When calculating limits, one should take into account the following basic rules:
1. The limit of the sum (difference) of functions is equal to the sum (difference) of the limits of the terms:
2. The limit of a product of functions is equal to the product of the limits of the factors:
3. The limit of the ratio of two functions is equal to the ratio of the limits of these functions:
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4. The constant factor can be taken beyond the limit sign:
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5. The limit of a constant is equal to the constant itself:
6. For continuous functions, the limit and function symbols can be swapped:
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Finding the limit of a function should begin by substituting the value into the expression for the function. Moreover, if the numerical value 0 or ¥ is obtained, then the desired limit has been found.
Example 2.1. Calculate the limit.
Solution.
.
Expressions of the form , , , , , are called uncertainties.
If you get an uncertainty of the form , then to find the limit you need to transform the function so as to reveal this uncertainty.
Uncertainty of form is usually obtained when the limit of the ratio of two polynomials is given. In this case, to calculate the limit, it is recommended to factor the polynomials and reduce by a common factor. This multiplier equal to zero at limit value X .
Example 2.2. Calculate the limit.
Solution.
Substituting , we get uncertainty:
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Let's factor the numerator and denominator:
;
Let's reduce by a common factor and get
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An uncertainty of the form is obtained when the limit of the ratio of two polynomials is given at . In this case, to calculate it, it is recommended to divide both polynomials by X in the senior degree.
Example 2.3. Calculate the limit.
Solution. When substituting ∞, we obtain an uncertainty of the form , so we divide all terms of the expression by x 3.
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It is taken into account here that .
When calculating the limits of a function containing roots, it is recommended to multiply and divide the function by its conjugate.
Example 2.4. Calculate limit
Solution.
When calculating limits to reveal uncertainty of the form or (1) ∞, the first and second remarkable limits are often used:
Many problems associated with the continuous growth of some quantity lead to the second remarkable limit.
Let's consider the example of Ya. I. Perelman, giving an interpretation of the number e in the problem about compound interest. In savings banks, interest money is added to the fixed capital annually. If the accession is made more often, then the capital grows faster, since a larger amount is involved in the formation of interest. Let's take a purely theoretical, very simplified example.
Let 100 deniers be deposited in the bank. units based on 100% per annum. If interest money is added to the fixed capital only after a year, then by this period 100 den. units will turn into 200 monetary units.
Now let's see what 100 denies will turn into. units, if interest money is added to fixed capital every six months. After six months, 100 den. units will grow by 100 × 1.5 = 150, and after another six months - by 150 × 1.5 = 225 (den. units). If the accession is done every 1/3 of the year, then after a year 100 den. units will turn into 100 × (1 +1/3) 3 "237 (den. units).
We will increase the terms for adding interest money to 0.1 year, to 0.01 year, to 0.001 year, etc. Then out of 100 den. units after a year it will be:
100 × (1 +1/10) 10 » 259 (den. units),
100 × (1+1/100) 100 » 270 (den. units),
100 × (1+1/1000) 1000 » 271 (den. units).
With an unlimited reduction in the terms for adding interest, the accumulated capital does not grow indefinitely, but approaches a certain limit equal to approximately 271. The capital deposited at 100% per annum cannot increase by more than 2.71 times, even if the accrued interest were added to the capital every just a second because
Example 2.5. Calculate the limit of a function
Solution.
Example 2.6. Calculate the limit of a function 
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Solution. Substituting we get the uncertainty:
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Using trigonometric formula, transform the numerator into a product:
As a result we get
The second one is taken into account here wonderful limit.
Example 2.7. Calculate the limit of a function
Solution.
.
To reveal uncertainty of the form or, you can use L'Hopital's rule, which is based on the following theorem.
Theorem. The limit of the ratio of two infinitesimal or infinitely large functions is equal to the limit of the ratio of their derivatives
Note that this rule can be applied several times in a row.
Example 2.8. Find
Solution. When substituting , we have an uncertainty of the form . Applying L'Hopital's rule, we get
Continuity of function
Important property function is continuity.
Definition. The function is considered continuous, if a small change in the value of the argument entails a small change in the value of the function.
Mathematically this is written as follows: when 
By and is meant the increment of variables, that is, the difference between the subsequent and previous values: , (Figure 2.3)
 Figure 2.3 – Increment of variables |
From the definition of a function continuous at the point it follows that 
. This equality means that three conditions are met: 
Solution. For function the point is suspicious for a discontinuity, let's check this and find one-sided limits
Hence, 
, Means - break point
Derivative of a function
Type and species uncertainty are the most common uncertainties that need to be disclosed when solving limits.
Most Limit problems encountered by students contain precisely 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:
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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:
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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 is not applicable 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's factor the numerator and denominator to reduce the fraction by a common factor tending to zero, and therefore make possible use Theorem 3.
Let us 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 using 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 - 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 function
If f′(x0)>0, then the function increases at the point x0. In the figure below the function is increasing as x
If f′(x0)<0, то функция убывает в точке x0 (интервал x1
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 )
