1 and 2 wonderful limits are examples of solutions. The first remarkable limit: theory and examples
There are several remarkable limits, but the most famous are the first and second remarkable limits. The remarkable thing about these limits is that they are widely used and with their help one can find other limits encountered in numerous problems. This is what we will do in the practical part of this lesson. To solve problems by reducing them to the first or second remarkable limit, there is no need to reveal the uncertainties contained in them, since the values of these limits have long been deduced by great mathematicians.
The first remarkable limit is called the limit of the ratio of the sine of an infinitesimal arc to the same arc, expressed in radian measure:
Let's move on to solving problems at the first remarkable limit. Note: if there is a trigonometric function under the limit sign, this is an almost sure sign that this expression can be reduced to the first remarkable limit.
Example 1. Find the limit.
Solution. Substitution instead x zero leads to uncertainty:
.
The denominator is sine, therefore, the expression can be brought to the first remarkable limit. Let's start the transformation:
.
The denominator is the sine of three X, but the numerator has only one X, which means you need to get three X in the numerator. For what? To introduce 3 x = a and get the expression .
And we come to a variation of the first remarkable limit:
because it doesn’t matter which letter (variable) in this formula stands instead of X.
We multiply X by three and immediately divide:
.
In accordance with the first remarkable limit noticed, we replace the fractional expression:
Now we can finally solve this limit:
.
Example 2. Find the limit.
Solution. Direct substitution again leads to the “zero divided by zero” uncertainty:
.
To get the first remarkable limit, it is necessary that the x under the sine sign in the numerator and just the x in the denominator have the same coefficient. Let this coefficient be equal to 2. To do this, imagine the current coefficient for x as below, performing operations with fractions, we obtain:
.
Example 3. Find the limit.
Solution. When substituting, we again get the uncertainty “zero divided by zero”:
.
You probably already understand that from the original expression you can get the first wonderful limit multiplied by the first wonderful limit. To do this, we decompose the squares of the x in the numerator and the sine in the denominator into identical factors, and in order to get the same coefficients for the x and sine, we divide the x in the numerator by 3 and immediately multiply by 3. We get:
.
Example 4. Find the limit.
Solution. Once again we get the uncertainty “zero divided by zero”:
.
We can obtain the ratio of the first two remarkable limits. We divide both the numerator and the denominator by x. Then, so that the coefficients for sines and xes coincide, we multiply the upper x by 2 and immediately divide by 2, and multiply the lower x by 3 and immediately divide by 3. We get:
Example 5. Find the limit.
Solution. And again the uncertainty of “zero divided by zero”:
We remember from trigonometry that tangent is the ratio of sine to cosine, and the cosine of zero is equal to one. We carry out the transformations and get:
.
Example 6. Find the limit.
Solution. The trigonometric function under the sign of a limit again suggests the use of the first remarkable limit. We represent it as the ratio of sine to cosine.
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, in 95% of cases, part-time students 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. The only important thing is 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 “two” and “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. Let’s 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 thing, 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 base 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 .
Now, with a calm soul, let’s move on to consider wonderful limits.
looks like .
Instead of the variable x, various functions can be present, the main thing is that they tend to 0.
It is necessary to calculate the limit
As you can see, this limit is very similar to the first remarkable one, but this is not entirely true. In general, if you notice sin in the limit, then you should immediately think about whether it is possible to use the first remarkable limit.
According to our rule No. 1, we substitute zero instead of x:
We get uncertainty.
Now let's try to organize the first wonderful limit ourselves. To do this, let's do a simple combination:
So we organize the numerator and denominator to highlight 7x. Now the familiar remarkable limit has already appeared. It is advisable to highlight it when deciding:
Let's substitute the solution to the first remarkable example and get:
Simplifying the fraction:
Answer: 7/3.
As you can see, everything is very simple.
Looks like 
, where e = 2.718281828... is an irrational number.
Various functions may be present instead of the variable x, the main thing is that they tend to .
It is necessary to calculate the limit
Here we see the presence of a degree under the sign of a limit, which means it is possible to use a second remarkable limit.
As always, we will use rule No. 1 - substitute x instead of:
It can be seen that at x the base of the degree is , and the exponent is 4x > , i.e. we obtain an uncertainty of the form:
Let's use the second wonderful limit to reveal our uncertainty, but first we need to organize it. As you can see, we need to achieve presence in the indicator, for which we raise the base to the power of 3x, and at the same time to the power of 1/3x, so that the expression does not change:
Don't forget to highlight our wonderful limit:
That's what they really are wonderful limits!
If you still have any questions about the first and second wonderful limits, then feel free to ask them in the comments.
We will answer everyone as much as possible.
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The first remarkable limit is the following equality:
\begin(equation)\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1 \end(equation)
Since for $\alpha\to(0)$ we have $\sin\alpha\to(0)$, they say that the first remarkable limit reveals an uncertainty of the form $\frac(0)(0)$. Generally speaking, in formula (1), instead of the variable $\alpha$, any expression can be placed under the sine sign and in the denominator, as long as two conditions are met:
- The expressions under the sine sign and in the denominator simultaneously tend to zero, i.e. there is uncertainty of the form $\frac(0)(0)$.
- The expressions under the sine sign and in the denominator are the same.
Corollaries from the first remarkable limit are also often used:
\begin(equation) \lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0) )\frac(\arcsin\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1 \end(equation)
Eleven examples are solved on this page. Example No. 1 is devoted to the proof of formulas (2)-(4). Examples No. 2, No. 3, No. 4 and No. 5 contain solutions with detailed comments. Examples No. 6-10 contain solutions with virtually no comments, because detailed explanations were given in previous examples. The solution uses some trigonometric formulas that can be found.
Let me note that the presence of trigonometric functions coupled with the uncertainty $\frac (0) (0)$ does not necessarily mean the application of the first remarkable limit. Sometimes simple trigonometric transformations are sufficient - for example, see.
Example No. 1
Prove that $\lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1$, $\lim_(\alpha\to(0))\frac(\arcsin\alpha )(\alpha)=1$, $\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$.
a) Since $\tg\alpha=\frac(\sin\alpha)(\cos\alpha)$, then:
$$ \lim_(\alpha\to(0))\frac(\tg(\alpha))(\alpha)=\left|\frac(0)(0)\right| =\lim_(\alpha\to(0))\frac(\sin(\alpha))(\alpha\cos(\alpha)) $$
Since $\lim_(\alpha\to(0))\cos(0)=1$ and $\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1$ , That:
$$ \lim_(\alpha\to(0))\frac(\sin(\alpha))(\alpha\cos(\alpha)) =\frac(\displaystyle\lim_(\alpha\to(0)) \frac(\sin(\alpha))(\alpha))(\displaystyle\lim_(\alpha\to(0))\cos(\alpha)) =\frac(1)(1) =1. $$
b) Let's make the change $\alpha=\sin(y)$. Since $\sin(0)=0$, then from the condition $\alpha\to(0)$ we have $y\to(0)$. In addition, there is a neighborhood of zero in which $\arcsin\alpha=\arcsin(\sin(y))=y$, so:
$$ \lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=\left|\frac(0)(0)\right| =\lim_(y\to(0))\frac(y)(\sin(y)) =\lim_(y\to(0))\frac(1)(\frac(\sin(y))( y)) =\frac(1)(\displaystyle\lim_(y\to(0))\frac(\sin(y))(y)) =\frac(1)(1) =1. $$
The equality $\lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=1$ has been proven.
c) Let's make the replacement $\alpha=\tg(y)$. Since $\tg(0)=0$, then the conditions $\alpha\to(0)$ and $y\to(0)$ are equivalent. In addition, there is a neighborhood of zero in which $\arctg\alpha=\arctg\tg(y))=y$, therefore, based on the results of point a), we will have:
$$ \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=\left|\frac(0)(0)\right| =\lim_(y\to(0))\frac(y)(\tg(y)) =\lim_(y\to(0))\frac(1)(\frac(\tg(y))( y)) =\frac(1)(\displaystyle\lim_(y\to(0))\frac(\tg(y))(y)) =\frac(1)(1) =1. $$
The equality $\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$ has been proven.
Equalities a), b), c) are often used along with the first remarkable limit.
Example No. 2
Calculate the limit $\lim_(x\to(2))\frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)( x+7))$.
Since $\lim_(x\to(2))\frac(x^2-4)(x+7)=\frac(2^2-4)(2+7)=0$ and $\lim_( x\to(2))\sin\left(\frac(x^2-4)(x+7)\right)=\sin(0)=0$, i.e. and both the numerator and denominator of the fraction simultaneously tend to zero, then here we are dealing with an uncertainty of the form $\frac(0)(0)$, i.e. done. In addition, it is clear that the expressions under the sine sign and in the denominator coincide (i.e., and is satisfied):
So, both conditions listed at the beginning of the page are met. It follows from this that the formula is applicable, i.e. $\lim_(x\to(2)) \frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)(x+ 7))=1$.
Answer: $\lim_(x\to(2))\frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)(x +7))=1$.
Example No. 3
Find $\lim_(x\to(0))\frac(\sin(9x))(x)$.
Since $\lim_(x\to(0))\sin(9x)=0$ and $\lim_(x\to(0))x=0$, then we are dealing with an uncertainty of the form $\frac(0 )(0)$, i.e. done. However, the expressions under the sine sign and in the denominator do not coincide. Here you need to adjust the expression in the denominator to the desired form. We need the expression $9x$ to be in the denominator, then it will become true. Essentially, we're missing a factor of $9$ in the denominator, which isn't that hard to enter—just multiply the expression in the denominator by $9$. Naturally, to compensate for multiplication by $9$, you will have to immediately divide by $9$:
$$ \lim_(x\to(0))\frac(\sin(9x))(x)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\sin(9x))(9x\cdot\frac(1)(9)) =9\lim_(x\to(0))\frac(\sin (9x))(9x)$$
Now the expressions in the denominator and under the sine sign coincide. Both conditions for the limit $\lim_(x\to(0))\frac(\sin(9x))(9x)$ are satisfied. Therefore, $\lim_(x\to(0))\frac(\sin(9x))(9x)=1$. And this means that:
$$ 9\lim_(x\to(0))\frac(\sin(9x))(9x)=9\cdot(1)=9. $$
Answer: $\lim_(x\to(0))\frac(\sin(9x))(x)=9$.
Example No. 4
Find $\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))$.
Since $\lim_(x\to(0))\sin(5x)=0$ and $\lim_(x\to(0))\tg(8x)=0$, here we are dealing with uncertainty of the form $\frac(0)(0)$. However, the form of the first remarkable limit is violated. A numerator containing $\sin(5x)$ requires a denominator of $5x$. In this situation, the easiest way is to divide the numerator by $5x$, and immediately multiply by $5x$. In addition, we will perform a similar operation with the denominator, multiplying and dividing $\tg(8x)$ by $8x$:
$$\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x) )$$
Reducing by $x$ and taking the constant $\frac(5)(8)$ outside the limit sign, we get:
$$ \lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x )) =\frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))( 8x)) $$
Note that $\lim_(x\to(0))\frac(\sin(5x))(5x)$ fully satisfies the requirements for the first remarkable limit. To find $\lim_(x\to(0))\frac(\tg(8x))(8x)$ the following formula is applicable:
$$ \frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))(8x )) =\frac(5)(8)\cdot\frac(\displaystyle\lim_(x\to(0))\frac(\sin(5x))(5x))(\displaystyle\lim_(x\to (0))\frac(\tg(8x))(8x)) =\frac(5)(8)\cdot\frac(1)(1) =\frac(5)(8). $$
Answer: $\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))=\frac(5)(8)$.
Example No. 5
Find $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)$.
Since $\lim_(x\to(0))(\cos(5x)-\cos^3(5x))=1-1=0$ (remember that $\cos(0)=1$) and $\lim_(x\to(0))x^2=0$, then we are dealing with uncertainty of the form $\frac(0)(0)$. However, in order to apply the first remarkable limit, you should get rid of the cosine in the numerator, moving on to sines (in order to then apply the formula) or tangents (in order to then apply the formula). This can be done with the following transformation:
$$\cos(5x)-\cos^3(5x)=\cos(5x)\cdot\left(1-\cos^2(5x)\right)$$ $$\cos(5x)-\cos ^3(5x)=\cos(5x)\cdot\left(1-\cos^2(5x)\right)=\cos(5x)\cdot\sin^2(5x).$$
Let's go back to the limit:
$$ \lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\cos(5x)\cdot\sin^2(5x))(x^2) =\lim_(x\to(0))\left(\cos (5x)\cdot\frac(\sin^2(5x))(x^2)\right) $$
The fraction $\frac(\sin^2(5x))(x^2)$ is already close to the form required for the first remarkable limit. Let's work a little with the fraction $\frac(\sin^2(5x))(x^2)$, adjusting it to the first remarkable limit (note that the expressions in the numerator and under the sine must match):
$$\frac(\sin^2(5x))(x^2)=\frac(\sin^2(5x))(25x^2\cdot\frac(1)(25))=25\cdot\ frac(\sin^2(5x))(25x^2)=25\cdot\left(\frac(\sin(5x))(5x)\right)^2$$
Let's return to the limit in question:
$$ \lim_(x\to(0))\left(\cos(5x)\cdot\frac(\sin^2(5x))(x^2)\right) =\lim_(x\to(0 ))\left(25\cos(5x)\cdot\left(\frac(\sin(5x))(5x)\right)^2\right)=\\ =25\cdot\lim_(x\to( 0))\cos(5x)\cdot\lim_(x\to(0))\left(\frac(\sin(5x))(5x)\right)^2 =25\cdot(1)\cdot( 1^2) =25. $$
Answer: $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=25$.
Example No. 6
Find the limit $\lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))$.
Since $\lim_(x\to(0))(1-\cos(6x))=0$ and $\lim_(x\to(0))(1-\cos(2x))=0$, then we are dealing with uncertainty $\frac(0)(0)$. Let us reveal it with the help of the first remarkable limit. To do this, let's move from cosines to sines. Since $1-\cos(2\alpha)=2\sin^2(\alpha)$, then:
$$1-\cos(6x)=2\sin^2(3x);\;1-\cos(2x)=2\sin^2(x).$$
Passing to sines in the given limit, we will have:
$$ \lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(2\sin^2(3x))(2\sin^2(x)) =\lim_(x\to(0))\frac(\sin^ 2(3x))(\sin^2(x))=\\ =\lim_(x\to(0))\frac(\frac(\sin^2(3x))((3x)^2)\ cdot(3x)^2)(\frac(\sin^2(x))(x^2)\cdot(x^2)) =\lim_(x\to(0))\frac(\left(\ frac(\sin(3x))(3x)\right)^2\cdot(9x^2))(\left(\frac(\sin(x))(x)\right)^2\cdot(x^ 2)) =9\cdot\frac(\displaystyle\lim_(x\to(0))\left(\frac(\sin(3x))(3x)\right)^2)(\displaystyle\lim_(x \to(0))\left(\frac(\sin(x))(x)\right)^2) =9\cdot\frac(1^2)(1^2) =9. $$
Answer: $\lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))=9$.
Example No. 7
Calculate the limit $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)$ subject to $\alpha\neq\ beta$.
Detailed explanations were given earlier, but here we simply note that again there is uncertainty $\frac(0)(0)$. Let's move from cosines to sines using the formula
$$\cos\alpha-\cos\beta=-2\sin\frac(\alpha+\beta)(2)\cdot\sin\frac(\alpha-\beta)(2).$$
Using this formula, we get:
$$ \lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)=\left|\frac(0)( 0)\right| =\lim_(x\to(0))\frac(-2\sin\frac(\alpha(x)+\beta(x))(2)\cdot\sin\frac(\alpha(x)-\ beta(x))(2))(x^2)=\\ =-2\cdot\lim_(x\to(0))\frac(\sin\left(x\cdot\frac(\alpha+\beta )(2)\right)\cdot\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x^2) =-2\cdot\lim_(x\to( 0))\left(\frac(\sin\left(x\cdot\frac(\alpha+\beta)(2)\right))(x)\cdot\frac(\sin\left(x\cdot\frac (\alpha-\beta)(2)\right))(x)\right)=\\ =-2\cdot\lim_(x\to(0))\left(\frac(\sin\left(x \cdot\frac(\alpha+\beta)(2)\right))(x\cdot\frac(\alpha+\beta)(2))\cdot\frac(\alpha+\beta)(2)\cdot\frac (\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x\cdot\frac(\alpha-\beta)(2))\cdot\frac(\alpha- \beta)(2)\right)=\\ =-\frac((\alpha+\beta)\cdot(\alpha-\beta))(2)\lim_(x\to(0))\frac(\ sin\left(x\cdot\frac(\alpha+\beta)(2)\right))(x\cdot\frac(\alpha+\beta)(2))\cdot\lim_(x\to(0)) \frac(\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x\cdot\frac(\alpha-\beta)(2)) =-\frac(\ alpha^2-\beta^2)(2)\cdot(1)\cdot(1) =\frac(\beta^2-\alpha^2)(2). $$
Answer: $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)=\frac(\beta^2-\ alpha^2)(2)$.
Example No. 8
Find the limit $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)$.
Since $\lim_(x\to(0))(\tg(x)-\sin(x))=0$ (remember that $\sin(0)=\tg(0)=0$) and $\lim_(x\to(0))x^3=0$, then here we are dealing with uncertainty of the form $\frac(0)(0)$. Let's break it down as follows:
$$ \lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(x))(\cos(x))-\sin(x))(x^3) =\lim_(x\to( 0))\frac(\sin(x)\cdot\left(\frac(1)(\cos(x))-1\right))(x^3) =\lim_(x\to(0)) \frac(\sin(x)\cdot\left(1-\cos(x)\right))(x^3\cdot\cos(x))=\\ =\lim_(x\to(0)) \frac(\sin(x)\cdot(2)\sin^2\frac(x)(2))(x^3\cdot\cos(x)) =\frac(1)(2)\cdot\ lim_(x\to(0))\left(\frac(\sin(x))(x)\cdot\left(\frac(\sin\frac(x)(2))(\frac(x)( 2))\right)^2\cdot\frac(1)(\cos(x))\right) =\frac(1)(2)\cdot(1)\cdot(1^2)\cdot(1 ) =\frac(1)(2). $$
Answer: $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\frac(1)(2)$.
Example No. 9
Find the limit $\lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))$.
Since $\lim_(x\to(3))(1-\cos(x-3))=0$ and $\lim_(x\to(3))(x-3)\tg\frac(x -3)(2)=0$, then there is uncertainty of the form $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to make a change of variable in such a way that the new variable tends to zero (note that in the formulas the variable $\alpha \to 0$). The easiest way is to introduce the variable $t=x-3$. However, for the sake of convenience of further transformations (this benefit can be seen in the course of the solution below), it is worth making the following replacement: $t=\frac(x-3)(2)$. I note that both replacements are applicable in this case, it’s just that the second replacement will allow you to work less with fractions. Since $x\to(3)$, then $t\to(0)$.
$$ \lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))=\left|\frac (0)(0)\right| =\left|\begin(aligned)&t=\frac(x-3)(2);\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\frac(1-\cos(2t))(2t\cdot\tg(t)) =\lim_(t\to(0))\frac(2\sin^ 2t)(2t\cdot\tg(t)) =\lim_(t\to(0))\frac(\sin^2t)(t\cdot\tg(t))=\\ =\lim_(t\ to(0))\frac(\sin^2t)(t\cdot\frac(\sin(t))(\cos(t))) =\lim_(t\to(0))\frac(\sin (t)\cos(t))(t) =\lim_(t\to(0))\left(\frac(\sin(t))(t)\cdot\cos(t)\right) =\ lim_(t\to(0))\frac(\sin(t))(t)\cdot\lim_(t\to(0))\cos(t) =1\cdot(1) =1. $$
Answer: $\lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))=1$.
Example No. 10
Find the limit $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2 )$.
Once again we are dealing with uncertainty $\frac(0)(0)$. Before proceeding to its expansion, it is convenient to make a change of variable in such a way that the new variable tends to zero (note that in the formulas the variable is $\alpha\to(0)$). The easiest way is to introduce the variable $t=\frac(\pi)(2)-x$. Since $x\to\frac(\pi)(2)$, then $t\to(0)$:
$$ \lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2) =\left|\frac(0)(0)\right| =\left|\begin(aligned)&t=\frac(\pi)(2)-x;\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\frac(1-\sin\left(\frac(\pi)(2)-t\right))(t^2) =\lim_(t\to(0 ))\frac(1-\cos(t))(t^2)=\\ =\lim_(t\to(0))\frac(2\sin^2\frac(t)(2))( t^2) =2\lim_(t\to(0))\frac(\sin^2\frac(t)(2))(t^2) =2\lim_(t\to(0))\ frac(\sin^2\frac(t)(2))(\frac(t^2)(4)\cdot(4)) =\frac(1)(2)\cdot\lim_(t\to( 0))\left(\frac(\sin\frac(t)(2))(\frac(t)(2))\right)^2 =\frac(1)(2)\cdot(1^2 ) =\frac(1)(2). $$
Answer: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2) =\frac(1)(2)$.
Example No. 11
Find the limits $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)$, $\lim_(x\to\frac(2\ pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)$.
In this case we don't have to use the first wonderful limit. Please note that both the first and second limits contain only trigonometric functions and numbers. Often in examples of this kind it is possible to simplify the expression located under the limit sign. Moreover, after the aforementioned simplification and reduction of some factors, the uncertainty disappears. I gave this example for only one purpose: to show that the presence of trigonometric functions under the limit sign does not necessarily mean the use of the first remarkable limit.
Since $\lim_(x\to\frac(\pi)(2))(1-\sin(x))=0$ (remember that $\sin\frac(\pi)(2)=1$ ) and $\lim_(x\to\frac(\pi)(2))\cos^2x=0$ (let me remind you that $\cos\frac(\pi)(2)=0$), then we have dealing with uncertainty of the form $\frac(0)(0)$. However, this does not mean that we will need to use the first wonderful limit. To reveal the uncertainty, it is enough to take into account that $\cos^2x=1-\sin^2x$:
$$ \lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x) =\left|\frac(0)(0)\right| =\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(1-\sin^2x) =\lim_(x\to\frac(\pi)( 2))\frac(1-\sin(x))((1-\sin(x))(1+\sin(x))) =\lim_(x\to\frac(\pi)(2) )\frac(1)(1+\sin(x)) =\frac(1)(1+1) =\frac(1)(2). $$
There is a similar solution in Demidovich’s solution book (No. 475). As for the second limit, as in the previous examples in this section, we have an uncertainty of the form $\frac(0)(0)$. Why does it arise? It arises because $\tg\frac(2\pi)(3)=-\sqrt(3)$ and $2\cos\frac(2\pi)(3)=-1$. We use these values to transform the expressions in the numerator and denominator. The goal of our actions is to write down the sum in the numerator and denominator as a product. By the way, often within a similar type it is convenient to change a variable, made in such a way that the new variable tends to zero (see, for example, examples No. 9 or No. 10 on this page). However, in this example there is no point in replacing, although if desired, replacing the variable $t=x-\frac(2\pi)(3)$ is not difficult to implement.
$$ \lim_(x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1) =\lim_(x\ to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cdot\left(\cos(x)+\frac(1)(2)\right )) =\lim_(x\to\frac(2\pi)(3))\frac(\tg(x)-\tg\frac(2\pi)(3))(2\cdot\left(\ cos(x)-\cos\frac(2\pi)(3)\right))=\\ =\lim_(x\to\frac(2\pi)(3))\frac(\frac(\sin \left(x-\frac(2\pi)(3)\right))(\cos(x)\cos\frac(2\pi)(3)))(-4\sin\frac(x+\frac (2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2)) =\lim_(x\to\frac(2\pi)(3 ))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac(x+\frac(2\pi)(3))(2)\ sin\frac(x-\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3))=\\ =\lim_(x\to\frac (2\pi)(3))\frac(2\sin\frac(x-\frac(2\pi)(3))(2)\cos\frac(x-\frac(2\pi)(3 ))(2))(-4\sin\frac(x+\frac(2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2) \cos(x)\cos\frac(2\pi)(3)) =\lim_(x\to\frac(2\pi)(3))\frac(\cos\frac(x-\frac(2 \pi)(3))(2))(-2\sin\frac(x+\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3 ))=\\ =\frac(1)(-2\cdot\frac(\sqrt(3))(2)\cdot\left(-\frac(1)(2)\right)\cdot\left( -\frac(1)(2)\right)) =-\frac(4)(\sqrt(3)). $$
As you can see, we didn't have to apply the first wonderful limit. Of course, you can do this if you want (see note below), but it is not necessary.
What is the solution using the first remarkable limit? show\hide
Using the first remarkable limit we get:
$$ \lim_(x\to\frac(2\pi)(3))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac (x+\frac(2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi )(3))=\\ =\lim_(x\to\frac(2\pi)(3))\left(\frac(\sin\left(x-\frac(2\pi)(3)\ right))(x-\frac(2\pi)(3))\cdot\frac(1)(\frac(\sin\frac(x-\frac(2\pi)(3))(2)) (\frac(x-\frac(2\pi)(3))(2)))\cdot\frac(1)(-2\sin\frac(x+\frac(2\pi)(3))( 2)\cos(x)\cos\frac(2\pi)(3))\right) =1\cdot(1)\cdot\frac(1)(-2\cdot\frac(\sqrt(3) )(2)\cdot\left(-\frac(1)(2)\right)\cdot\left(-\frac(1)(2)\right)) =-\frac(4)(\sqrt( 3)). $$
Answer: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.
Proof:
Let us first prove the theorem for the case of the sequence 
According to Newton's binomial formula:
Assuming we get
From this equality (1) it follows that as n increases, the number of positive terms on the right side increases. In addition, as n increases, the number decreases, so the values 
are increasing. Therefore the sequence 
increasing, and (2)*We show that it is bounded. Replace each parenthesis on the right side of the equality with one, the right side will increase, and we get the inequality
Let's strengthen the resulting inequality, replace 3,4,5, ..., standing in the denominators of the fractions, with the number 2: We find the sum in brackets using the formula for the sum of the terms of a geometric progression: Therefore 
(3)*
So, the sequence is bounded from above, and inequalities (2) and (3) are satisfied: 
Therefore, based on the Weierstrass theorem (criterion for the convergence of a sequence), the sequence 
monotonically increases and is limited, which means it has a limit, denoted by the letter e. Those. 
Knowing that the second remarkable limit is true for natural values of x, we prove the second remarkable limit for real x, that is, we prove that 
. Let's consider two cases:
1. Let Each value of x be enclosed between two positive integers: ,where is the integer part of x. => =>
If , then Therefore, according to the limit 
We have
Based on the criterion (about the limit of an intermediate function) of the existence of limits 
2. Let . Let's make the substitution − x = t, then
From these two cases it follows that 
for real x.
Consequences:
9 .) Comparison of infinitesimals. The theorem on replacing infinitesimals with equivalent ones in the limit and the theorem on the main part of infinitesimals.
Let the functions a( x) and b( x) – b.m. at x ® x 0 .
DEFINITIONS.
1)a( x) called infinitesimal higher order than b (x) If
Write down: a( x) = o(b( x)) .
2)a( x) And b( x)are called infinitesimals of the same order, If
where CÎℝ and C¹ 0 .
Write down: a( x) = O(b( x)) .
3)a( x) And b( x) are called equivalent , If
Write down: a( x) ~ b( x).
4)a( x) called infinitesimal of order k relative
absolutely infinitesimal b( x),
if infinitesimal a( x)And(b( x))k have the same order, i.e. If
where CÎℝ and C¹ 0 .
THEOREM 6 (on replacing infinitesimals with equivalent ones).
Let a( x), b( x), a 1 ( x), b 1 ( x)– b.m. at x ® x 0 . If a( x) ~ a 1 ( x), b( x) ~ b 1 ( x),
That 
Proof: Let a( x) ~ a 1 ( x), b( x) ~ b 1 ( x), Then
THEOREM 7 (about the main part of the infinitesimal).
Let a( x)And b( x)– b.m. at x ® x 0 , and b( x)– b.m. higher order than a( x).
= , a since b( x) – higher order than a( x), then, i.e. 
from it is clear that a( x) + b( x) ~ a( x)
10) Continuity of a function at a point (in the language of epsilon-delta, geometric limits) One-sided continuity. Continuity on an interval, on a segment. Properties of continuous functions.
1. Basic definitions
Let f(x) is defined in some neighborhood of the point x 0 .
DEFINITION 1. Function f(x) called continuous at a point x 0 if the equality is true
Notes.
1) By virtue of Theorem 5 §3, equality (1) can be written in the form
Condition (2) – definition of continuity of a function at a point in the language of one-sided limits.
2) Equality (1) can also be written as:
They say: “if a function is continuous at a point x 0, then the sign of the limit and the function can be swapped."
DEFINITION 2 (in e-d language).
Function f(x) called continuous at a point x 0 If"e>0 $d>0 such, What
if xОU( x 0 , d) (i.e. | x – x 0 | < d),
then f(x)ÎU( f(x 0), e) (i.e. | f(x) – f(x 0) | < e).
Let x, x 0 Î D(f) (x 0 – fixed, x – arbitrary)
Let's denote: D x= x – x 0 – argument increment
D f(x 0) = f(x) – f(x 0) – increment of function at pointx 0
DEFINITION 3 (geometric).
Function f(x) on called continuous at a point
x 0
if at this point an infinitesimal increment in the argument corresponds to an infinitesimal increment in the function, i.e.
Let the function f(x) is defined on the interval [ x 0 ; x 0 + d) (on the interval ( x 0 – d; x 0 ]).
DEFINITION. Function f(x) called continuous at a point
x 0 on right
(left
), if the equality is true
It's obvious that f(x) is continuous at the point x 0 Û f(x) is continuous at the point x 0 right and left.
DEFINITION. Function f(x) called continuous for an interval e ( a; b) if it is continuous at every point of this interval.
Function f(x) is called continuous on the segment [a; b] if it is continuous on the interval (a; b) and has one-way continuity at the boundary points(i.e. continuous at the point a on the right, at the point b- left).
11) Break points, their classification
DEFINITION. If function f(x) defined in some neighborhood of point x 0 , but is not continuous at this point, then f(x) called discontinuous at point x 0 , and the point itself x 0 called the break point functions f(x) .
Notes.
1) f(x) can be defined in an incomplete neighborhood of the point x 0 .
Then consider the corresponding one-sided continuity of the function.
2) From the definition of Þ point x 0 is the break point of the function f(x) in two cases:
a) U( x 0 , d)О D(f) , but for f(x) equality does not hold
b) U * ( x 0 , d)О D(f) .
For elementary functions, only case b) is possible.
Let x 0 – function break point f(x) .
DEFINITION. Point x 0 called break point I sort of if function f(x)has finite limits on the left and right at this point.
If these limits are equal, then point x 0 called removable break point , otherwise - jump point .
DEFINITION. Point x 0 called break point II sort of if at least one of the one-sided limits of the function f(x)at this point is equal¥ or doesn't exist.
12) Properties of functions continuous on an interval (theorems of Weierstrass (without proof) and Cauchy
Weierstrass's theorem
Let the function f(x) be continuous on the interval, then
1)f(x)is limited to
2) f(x) takes its smallest and largest value on the interval
Definition: The value of the function m=f is called the smallest if m≤f(x) for any x€ D(f).
The value of the function m=f is said to be greatest if m≥f(x) for any x € D(f).
The function can take on the smallest/largest value at several points of the segment.
f(x 3)=f(x 4)=max
Cauchy's theorem.
Let the function f(x) be continuous on the segment and x be the number contained between f(a) and f(b), then there is at least one point x 0 € such that f(x 0)= g