot 2+1)(\sqrt(2^2+5)+\sqrt(7\cdot 2 ^2-19)))(2+2)= -\frac(1)(6)\cdot\frac(7(3+3))(4)=-\frac(7)(4). $$
Answer: $\lim_(x\to 2)\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))=-\frac(7)( 4)$.
In the following example, consider the case where irrationalities will be present in both the numerator and the denominator of the fraction.
Example No. 3
Find $\lim_(x\to 5)\frac(\sqrt(x+4)-\sqrt(x^2-16))(\sqrt(x^2-3x+6)-\sqrt(5x-9 ))$.
Since $\lim_(x\to 5)(\sqrt(x+4)-\sqrt(x^2-16))=\sqrt(9)-\sqrt(9)=0$ and $\lim_( x\to 5)(\sqrt(x^2-3x+6)-\sqrt(5x-9))=\sqrt(16)-\sqrt(16)=0$, then we have an uncertainty of the form $\frac (0)(0)$. Since in this case the roots are present in both the denominator and the numerator, in order to get rid of uncertainty you will have to multiply by two brackets at once. First, to the expression $\sqrt(x+4)+\sqrt(x^2-16)$ conjugate to the numerator. And secondly, to the expression $\sqrt(x^2-3x+6)-\sqrt(5x-9)$ conjugate to the denominator.
$$ \lim_(x\to 5)\frac(\sqrt(x+4)-\sqrt(x^2-16))(\sqrt(x^2-3x+6)-\sqrt(5x-9 ))=\left|\frac(0)(0)\right|=\\ =\lim_(x\to 5)\frac((\sqrt(x+4)-\sqrt(x^2-16) )(\sqrt(x+4)+\sqrt(x^2-16))(\sqrt(x^2-3x+6)+\sqrt(5x-9)))((\sqrt(x^2 -3x+6)-\sqrt(5x-9))(\sqrt(x^2-3x+6)+\sqrt(5x-9))(\sqrt(x+4)+\sqrt(x^2 -16))) $$ $$ -x^2+x+20=0;\\ \begin(aligned) & D=1^2-4\cdot(-1)\cdot 20=81;\\ & x_1=\frac(-1-\sqrt(81))(-2)=\frac(-10)(-2)=5;\\ & x_2=\frac(-1+\sqrt(81))( -2)=\frac(8)(-2)=-4. \end(aligned) \\ -x^2+x+20=-1\cdot(x-5)(x-(-4))=-(x-5)(x+4). $$
For the expression $x^2-8x+15$ we get:
$$ x^2-8x+15=0;\\ \begin(aligned) & D=(-8)^2-4\cdot 1\cdot 15=4;\\ & x_1=\frac(-(- 8)-\sqrt(4))(2)=\frac(6)(2)=3;\\ & x_2=\frac(-(-8)+\sqrt(4))(2)=\frac (10)(2)=5. \end(aligned)\\ x^2+8x+15=1\cdot(x-3)(x-5)=(x-3)(x-5). $$
Substituting the resulting expansions $-x^2+x+20=-(x-5)(x+4)$ and $x^2+8x+15=(x-3)(x-5)$ into the limit under consideration, will have:
$$ \lim_(x\to 5)\frac((-x^2+x+20)(\sqrt(x^2-3x+6)+\sqrt(5x-9)))((x^2 -8x+15)(\sqrt(x+4)+\sqrt(x^2-16)))= \lim_(x\to 5)\frac(-(x-5)(x+4)(\ sqrt(x^2-3x+6)+\sqrt(5x-9)))((x-3)(x-5)(\sqrt(x+4)+\sqrt(x^2-16)) )=\\ =\lim_(x\to 5)\frac(-(x+4)(\sqrt(x^2-3x+6)+\sqrt(5x-9)))((x-3) (\sqrt(x+4)+\sqrt(x^2-16)))= \frac(-(5+4)(\sqrt(5^2-3\cdot 5+6)+\sqrt(5 \cdot 5-9)))((5-3)(\sqrt(5+4)+\sqrt(5^2-16)))=-6. $$
Answer: $\lim_(x\to 5)\frac(\sqrt(x+4)-\sqrt(x^2-16))(\sqrt(x^2-3x+6)-\sqrt(5x-9 ))=-6$.
In the next (second) part, we will consider a couple more examples in which the conjugate expression will have a different form than in the previous problems. The main thing to remember is that the purpose of using a conjugate expression is to get rid of the irrationality that causes uncertainty.
Elementary functions and their graphs.
The main elementary functions are: power function, exponential function, logarithmic function, trigonometric functions and inverse trigonometric functions, as well as a polynomial and a rational function, which is the ratio of two polynomials.
Elementary functions also include those functions that are obtained from elementary ones by applying the basic four arithmetic operations and forming a complex function.
Graphs of elementary functions
| Straight line- graph of a linear function y = ax + b. The function y monotonically increases for a > 0 and decreases for a< 0. При b = 0 прямая линия проходит через начало координат т. 0 (y = ax - прямая пропорциональность) | |
 | Parabola- graph of the quadratic trinomial function y = ax 2 + bx + c. It has a vertical axis of symmetry. If a > 0, has a minimum if a< 0 - максимум. Точки пересечения (если они есть) с осью абсцисс - корни соответствующего квадратного уравнения ax 2 + bx +c =0 |
 | Hyperbola- graph of the function. When a > O it is located in the I and III quarters, when a< 0 - во II и IV. Асимптоты - оси координат. Ось симметрии - прямая у = х(а >0) or y - - x(a< 0). |
 | Exponential function. Exhibitor(exponential function to base e) y = e x. (Another spelling y = exp(x)). Asymptote is the abscissa axis. |
 | Logarithmic function y = log a x(a > 0) |
 | y = sinx. Sine wave- periodic function with period T = 2π |
Function limit.
The function y=f(x) has a number A as a limit as x tends to a, if for any number ε › 0 there is a number δ › 0 such that | y – A | ‹ ε if |x - a| ‹ δ,
or lim y = A
Continuity of function.
The function y=f(x) is continuous at the point x = a if lim f(x) = f(a), i.e.
the limit of a function at a point x = a is equal to the value of the function at a given point.
Finding the limits of functions.
Basic theorems on the limits of functions.
1. The limit of a constant value is equal to this constant value:
2. The limit of an algebraic sum is equal to the algebraic sum of the limits of these functions:
lim (f + g - h) = lim f + lim g - lim h
3. The limit of the product of several functions is equal to the product of the limits of these functions:
lim (f * g* h) = lim f * lim g * lim h
4. The limit of the quotient of two functions is equal to the quotient of the limits of these functions if the limit of the denominator is not equal to 0:
lim------- = ----------
The first remarkable limit: lim --------- = 1
Second remarkable limit: lim (1 + 1/x) x = e (e = 2, 718281..)
Examples of finding the limits of functions.
5.1. Example:
Any limit consists of three parts:
1) The well-known limit icon.
2) Entries under the limit icon. The entry reads “X tends to one.” Most often it is x, although instead of “x” there can be any other variable. In place of one there can be absolutely any number, as well as infinity 0 or .
3) Functions under the limit sign, in this case .
The recording itself reads like this: “the limit of a function as x tends to unity.”
A very important question - what does the expression “x” mean? strives to one"? The expression "x" strives to one” should be understood as follows: “x” consistently takes on the values which approach unity infinitely close and practically coincide with it.
How to solve the above example? Based on the above, you just need to substitute one into the function under the limit sign:
So the first rule : When given a limit, you first simply plug the number into the function.
5.2. Example with infinity:
Let's figure out what it is? This is the case when it increases without limit.
So: if , then the function tends to minus infinity:
According to our first rule, instead of “X” we substitute in the function infinity and we get the answer.
5.3. Another example with infinity:
Again we begin to increase to infinity, and look at the behavior of the function.
Conclusion: the function increases unlimitedly
5.4. A series of examples:
Try to mentally analyze the following examples yourself and solve the simplest types of limits:
, , , , 
, , , , 
,
What do you need to remember and understand from the above?
When given any limit, first simply plug the number into the function. At the same time, you must understand and immediately solve the simplest limits, such as 
,
,
etc.
6. Limits with uncertainty of type and a method for solving them.
Now we will consider the group of limits when , and the function is a fraction whose numerator and denominator contain polynomials.
6.1. Example:
Calculate limit 
According to our rule, we 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 = 1, and the answer is ready, but in the general case this is not at all the case, and you need 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 uncertainty you need to divide the numerator and denominator by in the senior degree.
Thus, the answer is not 1.
Example
Find the limit 
Again in the numerator and denominator we find in the highest degree: 
Maximum degree in numerator: 3
Maximum degree in denominator: 4
Choose greatest value, in this case four.
According to our algorithm, to reveal uncertainty, we divide the numerator and denominator by .
Example
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 the uncertainty, it is necessary to divide the numerator and denominator by . The final solution might look like this:
Divide the numerator and denominator by
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