Methods for solving limits. Uncertainties.
Function growth order. Replacement method
Example 4
Find the limit 
This is a simpler example for a do-it-yourself solution. In the proposed example, again, uncertainty (of a higher order of growth than the root).
If "x" tends to "minus infinity"
The ghost of "minus infinity" has long been hovering in this article. Consider limits with polynomials in which . The principles and methods of solution will be exactly the same as in the first part of the lesson, with the exception of a number of nuances.
Consider 4 chips that will be required to solve practical tasks:
1) Calculate the limit 
The value of the limit depends only on the term because it has the highest order of growth. If , then infinitely large modulo negative number to the power of EVEN, in this case - in the fourth, is equal to "plus infinity": . Constant ("two") positive, That's why: 
2) Calculate the limit 
Here is the senior degree again even, That's why: . But there is a "minus" in front ( negative constant –1), therefore: 
3) Calculate the limit 
The value of the limit depends only on . As you remember from school, "minus" "pops out" from under the odd degree, so infinitely large modulo negative number to an ODD power equals "minus infinity", in this case: .
Constant ("four") positive, Means: 
4) Calculate the limit
The first guy in the village has again odd degree, moreover, in the bosom negative constant, which means: Thus:
.
Example 5
Find the limit 
Using the above points, we conclude that there is uncertainty here. The numerator and denominator are of the same order of growth, which means that in the limit a finite number will be obtained. We learn the answer by discarding all the fry: 
The solution is trivial: 
Example 6
Find the limit 
This is a do-it-yourself example. Full solution and answer at the end of the lesson.
And now, perhaps the most subtle of the cases:
Example 7
Find the limit 
Considering the senior terms, we come to the conclusion that there is uncertainty here. The numerator is of a higher order of growth than the denominator, so we can immediately say that the limit is infinity. But what kind of infinity, "plus" or "minus"? The reception is the same - in the numerator and denominator we will get rid of the little things: 
We decide: 
Divide the numerator and denominator by
Example 15
Find the limit
This is a do-it-yourself example. An approximate sample of finishing at the end of the lesson.
A couple more interesting examples on the topic of variable substitution:
Example 16
Find the limit
Substituting one into the limit results in uncertainty. The replacement of the variable is already suggesting, but first we convert the tangent using the formula. Indeed, why do we need a tangent?
Note that , therefore . If it is not entirely clear, look at the sine values \u200b\u200bin trigonometric table. Thus, we immediately get rid of the factor , in addition, we get the more familiar uncertainty 0:0. It would be nice if our limit also tended to zero.
Let's replace:
If , then
Under the cosine we have "x", which also needs to be expressed through "te".
From the replacement we express: .
We complete the solution: 
(1) Performing the substitution
(2) Expand the brackets under the cosine.
(4) To organize first wonderful limit, artificially multiply the numerator by and the reciprocal of .
Task for independent solution:
Example 17
Find the limit
Full solution and answer at the end of the lesson.
These were simple tasks in their class; in practice, everything is worse, and, in addition to reduction formulas, one has to use different trigonometric formulas, as well as other tricks. In the article Complex Limits, I analyzed a couple of real examples =)
On the eve of the holiday, we will finally clarify the situation with one more common uncertainty:
Elimination of uncertainty "one to the power of infinity"
This uncertainty is “served” second wonderful limit, and in the second part of that lesson, we looked in great detail at standard examples of solutions that are found in practice in most cases. Now the picture with exhibitors will be completed, in addition, the final tasks of the lesson will be devoted to the limits-"tricks" in which it seems that it is necessary to apply the 2nd wonderful limit, although this is not at all the case.
The disadvantage of the two working formulas of the 2nd remarkable limit is that the argument must tend to "plus infinity" or to zero. But what if the argument tends to a different number?
The universal formula comes to the rescue (which is actually a consequence of the second remarkable limit):
Uncertainty can be eliminated by the formula:
Somewhere like I already explained what the square brackets mean. Nothing special, brackets are just brackets. Usually they are used to clearly highlight a mathematical notation.
Let's highlight the essential points of the formula:
1) It's about only about uncertainty and no other.
2) Argument "x" can tend to arbitrary value(and not only to zero or ), in particular, to "minus infinity" or to anyone final number.
Using this formula, you can solve all the examples of the lesson Remarkable Limits, which belong to the 2nd remarkable limit. For example, let's calculate the limit:
In this case 
, and according to the formula 
:
True, I don’t advise you to do this, in the tradition, you still use the “usual” design of the solution, if it can be applied. However using the formula is very convenient to check"classical" examples to the 2nd wonderful limit.
Function limit- number a will be the limit of some variable value if, in the process of its change, this variable approaches indefinitely a.
Or in other words, the number A is the limit of the function y=f(x) at the point x0, if for any sequence of points from the domain of definition of the function , not equal to x0, and which converges to the point x 0 (lim x n = x0), the sequence of corresponding values of the function converges to the number A.
Graph of a function whose limit with an argument that tends to infinity is L:
Meaning A is limit (limit value) of the function f(x) at the point x0 if for any sequence of points 
, which converges to x0, but which does not contain x0 as one of its elements (i.e. in the punctured neighborhood x0), the sequence of function values 
converges to A.
The limit of a function according to Cauchy.
Meaning A will be function limit f(x) at the point x0 if for any forward taken non-negative number ε a non-negative corresponding number will be found δ = δ(ε) such that for each argument x, satisfying the condition 0 < | x - x0 | < δ , the inequality | f(x) A |< ε .
It will be very simple if you understand the essence of the limit and the basic rules for finding it. That the limit of the function f(x) at x aspiring to a equals A, is written like this:
Moreover, the value to which the variable tends x, can be not only a number, but also infinity (∞), sometimes +∞ or -∞, or there may be no limit at all.
To understand how find the limits of a function, it is best to see examples of solutions.
We need to find the limits of the function f(x) = 1/x at:
x→ 2, x→ 0, x→ ∞.
Let's find the solution of the first limit. To do this, you can simply substitute x the number to which it aspires, i.e. 2, we get:
Find the second limit of the function. Here, substitute in pure form 0 instead of x it is impossible, because cannot be divided by 0. But we can take values close to zero, for example, 0.01; 0.001; 0.0001; 0.00001 and so on, with the value of the function f(x) will increase: 100; 1000; 10000; 100000 and so on. Thus, it can be understood that when x→ 0 the value of the function that is under the limit sign will increase indefinitely, i.e. strive for infinity. Which means:
Regarding the third limit. The same situation as in the previous case, it is impossible to substitute ∞ in its purest form. We need to consider the case of unlimited increase x. We alternately substitute 1000; 10000; 100000 and so on, we have that the value of the function f(x) = 1/x will decrease: 0.001; 0.0001; 0.00001; and so on, tending to zero. That's why:
It is necessary to calculate the limit of the function 
Starting to solve the second example, we see the uncertainty. From here we find the highest degree of the numerator and denominator - this is x 3, we take it out of brackets in the numerator and denominator and then reduce it by it:
Answer 
The first step in finding this limit, substitute the value 1 instead of x, resulting in the uncertainty . To solve it, we decompose the numerator into factors , we will do this by finding the roots of the quadratic equation x 2 + 2x - 3:
D \u003d 2 2 - 4 * 1 * (-3) \u003d 4 +12 \u003d 16→ √ D=√16 = 4
x 1,2 = (-2± 4) / 2→ x 1 \u003d -3;x2= 1.
So the numerator would be:
Answer 
This is the definition of its specific value or a specific area where the function falls, which is limited by the limit.
To decide the limits, follow the rules:
Having understood the essence and main limit decision rules, you will get a basic understanding of how to solve them.
Type and form uncertainty are the most common uncertainties that need to be addressed when solving limits.
Most of the tasks on the limits that come across to students just carry such uncertainties. To reveal them, or, more precisely, avoid ambiguities, there are several artificial methods for transforming the form of an 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 of quadratic equations and abbreviated multiplication formulas.
Species indeterminacy
Example 1
n is equal to 2. Therefore, we divide the numerator and denominator by term by:
.
Comment on the right side of the expression. The arrows and numbers indicate what the fractions tend to after substitution instead of n infinity values. Here, as in example 2, the degree n there is more in the denominator than in the numerator, as a result of which the whole fraction tends to an infinitesimal value or "super small number".
We get the answer: the limit of this function with a variable tending to infinity is .
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 course of the solution. In the numerator, we drive "x" under the root of the third degree, and so that its initial degree (1) remains unchanged, we assign it the same degree as the root, that is, 3. There are no arrows and additional numbers in this entry, so try mentally, but by analogy with the previous example, determine what the expressions in the numerator and denominator tend to after substituting infinity for "x".
We got the answer: the limit of this function with a variable tending to infinity is equal to zero.
Species indeterminacy
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 is a square trinomial, which we factorize by solving a quadratic equation (again a reference to solving quadratic equations):
Let's write down the expression obtained as a result of transformations and find the limit of the function:
Example 4 Uncover uncertainty and find the limit
Solution. The quotient limit theorem does not apply here, because
Therefore, we transform the fraction identically: by 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 Uncover uncertainty and find the limit
Solution. Direct value substitution x= 0 into a given function leads to an indeterminacy of the form 0/0. To reveal it, we perform identical transformations and, as a result, we obtain the desired limit: 
Example 6 Calculate 
Solution: use the limit theorems
Answer: 11
Example 7 Calculate 
Solution: in this example, the limits of the numerator and denominator at are 0:
; 
. We have obtained, therefore, the quotient limit theorem 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.
We expand the square trinomial in the numerator by the formula, where x 1 and x 2 are the roots of the trinomial. Factoring and denominator, reduce the fraction by (x-2), then apply Theorem 3.
Answer:
Example 8 Calculate
Solution: For , the numerator and denominator tend to infinity, so when applying Theorem 3 directly, we obtain the expression , which represents the uncertainty. To get rid of this kind of uncertainty, 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: The numerator and denominator tend to infinity. We divide the numerator and denominator by the highest power of the argument, i.e. x 5:
= 
The numerator of a fraction tends to 1, the denominator to 0, so the fraction tends to infinity.
Answer:
Example 11. Calculate
Solution: The numerator and denominator tend to infinity. We divide the numerator and denominator by the highest power of the argument, i.e. x 7:
Answer: 0
Derivative.
The derivative of the function y = f(x) with respect to the argument x the limit of the ratio of its increment y to the increment x of the argument x is called when the increment of the argument tends to zero: . If this limit is finite, then the function y = f(x) is called differentiable at the point x. If this limit exists, then we say that the function y = f(x) has an infinite derivative at x.
Derivatives of basic elementary functions:
1. (const)=0 9. 
3. 11. 
4. 
12. 
5. 13. 
6. 
14. 
Differentiation rules:
a) 
V) 
Example 1 Find the derivative of a function 
Solution: If we find the derivative of the second term by the rule of differentiation of a fraction, 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) is 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 touch point, f′(x0) is the value of the derivative of the function f(x) at the touch 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 the given point.
4. Function Ascending and Decreasing
If f′(x0)>0, then the function increases at the point x0. In the figure below, the function is increasing at x
If f′(x0)<0, то функция убывает в точке x0 (интервал x1
5. Local extrema of the function
The function f(x) has local maximum at a point x1 if there exists 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 a point x2 if there exists a neighborhood of the point x2 such that for all x from this neighborhood the inequality f(x2)≤f(x) holds.
6. Critical points
The 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) is increasing (f′(x)>0) for all x in some interval (a,x1] and decreasing (f′(x)<0) для всех x в интервале и возрастает (f′(x)>0) for all x from the interval )