Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
2. Calculate the limit of a number sequence:
3. Calculate the limit of a number sequence:
4. Calculate the limit of a number sequence:
5. Calculate the limit of a number sequence:
6. Calculate the limit of a number sequence:
7. Calculate the limit of a number sequence:
8. Calculate the limit of a number sequence:
9. Calculate the limit of a number sequence:
10. Calculate the limit of a number sequence:
11. Calculate the limit of a number sequence:
1) From the numerator and denominator, select the factor that makes the greatest contribution and reduce by it
2) In this type of example, you need to remove the factor to the greatest extent from under the root in the denominator
3) It is necessary to expand to the greatest common factorial
4) In this example it grows much faster, so we single it out as the biggest factor
5) The quantities and tend to zero at . Based on this we calculate the limit
The solution to most of these examples is to find the dominant factor. If it is in the numerator, then the boundary goes to infinity, in the denominator - to zero. And only when both here and there can you reduce the fraction by this factor and get the limit in the form of a constant.
Exercise:
1. Analyze the solutions of the considered examples
2. Calculate the following limits:
Section 2. Beginnings of mathematical analysis
(Independent work 48 hours.)
2.1. Implicit function derivative (4 hours).
Example 1. Find the derivative of an implicit function
Solution. Since y is a function of X, then we will consider y 2 as a complex function of X. Hence, . Differentiating by X both sides of this equation, we obtain, i.e.
Example 2. Find the derivative of an implicit function
Solution. Differentiating by X
Example 3. Find the derivative of an implicit function
Solution. Differentiating by X both sides of this equation, we get
1. Find the derivative f ’(x).
2. Find stationary points of this function, i.e. points at which
3. Find the second derivative f ’’(x).
4. Investigate the sign of the second derivative at each of the stationary points. If the second derivative turns out to be negative, then the function at such a point has a maximum, and if it is positive, then it has a minimum. If the second derivative is equal to zero, then the extremum of the function must be sought using the first derivative.
5. Calculate the values of the function at the extremum points.
Example. Examine the extremum using the second derivative of the function: f(x) = x 2 – 2x - 3.
Solution: Find the derivative: f ‘(x) = 2x - 2.
Solving the equation f ’(x) = 0, we obtain a stationary point x =1. Let us now find the second derivative: f ’’(x) = 2.
Since the second derivative at) = x 2 – 2x - 3. at the stationary point is positive, f’’(1) = 2 > 0, then at x = 1 the function has a minimum: f min = f(1) = -4.
Answer: The minimum point has coordinates (1; -4).
Tasks.
1. Consider and analyze the considered solutions to examples on these topics.
2. Investigate for extremum using the second derivative of the function:
a) f(x) = 1 – x 4;
b) f(x) = x 3 - 1;
2.3. Application of the derivative to solving physical problems (11 hours).
2.4. Compiling crossnumbers on the topic “Definite Integral”
2.5 Calculation of body volume and arc length of a curve (12 hours)
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: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3))$.
Definition of limits of sequence and function, properties of limits, first and second remarkable limits, examples.
Constant number A called limit sequences(x n ), if for any arbitrarily small positive number ε > 0 there is a number N such that all values x n, for which n>N, satisfy the inequality
Write it down as follows: or x n → a.
Inequality (6.1) is equivalent to the double inequality
a - ε< x n < a + ε которое означает, что точки x n, starting from some number n>N, lie inside the interval (a-ε , a+ε), i.e. fall into any small ε-neighborhood of the point A.
A sequence having a limit is called convergent, otherwise - divergent.
The concept of a function limit is a generalization of the concept of a sequence limit, since the limit of a sequence can be considered as the limit of a function x n = f(n) of an integer argument n.
Let the function f(x) be given and let a - limit point domain of definition of this function D(f), i.e. such a point, any neighborhood of which contains points of the set D(f) other than a. Dot a may or may not belong to the set D(f).
Definition 1. The constant number A is called limit functions f(x) at x→ a, if for any sequence (x n ) of argument values tending to A, the corresponding sequences (f(x n)) have the same limit A.
This definition is called determining the limit of a function according to Heine, or " in sequence language”.
Definition 2. The constant number A is called limit functions f(x) at x→a, if, given an arbitrary, arbitrarily small positive number ε, one can find such δ >0 (depending on ε) that for all x, lying in the ε-neighborhood of the number A, i.e. For x, satisfying the inequality
0 < x-a < ε , значения функции f(x) будут лежать в
ε-окрестности числа А, т.е. |f(x)-A| < ε
This definition is called by defining the limit of a function according to Cauchy, or “in the language ε - δ"
Definitions 1 and 2 are equivalent. If the function f(x) as x → a has limit, equal to A, this is written in the form
In the event that the sequence (f(x n)) increases (or decreases) without limit for any method of approximation x to your limit A, then we will say that the function f(x) has infinite limit, and write it in the form:
A variable (i.e. a sequence or function) whose limit is zero is called infinitely small.
A variable whose limit is equal to infinity is called infinitely large.
To find the limit in practice, the following theorems are used.
Theorem 1 . If every limit exists
(6.4)
(6.5)
(6.6)
Comment. Expressions of the form 0/0, ∞/∞, ∞-∞ 0*∞ are uncertain, for example, the ratio of two infinitesimal or infinitely large quantities, and finding a limit of this type is called “uncertainty disclosure.”
Theorem 2.
those. one can go to the limit based on the power with a constant exponent, in particular, 
Theorem 3.
(6.11)
Where e» 2.7 - base of natural logarithm. Formulas (6.10) and (6.11) are called the first remarkable limit and the second remarkable limit.
The consequences of formula (6.11) are also used in practice:
(6.12)
(6.13)
(6.14)
in particular the limit,
If x → a and at the same time x > a, then write x →a + 0. If, in particular, a = 0, then instead of the symbol 0+0 write +0. Similarly, if x→a and at the same time x 
. The function f(x) is called continuous at the point x 0 if limit
(6.15)
Condition (6.15) can be rewritten as:
that is, passage to the limit under the sign of a function is possible if it is continuous at a given point.
If equality (6.15) is violated, then we say that at x = x o function f(x) It has gap Consider the function y = 1/x. The domain of definition of this function is the set R, except for x = 0. The point x = 0 is a limit point of the set D(f), since in any neighborhood of it, i.e. in any open interval containing the point 0, there are points from D(f), but it itself does not belong to this set. The value f(x o)= f(0) is not defined, so at the point x o = 0 the function has a discontinuity.
The function f(x) is called continuous on the right at the point x o if the limit
And continuous on the left at the point x o, if the limit
Continuity of a function at a point x o is equivalent to its continuity at this point both to the right and to the left.
In order for the function to be continuous at the point x o, for example, on the right, it is necessary, firstly, that there be a finite limit, and secondly, that this limit be equal to f(x o). Therefore, if at least one of these two conditions is not met, then the function will have a discontinuity.
1. If the limit exists and is not equal to f(x o), then they say that function f(x) at the point x o has rupture of the first kind, or leap.
2. If the limit is +∞ or -∞ or does not exist, then they say that in point x o the function has a discontinuity second kind.
For example, the function y = ctg x as x → +0 has a limit equal to +∞, which means that at the point x=0 it has a discontinuity of the second kind. Function y = E(x) (integer part of x) at points with whole abscissas has discontinuities of the first kind, or jumps.
A function that is continuous at every point in the interval is called continuous V . A continuous function is represented by a solid curve.
Many problems associated with the continuous growth of some quantity lead to the second remarkable limit. Such tasks, for example, include: growth of deposits according to the law of compound interest, growth of the country's population, decay of radioactive substances, proliferation of bacteria, etc.
Let's consider example of Ya. I. Perelman, giving an interpretation of the number e in the compound interest problem. Number e there is a limit 
. 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 denize 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 second because the limit
Example 3.1. Using the definition of the limit of a number sequence, prove that the sequence x n =(n-1)/n has a limit equal to 1.
Solution. We need to prove that, no matter what ε > 0 we take, for it there is a natural number N such that for all n > N the inequality |x n -1|< ε
Take any ε > 0. Since x n -1 =(n+1)/n - 1= 1/n, then to find N it is enough to solve the inequality 1/n<ε. Отсюда n>1/ε and, therefore, N can be taken to be the integer part of 1/ε N = E(1/ε). We have thereby proven that the limit .
Example 3.2. Find the limit of a sequence given by a common term
.
Solution. Let's apply the limit of the sum theorem and find the limit of each term. As n → ∞, the numerator and denominator of each term tend to infinity, and we cannot directly apply the quotient limit theorem. Therefore, first we transform x n, dividing the numerator and denominator of the first term by n 2, and the second on n. Then, applying the limit of the quotient and the limit of the sum theorem, we find:
Example 3.3. 
. Find .
Here we used the limit of degree theorem: the limit of a degree is equal to the degree of the limit of the base.
Example 3.4. Find ( 
).
Solution. It is impossible to apply the limit of difference theorem, since we have an uncertainty of the form ∞-∞. Let's transform the general term formula:
Example 3.5. The function f(x)=2 1/x is given. Prove that there is no limit.
Solution. Let's use definition 1 of the limit of a function through a sequence. Let us take a sequence ( x n ) converging to 0, i.e. Let us show that the value f(x n)= behaves differently for different sequences. Let x n = 1/n. Obviously, then the limit 
Let us now choose as x n a sequence with a common term x n = -1/n, also tending to zero. 
Therefore there is no limit.
Example 3.6. Prove that there is no limit.
Solution. Let x 1 , x 2 ,..., x n ,... be a sequence for which
. How does the sequence (f(x n)) = (sin x n) behave for different x n → ∞
If x n = p n, then sin x n = sin (p n) = 0 for all n and the limit If
x n =2 p n+ p /2, then sin x n = sin(2 p n+ p /2) = sin p /2 = 1 for all n and therefore the limit. So it doesn't exist.
Limits give all mathematics students a lot of trouble. To solve a limit, sometimes you have to use a lot of tricks and choose from a variety of solution methods exactly the one that is suitable for a particular example.
In this article we will not help you understand the limits of your capabilities or comprehend the limits of control, but we will try to answer the question: how to understand limits in higher mathematics? Understanding comes with experience, so at the same time we will give several detailed examples of solving limits with explanations.
The concept of limit in mathematics
The first question is: what is this limit and the limit of what? We can talk about the limits of numerical sequences and functions. We are interested in the concept of the limit of a function, since this is what students most often encounter. But first, the most general definition of a limit:
Let's say there is some variable value. If this value in the process of change unlimitedly approaches a certain number a , That a – the limit of this value.
For a function defined in a certain interval f(x)=y such a number is called a limit A , which the function tends to when X , tending to a certain point A . Dot A belongs to the interval on which the function is defined.
It sounds cumbersome, but it is written very simply:
Lim- from English limit- limit.
There is also a geometric explanation for determining the limit, but here we will not delve into the theory, since we are more interested in the practical rather than the theoretical side of the issue. When we say that X tends to some value, this means that the variable does not take on the value of a number, but approaches it infinitely close.
Let's give a specific example. The task is to find the limit.
To solve this example, we substitute the value x=3 into a function. We get:
By the way, if you are interested in basic operations on matrices, read a separate article on this topic.
In the examples X can tend to any value. It can be any number or infinity. Here's an example when X tends to infinity:
Intuitively, the larger the number in the denominator, the smaller the value the function will take. So, with unlimited growth X meaning 1/x will decrease and approach zero.
As you can see, to solve the limit, you just need to substitute the value to strive for into the function X . However, this is the simplest case. Often finding the limit is not so obvious. Within the limits there are uncertainties of the type 0/0 or infinity/infinity . What to do in such cases? Resort to tricks!
Uncertainties within
Uncertainty of the form infinity/infinity
Let there be a limit:
If we try to substitute infinity into the function, we will get infinity in both the numerator and the denominator. In general, it is worth saying that there is a certain element of art in resolving such uncertainties: you need to notice how you can transform the function in such a way that the uncertainty goes away. In our case, we divide the numerator and denominator by X in the senior degree. What will happen?
From the example already discussed above, we know that terms containing x in the denominator will tend to zero. Then the solution to the limit is:
To resolve type uncertainties infinity/infinity divide the numerator and denominator by X to the highest degree.
By the way! For our readers there is now a 10% discount on any type of work
Another type of uncertainty: 0/0
As always, substituting values into the function x=-1 gives 0 in the numerator and denominator. Look a little more closely and you will notice that we have a quadratic equation in the numerator. Let's find the roots and write:
Let's reduce and get:
So, if you are faced with type uncertainty 0/0 – factor the numerator and denominator.
To make it easier for you to solve examples, we present a table with the limits of some functions:
L'Hopital's rule within
Another powerful way to eliminate both types of uncertainty. What is the essence of the method?
If there is uncertainty in the limit, take the derivative of the numerator and denominator until the uncertainty disappears.
L'Hopital's rule looks like this:
Important point : the limit in which the derivatives of the numerator and denominator stand instead of the numerator and denominator must exist.
And now - a real example:
There is typical uncertainty 0/0 . Let's take the derivatives of the numerator and denominator:
Voila, uncertainty is resolved quickly and elegantly.
We hope that you will be able to usefully apply this information in practice and find the answer to the question “how to solve limits in higher mathematics.” If you need to calculate the limit of a sequence or the limit of a function at a point, and there is absolutely no time for this work, contact a professional student service for a quick and detailed solution.
Elementary functions and their graphs.
The main elementary functions are: power function, exponential function, logarithmic function, trigonometric functions and inverse trigonometric functions, as well as a polynomial and a rational function, which is the ratio of two polynomials.
Elementary functions also include those functions that are obtained from elementary ones by applying the basic four arithmetic operations and forming a complex function.
Graphs of elementary functions
| Straight line- graph of a linear function y = ax + b. The function y monotonically increases for a > 0 and decreases for a< 0. При b = 0 прямая линия проходит через начало координат т. 0 (y = ax - прямая пропорциональность) | |
 | Parabola- graph of the quadratic trinomial function y = ax 2 + bx + c. It has a vertical axis of symmetry. If a > 0, has a minimum if a< 0 - максимум. Точки пересечения (если они есть) с осью абсцисс - корни соответствующего квадратного уравнения ax 2 + bx +c =0 |
 | Hyperbola- graph of the function. When a > O it is located in the I and III quarters, when a< 0 - во II и IV. Асимптоты - оси координат. Ось симметрии - прямая у = х(а >0) or y - - x(a< 0). |
 | Exponential function. Exhibitor(exponential function to base e) y = e x. (Another spelling y = exp(x)). Asymptote is the abscissa axis. |
| Logarithmic function y = log a x(a > 0) | |
 | y = sinx. Sine wave- periodic function with period T = 2π |
Function limit.
The function y=f(x) has a number A as a limit as x tends to a, if for any number ε › 0 there is a number δ › 0 such that | y – A | ‹ ε if |x - a| ‹ δ,
or lim y = A
Continuity of function.
The function y=f(x) is continuous at the point x = a if lim f(x) = f(a), i.e.
the limit of a function at a point x = a is equal to the value of the function at a given point.
Finding the limits of functions.
Basic theorems on the limits of functions.
1. The limit of a constant value is equal to this constant value:
2. The limit of an algebraic sum is equal to the algebraic sum of the limits of these functions:
lim (f + g - h) = lim f + lim g - lim h
3. The limit of the product of several functions is equal to the product of the limits of these functions:
lim (f * g* h) = lim f * lim g * lim h
4. The limit of the quotient of two functions is equal to the quotient of the limits of these functions if the limit of the denominator is not equal to 0:
lim------- = ----------
The first remarkable limit: lim --------- = 1
Second remarkable limit: lim (1 + 1/x) x = e (e = 2, 718281..)
Examples of finding the limits of functions.
5.1. Example:
Any limit consists of three parts:
1) The well-known limit icon.
2) Entries under the limit icon. The entry reads “X tends to one.” Most often it is x, although instead of “x” there can be any other variable. In place of one there can be absolutely any number, as well as infinity 0 or .
3) Functions under the limit sign, in this case .
The recording itself reads like this: “the limit of a function as x tends to unity.”
A very important question - what does the expression “x” mean? strives to one"? The expression "x" strives to one” should be understood as follows: “x” consistently takes on the values which approach unity infinitely close and practically coincide with it.
How to solve the above example? Based on the above, you just need to substitute one into the function under the limit sign:
So the first rule : When given a limit, you first simply plug the number into the function.
5.2. Example with infinity:
Let's figure out what it is? This is the case when it increases without limit.
So: if , then the function tends to minus infinity:
According to our first rule, instead of “X” we substitute in the function infinity and we get the answer.
5.3. Another example with infinity:
Again we begin to increase to infinity, and look at the behavior of the function.
Conclusion: the function increases unlimitedly
5.4. A series of examples:
Try to mentally analyze the following examples yourself and solve the simplest types of limits:
, , , , 
, , , , 
,
What do you need to remember and understand from the above?
When given any limit, first simply plug the number into the function. At the same time, you must understand and immediately solve the simplest limits, such as 
,
,
etc.
6. Limits with uncertainty of type and a method for solving them.
Now we will consider the group of limits when , and the function is a fraction whose numerator and denominator contain polynomials.
6.1. Example:
Calculate limit 
According to our rule, we try to substitute infinity into the function. What do we get at the top? Infinity. And what happens below? Also infinity. Thus, we have what is called species uncertainty. One might think that = 1, and the answer is ready, but in the general case this is not at all the case, and you need to apply some solution technique, which we will now consider.
How to solve limits of this type?
First we look at the numerator and find the highest power:
The leading power in the numerator is two.
Now we look at the denominator and also find it to the highest power:
The highest degree of the denominator is two.
Then we choose the highest power of the numerator and denominator: in this example, they are the same and equal to two.
So, the solution method is as follows: to reveal uncertainty you need to divide the numerator and denominator by in the senior degree.
Thus, the answer is not 1.
Example
Find the limit 
Again in the numerator and denominator we find in the highest degree: 
Maximum degree in numerator: 3
Maximum degree in denominator: 4
Choose greatest value, in this case four.
According to our algorithm, to reveal uncertainty, we divide the numerator and denominator by .
Example
Find the limit
Maximum degree of “X” in the numerator: 2
Maximum degree of “X” in the denominator: 1 (can be written as)
To reveal the uncertainty, it is necessary to divide the numerator and denominator by . The final solution might look like this:
Divide the numerator and denominator by
