Lim x tends to zero. Limits. Examples of solutions. Calculate the limit of a number sequence
Usually the second remarkable limit is written in this form:
\begin(equation) \lim_(x\to\infty)\left(1+\frac(1)(x)\right)^x=e\end(equation)
The number $e$ indicated on the right side of equality (1) is irrational. The approximate value of this number is: $e\approx(2(,)718281828459045)$. If we make the replacement $t=\frac(1)(x)$, then formula (1) can be rewritten as follows:
\begin(equation) \lim_(t\to(0))\biggl(1+t\biggr)^(\frac(1)(t))=e\end(equation)
As for the first remarkable limit, it does not matter which expression stands in place of the variable $x$ in formula (1) or instead of the variable $t$ in formula (2). The main thing is to fulfill two conditions:
- The base of the degree (i.e., the expression in brackets of formulas (1) and (2)) should tend to unity;
- The exponent (i.e. $x$ in formula (1) or $\frac(1)(t)$ in formula (2)) must tend to infinity.
The second remarkable limit is said to reveal the uncertainty of $1^\infty$. Please note that in formula (1) we do not specify which infinity ($+\infty$ or $-\infty$) we are talking about. In any of these cases, formula (1) is correct. In formula (2), the variable $t$ can tend to zero both on the left and on the right.
I note that there are also several useful consequences from the second remarkable limit. Examples of the use of the second remarkable limit, as well as its consequences, are very popular among compilers of standard standard calculations and tests.
Example No. 1
Calculate the limit $\lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7)$.
Let us immediately note that the base of the degree (i.e. $\frac(3x+1)(3x-5)$) tends to unity:
$$ \lim_(x\to\infty)\frac(3x+1)(3x-5)=\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(3+\frac(1)(x))(3-\frac(5)(x)) =\frac(3+0)(3-0) = 1. $$
In this case, the exponent (expression $4x+7$) tends to infinity, i.e. $\lim_(x\to\infty)(4x+7)=\infty$.
The base of the degree tends to unity, the exponent tends to infinity, i.e. we are dealing with uncertainty $1^\infty$. Let's apply a formula to reveal this uncertainty. At the base of the power of the formula is the expression $1+\frac(1)(x)$, and in the example we are considering, the base of the power is: $\frac(3x+1)(3x-5)$. Therefore, the first action will be a formal adjustment of the expression $\frac(3x+1)(3x-5)$ to the form $1+\frac(1)(x)$. First, add and subtract one:
$$ \lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(3x+1)(3x-5)-1\right)^(4x+7) $$
Please note that you cannot simply add a unit. If we are forced to add one, then we also need to subtract it so as not to change the value of the entire expression. To continue the solution, we take into account that
$$ \frac(3x+1)(3x-5)-1 =\frac(3x+1)(3x-5)-\frac(3x-5)(3x-5) =\frac(3x+1- 3x+5)(3x-5) =\frac(6)(3x-5). $$
Since $\frac(3x+1)(3x-5)-1=\frac(6)(3x-5)$, then:
$$ \lim_(x\to\infty)\left(1+ \frac(3x+1)(3x-5)-1\right)^(4x+7) =\lim_(x\to\infty)\ left(1+\frac(6)(3x-5)\right)^(4x+7) $$
Let's continue the adjustment. In the expression $1+\frac(1)(x)$ of the formula, the numerator of the fraction is 1, and in our expression $1+\frac(6)(3x-5)$ the numerator is $6$. To get $1$ in the numerator, drop $6$ into the denominator using the following conversion:
$$ 1+\frac(6)(3x-5) =1+\frac(1)(\frac(3x-5)(6)) $$
Thus,
$$ \lim_(x\to\infty)\left(1+\frac(6)(3x-5)\right)^(4x+7) =\lim_(x\to\infty)\left(1+ \frac(1)(\frac(3x-5)(6))\right)^(4x+7) $$
So, the basis of the degree, i.e. $1+\frac(1)(\frac(3x-5)(6))$, adjusted to the form $1+\frac(1)(x)$ required in the formula. Now let's start working with the exponent. Note that in the formula the expressions in the exponents and in the denominator are the same:
This means that in our example, the exponent and the denominator must be brought to the same form. To get the expression $\frac(3x-5)(6)$ in the exponent, we simply multiply the exponent by this fraction. Naturally, to compensate for such a multiplication, you will have to immediately multiply by the reciprocal fraction, i.e. by $\frac(6)(3x-5)$. So we have:
$$ \lim_(x\to\infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(4x+7) =\lim_(x\to\ infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\frac(3x-5)(6)\cdot\frac(6)(3x-5 )\cdot(4x+7)) =\lim_(x\to\infty)\left(\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\ frac(3x-5)(6))\right)^(\frac(6\cdot(4x+7))(3x-5)) $$
Let us separately consider the limit of the fraction $\frac(6\cdot(4x+7))(3x-5)$ located in the power:
$$ \lim_(x\to\infty)\frac(6\cdot(4x+7))(3x-5) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(6\cdot\left(4+\frac(7)(x)\right))(3-\frac(5)(x)) =6\cdot\ frac(4)(3) =8. $$
Answer: $\lim_(x\to(0))\biggl(\cos(2x)\biggr)^(\frac(1)(\sin^2(3x)))=e^(-\frac(2) (9))$.
Example No. 4
Find the limit $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)$.
Since for $x>0$ we have $\ln(x+1)-\ln(x)=\ln\left(\frac(x+1)(x)\right)$, then:
$$ \lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right) =\lim_(x\to+\infty)\left(x\cdot\ln\ left(\frac(x+1)(x)\right)\right) $$
Expanding the fraction $\frac(x+1)(x)$ into the sum of fractions $\frac(x+1)(x)=1+\frac(1)(x)$ we get:
$$ \lim_(x\to+\infty)\left(x\cdot\ln\left(\frac(x+1)(x)\right)\right) =\lim_(x\to+\infty)\left (x\cdot\ln\left(1+\frac(1)(x)\right)\right) =\lim_(x\to+\infty)\left(\ln\left(\frac(x+1) (x)\right)^x\right) =\ln(e) =1. $$
Answer: $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)=1$.
Example No. 5
Find the limit $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))$.
Since $\lim_(x\to(2))(3x-5)=6-5=1$ and $\lim_(x\to(2))\frac(2x)(x^2-4)= \infty$, then we are dealing with uncertainty of the form $1^\infty$. Detailed explanations are given in example No. 2, but here we will limit ourselves short solution. Making the replacement $t=x-2$, we get:
$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=x-2 ;\;x=t+2\\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\biggl(1+3t\biggr)^(\frac(2t+4)(t^2+4t))=\\ =\lim_(t\to(0) )\biggl(1+3t\biggr)^(\frac(1)(3t)\cdot 3t\cdot\frac(2t+4)(t^2+4t)) =\lim_(t\to(0) )\left(\biggl(1+3t\biggr)^(\frac(1)(3t))\right)^(\frac(6\cdot(t+2))(t+4)) =e^ 3. $$
Can be solved this example and in another way, using the replacement: $t=\frac(1)(x-2)$. Of course, the answer will be the same:
$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=\frac( 1)(x-2);\;x=\frac(2t+1)(t)\\&t\to\infty\end(aligned)\right| =\lim_(t\to\infty)\left(1+\frac(3)(t)\right)^(t\cdot\frac(4t+2)(4t+1))=\\ =\lim_ (t\to\infty)\left(1+\frac(1)(\frac(t)(3))\right)^(\frac(t)(3)\cdot\frac(3)(t) \cdot\frac(t\cdot(4t+2))(4t+1)) =\lim_(t\to\infty)\left(\left(1+\frac(1)(\frac(t)( 3))\right)^(\frac(t)(3))\right)^(\frac(6\cdot(2t+1))(4t+1)) =e^3. $$
Answer: $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))=e^3$.
Example No. 6
Find the limit $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) $.
Let's find out what the expression $\frac(2x^2+3)(2x^2-4)$ tends to under the condition $x\to\infty$:
$$ \lim_(x\to\infty)\frac(2x^2+3)(2x^2-4) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(2+\frac(3)(x^2))(2-\frac(4)(x^2)) =\frac(2+0)(2 -0)=1. $$
Thus, in the given limit we are dealing with an uncertainty of the form $1^\infty$, which we will reveal using the second remarkable limit:
$$ \lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(2x^2+3)(2x^2-4)-1\right)^(3x)=\\ =\lim_(x\to \infty)\left(1+\frac(7)(2x^2-4)\right)^(3x) =\lim_(x\to\infty)\left(1+\frac(1)(\frac (2x^2-4)(7))\right)^(3x)=\\ =\lim_(x\to\infty)\left(1+\frac(1)(\frac(2x^2-4 )(7))\right)^(\frac(2x^2-4)(7)\cdot\frac(7)(2x^2-4)\cdot 3x) =\lim_(x\to\infty) \left(\left(1+\frac(1)(\frac(2x^2-4)(7))\right)^(\frac(2x^2-4)(7))\right)^( \frac(21x)(2x^2-4)) =e^0 =1. $$
Answer: $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x)=1$.
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Function limit
We have to calculate the limit of a function in mathematics quite often. When analyzing a function to construct its graph, finding the limit of the function at infinity allows you to find the asymptotes of the graph, and at the breakpoints, the value of the limit determines the discontinuity of the function and determines the type of breakpoints. Also, when calculating the sum of a series a necessary condition its convergence is the condition that the limit at infinity is equal to zero on the ratio of the terms of the series.
When solving problems of finding limits, you should remember some limits so as not to recalculate them each time. Combining these known limits, we will find new limits using the properties indicated in § 4. For convenience, we present the most frequently encountered limits: Limits 1 lim x - a x a 2 lim 1 = 0 3 lim x- ± co X ± 00 4 lim -L, = oo X->o\X\ 5 lim sin*- l X -о X 6 lim f(x) = f(a), if f (x) is continuous x a If it is known that the function is continuous, then instead of finding the limit, we calculate the value of the function. Example 1. Find lim (x*-6l:+ 8). Since there are many - X->2
member function is continuous, then lim (x*-6x4- 8) = 2*-6-2 + 8 = 4. x-+2 x*_2x 4-1 Example 2. Find lim -r. . First, we find the limit of the denominator: lim [xr-\-bx)= 12 + 5-1 =6; it is not equal to X-Y1 zero, which means we can apply property 4 § 4, then x™i *" + &* ~~ lim (x2 bx) - 12 + 5-1 ""6 1. The limit of the denominator X X is equal to zero, therefore, property 4 of § 4 cannot be applied. Since the numerator is a constant number, and the denominator is [x2x) -> -0 for x - 1, then the whole fraction increases indefinitely. absolute value, i.e. lim "1 X-*- - 1 x* + x Example 4. Find lim \-ll*"!"" "The limit of the denominator is zero: lim (xr-6lg+ 8) = 2*-6 -2 + 8 = 0, so X property 4 § 4 is not applicable. But the limit of the numerator is also zero: lim (x2 - 5d; + 6) = 22 - 5-2-f 6 = 0. So, the limits of the numerator and denominator are both equal to zero. However, the number 2 is the root of both the numerator and the denominator, so the fraction can be reduced by the difference x-2 (according to Bezout’s theorem, x*-5x + 6 (x-2) (x-3)). x-3 x"-6x + 8~ (x-2) (x-4) ~~ x-4 "therefore, xr--f- 6 g x-3 -1 1 Example 5. Find lim xn (n integer , positive). n is an integer, positive). lim *n= + oo (if n is even). *-* -о In the case of an odd degree, the absolute value of the product increases, but it remains negative, i.e. lim xn = - oo (for n odd). p -- 00 Example 7. Find lim . x x-*- co * If m>pu then we can write: m = n + kt where k>0. Therefore xm b lim -=- = lim -=-= lim x. UP Yn x -x> A x yu We came to example 6. If ti uTL xm I lim lim lim t. X - O x-* yu L X ->co Here the numerator remains constant, and the denominator increases in absolute value, therefore lim -ь = 0. Х-*оо X* It is recommended to remember the result of this example in the following form: The power function grows faster, the larger the exponent. $хв_Зхг + 7
Examples
Example 8. Find lim g L -g-=. In this example x-*® "J* "G bX -ox-o and the numerator and denominator increase without limit. Let us divide both the numerator and denominator by the highest power of x, i.e. on xb, then 3 7_ Example 9. Find lira. Performing transformations, we obtain lira ^ = lim X CO + 3 7 3 Since lim -5 = 0, lim -, = 0, then the limit of the denominator. rad-*® X X-+-CD X is zero, while the limit of the numerator is 1. Consequently, the entire fraction increases without limit, i.e. t. 7x hm X-+ ω Example 10. Find lim Let's calculate the limit S denominator, remembering that the cos*-function is continuous: lira (2 + cos x) = 2 + cozy =2. Then x->- S lim (l-fsin*) Example 15. Find lim *<*-e>2 and lim e "(X"a)\ Polo X-+ ± co X ± CO press (l: - a)2 = z; since (l;-a)2 always grows non-negatively and without limit with x, then for x - ±oo the new variable z-*oc. Therefore we obtain qt £<*-«)* = X ->± 00 s=lim ег = oo (see note to §5). g -*■ co Similarly lim e~(X-a)2 = lim e~z=Q, since x ± oo g m - (x- a)z decreases without limit as x ->±oo (see note to §
In this article, you will learn about how to solve limits?
Solving limits is one of the important sections of mathematical and computational analysis. Many students and university students cope with this problem freely, when others constantly ask the same question: “How to solve limits?” Finding limits is a hot topic. There are many ways to solve limits. Identical limits can be found according to L'Hopital's law without its help. However, first we need to understand what a limit is?
The limit has three parts
The first is the well-known lim icon, the second is what is written under it.
For example: x -> 1. This entry will read like this (x tends to 1).
The third part is the function itself, which comes after the lim sign.
I would like to clarify that the value of x tends to 1, this is the value x, at which X takes on certain values that are close to unity or almost coincide with it.
Solving limits is easy if you understand them.
First rule for solving limits
If a function is provided to us, simply substitute the number into the function. These are elementary limits that actually occur in examples and very often.
Are there limits where x->? Then infinity is the function where x increases infinitely. The value of such a function is (1-x). To solve this limit, we need to follow our first rule and substitute the value (1's) into the function and get the answer.
From the above, in order to learn how to solve the most difficult limits, You must remember the rules for solving elementary limits.
- Rule one: Given a function, we substitute the number into the function.
- Rule two: Given infinity, we substitute (1's) into the function.
Once you understand this, you will immediately begin to notice elementary limits and be able to solve them.So we learned how to solve easy limits. Now let's look at solving more complex limits.
There are many limits with? One such option is the view limit?/?
Such a function is possible when x->?, and the limit is expressed as a fraction.
Many people wonder whether it is easy to solve such a limit?
The first thing you need to remember is that you need to find the x in the numerator by seniority, i.e. to the greatest extent of all the x's that are in the numerator.
lim+(x->?)?((2x^2-3x-4)/(3x^2+1+x))^ ?
We see that the highest degree in the numerator is 2
Now, we need to do the same only with the denominator. The highest power in the denominator is also 2.
Principle: In order to resolve this function, we must divide both the dividend and the divisor by x to the highest power in the limit. If it were equal to 2. If the degree of the numerator was equal to 4, and the denominator was 2, then we would choose 4. Because this is the highest degree in the function given to us. See how quickly we learned to solve the limits of the species?/?
Now let's look at solving the most difficult limits. This is the 0/0 view.
Such limits remind us very much of the solution of limits of the form infinity to infinity. But there is a difference that is important to remember when deciding. When x tends to infinity, it increases infinitely, but here it is equal to 0, i.e. finite number.
To enable such a function, we should, and factor the numerators and denominator. To get an elementary discriminant, known to us since 6th grade. We calculate the discriminant and substitute the answers into our function. We find the final answer.
Rule: If in the numerator or denominator a certain number can be taken out of a given bracket, then without thinking, we definitely take it out.
There are many different ways to solve more complex limits. One of them is the replacement method. Replacing any variable is easier than constantly factoring. Very often this method is used to complex limit make the first wonderful limit.
Let's take a closer look at an example
Example: lim+(x->0)?(arctg4x/7x)^ ?
Solution: We see that our function is represented by the uncertainty 0/0, which we have already passed
lim+(x->0)?(arctg4x/7x)^ ? = 0/0
We see the arctangent in the limit, a bad function that we need to get rid of. It will be very comfortable for us if we turn the arctangent into one simple and easy letter.
Let's make a replacement: replace arctg with y. And in the process of solving we will refer to the arctangent as y. If our x tends to zero, we replaced the arctangent with y, then we write down that y also tends to zero. All that remains for us in the denominator is to express x through y. To do this, we add tg to both sides of the equality
The expressions will take the following form:
tg (arctg4x)=tgy
On the left side we remove two functions; they are reciprocal and disappear.
We are left with:
4х = tgy, hence: x= tgy/4
And now the most basic things remain:
lim+(x->0)?(y/(7*tgy/4))^ ?
Go ahead. Within limits there is not only one wonderful limit, but there are two of them. Now we will not only understand the concept of the second remarkable limit, but also learn how to solve it. A second remarkable limit exists for solving uncertainties of the form 1^? In mathematics it is written as a(x) ->? This type of this function is the simplest, there are functions that are more complex, the most important thing is that it tends to infinity.
It should be remembered that as soon as our limit turns out to be a degree, this is the main sign that such an expression will help us solve the second remarkable limit. Now we will dwell in more detail on an example that occurs very often; I advise you to study it in detail.
We are given a limit: lim+(x->?)?((x-2)/(x+1))^(2x+3) ?
This limit of the form (?/?)^ ?The second remarkable limit does not solve this type, as we know, it solves the form 1^?, for this our function must be transformed into another form. We see x+1 in the denominator, which means there should also be x+1 in the numerator
lim+(x->?)?((x+1-3)/(x+1))^(2x+3) ?
Now we It is necessary to divide the numerator by the denominator term by term. Then our foundation will be similar to our uncertainty, but there is a minus sign that interferes with us. We make a fraction with three floors and see our uncertainty?/?. And we already know how to calculate such a function. Divide both sides of the fraction by x, and you're done. We have the answer.
I want to congratulate you, dear readers, you have learned to solve limits. I hope my article was informative, entertaining and interesting!
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