If the limit tends to the number. Limits
Let's look at some illustrative examples.
Let x be a numerical variable, X the area of its change. If each number x belonging to X is associated with a certain number y, then they say that a function is defined on the set X, and write y = f(x).
The X set in this case is a plane consisting of two coordinate axes – 0X and 0Y. For example, let's depict the function y = x 2. The 0X and 0Y axes form X - the area of its change. The figure clearly shows how the function behaves. In this case, they say that the function y = x 2 is defined on the set X.
The set Y of all partial values of a function is called the set of values f(x). In other words, the set of values is the interval along the 0Y axis where the function is defined. The depicted parabola clearly shows that f(x) > 0, because x2 > 0. Therefore, the range of values will be . We look at many values by 0Y.
The set of all x is called the domain of f(x). We look at many definitions by 0X and in our case the range of acceptable values is [-; +].
A point a (a belongs to or X) is called a limit point of the set X if in any neighborhood of the point a there are points of the set X different from a.
The time has come to understand what is the limit of a function?
The pure b to which the function tends as x tends to the number a is called limit of the function. This is written as follows:
For example, f(x) = x 2. We need to find out what the function tends to (is not equal to) at x 2. First, we write down the limit:
Let's look at the graph.
Let's draw a line parallel to the 0Y axis through point 2 on the 0X axis. It will intersect our graph at point (2;4). Let us drop a perpendicular from this point to the 0Y axis and get to point 4. This is what our function strives for at x 2. If we now substitute the value 2 into the function f(x), the answer will be the same.
Now before we move on to calculation of limits, let us introduce basic definitions.
Introduced by the French mathematician Augustin Louis Cauchy in the 19th century.
Suppose the function f(x) is defined on a certain interval that contains the point x = A, but it is not at all necessary that the value of f(A) be defined.
Then, according to Cauchy's definition, limit of the function f(x) will be a certain number B with x tending to A if for every C > 0 there is a number D > 0 for which
Those. if the function f(x) at x A is limited by limit B, this is written as
Sequence limit a certain number A is called if for any arbitrarily small positive number B > 0 there is a number N for which all values in the case n > N satisfy the inequality
This limit looks like .
A sequence that has a limit will be called convergent; if not, we will call it divergent.
As you have already noticed, limits are indicated by the lim icon, under which some condition for the variable is written, and then the function itself is written. Such a set will be read as “the limit of a function subject to...”. For example:
- the limit of the function as x tends to 1.
The expression “approaching 1” means that x successively takes on values that approach 1 infinitely close.
Now it becomes clear that to calculate this limit it is enough to substitute the value 1 for x:
In addition to a specific numerical value, x can also tend to infinity. For example:
The expression x means that x is constantly increasing and approaching infinity without limit. Therefore, substituting infinity for x, it becomes obvious that the function 1-x will tend to , but with the opposite sign:
Thus, calculation of limits comes down to finding its specific value or a certain area in which the function limited by the limit falls.
Based on the above, it follows that when calculating limits it is important to use several rules:
Understanding essence of limit and basic rules limit calculations, you'll gain key insight into how to solve them. If any limit causes you difficulties, then write in the comments and we will definitely help you.
Note: Jurisprudence is the science of laws, which helps in conflicts and other life difficulties.
When solving problems of 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 there are many - X->2
member function is continuous, then lim (x*-6x4- 8) = 2*-6-2 + 8 = 4. x-+2 x*_2x 4-1 Example 2. Find lim -r. . 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 is [x2x) -> -0 for x - - 1, then the entire fraction increases indefinitely 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 L X ->co Here the numerator remains constant, and the denominator increases in absolute value, therefore lim -ь = 0. Х-*оо X* It is recommended to remember the result of this example in the following form: The power function grows faster, the larger the exponent. $хв_Зхг + 7
Examples
Example 8. Find lim g L -g-=. In this example x-*® "J* "G bX -ox-o and the numerator and denominator increase without limit. Let us divide both the numerator and 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-+ ω Example 10. Find lim Let's calculate the limit S 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 (l;-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 §
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:
- The base of the degree (i.e., the expression in brackets of formulas (1) and (2)) should tend to unity;
- 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 a 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: $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x)=1$.
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 this or that 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. There lived in the 19th century a Frenchman, Augustin Louis Cauchy, who laid the foundations of mathematical analysis and gave strict definitions, the definition of a limit, in particular. It must be said that this same Cauchy was, is, and will be in the nightmares of all students of physics and mathematics, since he proved a huge number of theorems of mathematical analysis, and each 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 just an example of why to shaggy grandma...
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 one.” Most often - exactly, although instead of “X” in practice there are other variables. In practical tasks, the place of one can be absolutely any number, as well as infinity ().
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.”
Let's look at the next important question - what does the expression “x” mean? strives to one"? And what does “strive” even mean?
The concept of a limit is a concept, so to speak, dynamic. Let's build a sequence: first , then , , …, 
, ….
That is, 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 any limit, first we simply try to plug the number into the function.
We have considered the simplest limit, but these also occur in practice, and not so rarely!
Example with infinity:
Let's figure out what it is? This is the case when it increases without limit, that is: first, then, then, then and so on ad infinitum.
What happens to the function at this time?
, , 
, …
So: if , then the function tends to minus infinity:
Roughly speaking, according to our first rule, instead of “X” we substitute infinity into the function and get the answer.
Another example with infinity:
Again we begin to increase to infinity, and look at the behavior of the function: 
Conclusion: when the function increases without limit:
And another series of examples:
Please try to mentally analyze the following for yourself and remember the simplest types of limits:
, , , , 
, , , , 
,
If you have doubts anywhere, you can pick up a calculator and practice a little.
In the event that , try to construct the sequence , , . If , then , , .
Note: strictly speaking, this approach to constructing sequences of several numbers is incorrect, but for understanding the simplest examples it is quite suitable.
Also pay attention to the following thing. Even if a limit is given with a large number at the top, or even with a million: , then it’s all the same 
, since sooner or later “X” will take on such gigantic values that a million compared to them will be a real microbe.
What do you need to remember and understand from the above?
1) When given any limit, first we simply try to substitute the 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 whose numerator and denominator contain polynomials
Example:
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 , and 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: in order to reveal the uncertainty, it is necessary to divide the numerator and denominator by the highest power.
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 2
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 .
The complete assignment might look like this:
Divide the numerator and denominator by
Example 3
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
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.
Limits with uncertainty of type and method for solving them
The next group of limits is somewhat similar to the limits just considered: the numerator and denominator contain polynomials, but “x” no longer tends to infinity, but to finite number.
Example 4
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 disclose 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. If these things have been forgotten, then visit the page Mathematical formulas and tables and read the teaching material Hot formulas for school mathematics course. By the way, it is best to print it out; it is required very often, and information is absorbed better from paper.
So, let's solve our limit 
Factor the numerator and denominator
In order to factor 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 function of extracting the square root is on the simplest calculator.
! If the root is not extracted in its entirety (a fractional number with a comma is obtained), it is very likely that the discriminant was calculated incorrectly or there was a typo in the task.
Next we find the roots: 
Thus:
All. The numerator is factorized.
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 into the expression that remains under the limit sign:
Naturally, in a test, test, or exam, the solution is never described in such detail. In the final version, the design should look something like this:
Let's factorize the numerator. 
Example 5
Calculate limit 
First, the “finish” version of the solution
Let's factor the numerator and denominator.
Numerator:
Denominator: 
, 
What is important in this example?
Firstly, you must have a good understanding of how the numerator is revealed, first we took 2 out of brackets, and then used the formula for the difference of squares. This is the formula you need to know and see.
From the above article you can find out what the limit is and what it is eaten with - this is VERY important. Why? You may not understand what determinants are and successfully solve them, you may not understand at all what a derivative is and find them with an “A”. But if you don’t understand what a limit is, then solving practical tasks will be difficult. It would also be a good idea to familiarize yourself with the sample solutions and my design recommendations. All information is presented in a simple and accessible form.
And for the purposes of this lesson we will need the following teaching materials: Wonderful Limits And Trigonometric formulas. They can be found on the page. It is best to print out the manuals - it is much more convenient, and besides, you will often have to refer to them offline.
What is so special about remarkable limits? The remarkable thing about these limits is that they were proven by the greatest minds of famous mathematicians, and grateful descendants do not have to suffer from terrible limits with a pile of trigonometric functions, logarithms, powers. That is, when finding the limits, we will use ready-made results that have been proven theoretically.
There are several wonderful limits, but in practice, part-time students in 95% of cases have two wonderful limits: The first wonderful limit, Second wonderful limit. It should be noted that these are historically established names, and when, for example, they talk about “the first remarkable limit,” they mean by this a very specific thing, and not some random limit taken from the ceiling.
The first wonderful limit
Consider the following limit: (instead of the native letter “he” I will use the Greek letter “alpha”, this is more convenient from the point of view of presenting the material).
According to our rule for finding limits (see article Limits. Examples of solutions) we try to substitute zero into the function: in the numerator we get zero (the sine of zero is zero), and in the denominator, obviously, there is also zero. Thus, we are faced with an uncertainty of the form, which, fortunately, does not need to be disclosed. In the course of mathematical analysis, it is proven that:
This mathematical fact is called The first wonderful limit. I won’t give an analytical proof of the limit, but we’ll look at its geometric meaning in the lesson about infinitesimal functions.
Often in practical tasks functions can be arranged differently, this does not change anything:
- the same first wonderful limit.
But you cannot rearrange the numerator and denominator yourself! If a limit is given in the form , then it must be solved in the same form, without rearranging anything.
In practice, not only a variable, but also an elementary function or a complex function can act as a parameter. It is only important that it tends to zero.
Examples:
, , 
, 
Here , , , 
, and everything is good - the first wonderful limit is applicable.
But the following entry is heresy:
Why? Because the polynomial does not tend to zero, it tends to five.
By the way, a quick question: what is the limit? 
? The answer can be found at the end of the lesson.
In practice, not everything is so smooth; almost never a student is offered to solve a free limit and get an easy pass. Hmmm... I’m writing these lines, and a very important thought came to mind - after all, it’s better to remember “free” mathematical definitions and formulas by heart, this can provide invaluable help in the test, when the question will be decided between a “two” and a “three”, and the teacher decides to ask the student some simple question or offer to solve a simple example (“maybe he (s) still knows what?!”).
Let's move on to consider practical examples:
Example 1
Find the limit
If we notice a sine in the limit, then this should immediately lead us to think about the possibility of applying the first remarkable limit.
First, we try to substitute 0 into the expression under the limit sign (we do this mentally or in a draft):
So we have an uncertainty of the form be sure to indicate in making a decision. The expression under the limit sign is similar to the first wonderful limit, but this is not exactly it, it is under the sine, but in the denominator.
In such cases, we need to organize the first remarkable limit ourselves, using an artificial technique. The line of reasoning could be as follows: “under the sine we have , which means that we also need to get in the denominator.”
And this is done very simply:
That is, the denominator is artificially multiplied in this case by 7 and divided by the same seven. Now our recording has taken on a familiar shape.
When the task is drawn up by hand, it is advisable to mark the first remarkable limit with a simple pencil:
What happened? In fact, our circled expression turned into a unit and disappeared in the work: 
Now all that remains is to get rid of the three-story fraction: 
Who has forgotten the simplification of multi-level fractions, please refresh the material in the reference book Hot formulas for school mathematics course .
Ready. Final answer:
If you don’t want to use pencil marks, then the solution can be written like this:
“
Let's use the first wonderful limit 
“
Example 2
Find the limit
Again we see a fraction and a sine in the limit. We try to substitute zero into the numerator and denominator:
Indeed, we have uncertainty and, therefore, we need to try to organize the first wonderful limit. At the lesson Limits. Examples of solutions we considered the rule that when we have uncertainty, we need to factorize the numerator and denominator. Here it’s the same, we’ll represent the degrees as a product (multipliers):
Similar to the previous example, we draw a pencil around the remarkable limits (here there are two of them), and indicate that they tend to unity:
Actually, the answer is ready:
In the following examples, I will not do art in Paint, I think how to correctly draw up a solution in a notebook - you already understand.
Example 3
Find the limit
We substitute zero into the expression under the limit sign:
An uncertainty has been obtained that needs to be disclosed. If there is a tangent in the limit, then it is almost always converted into sine and cosine using the well-known trigonometric formula (by the way, they do approximately the same thing with cotangent, see methodological material Hot trigonometric formulas On the page Mathematical formulas, tables and reference materials).
In this case:
The cosine of zero is equal to one, and it’s easy to get rid of it (don’t forget to mark that it tends to one):
Thus, if in the limit the cosine is a MULTIPLIER, then, roughly speaking, it needs to be turned into a unit, which disappears in the product.
Here everything turned out simpler, without any multiplications and divisions. The first remarkable limit also turns into one and disappears in the product:
As a result, infinity is obtained, and this happens.
Example 4
Find the limit
Let's try to substitute zero into the numerator and denominator:
The uncertainty is obtained (the cosine of zero, as we remember, is equal to one)
We use the trigonometric formula. Take note! For some reason, limits using this formula are very common.
Let us move the constant factors beyond the limit icon:
Let's organize the first wonderful limit:
Here we have only one remarkable limit, which turns into one and disappears in the product:
Let's get rid of the three-story structure:
The limit is actually solved, we indicate that the remaining sine tends to zero:
Example 5
Find the limit 
This example is more complicated, try to figure it out yourself:
Some limits can be reduced to the 1st remarkable limit by changing a variable, you can read about this a little later in the article Methods for solving limits.
Second wonderful limit
In the theory of mathematical analysis it has been proven that:
This fact is called second wonderful limit.
Reference: 
is an irrational number.
The parameter can be not only a variable, but also a complex function. The only important thing is that it strives for infinity.
Example 6
Find the limit
When the expression under the limit sign is in a degree, this is the first sign that you need to try to apply the second wonderful limit.
But first, as always, we try to substitute an infinitely large number into the expression, the principle by which this is done is discussed in the lesson Limits. Examples of solutions.
It is easy to notice that when the basis of the degree is , and the exponent is , that is, there is uncertainty of the form:
This uncertainty is precisely revealed with the help of the second remarkable limit. But, as often happens, the second wonderful limit does not lie on a silver platter, and it needs to be artificially organized. You can reason as follows: in this example the parameter is , which means that we also need to organize in the indicator. To do this, we raise the base to the power, and so that the expression does not change, we raise it to the power:
When the task is completed by hand, we mark with a pencil:
Almost everything is ready, the terrible degree has turned into a nice letter:
In this case, we move the limit icon itself to the indicator:
Example 7
Find the limit
Attention! This type of limit occurs very often, please study this example very carefully.
Let's try to substitute an infinitely large number into the expression under the limit sign:
The result is uncertainty. But the second remarkable limit applies to the uncertainty of the form. What to do? We need to convert the base of the degree. We reason like this: in the denominator we have , which means that in the numerator we also need to organize .
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