The first remarkable limit is often used to calculate the limits containing sine, arcsine, tangent, arctangent and the resulting uncertainties of zero divided by zero.
Formula
The formula for the first remarkable limit is: $$ \\ lim _ (\\ alpha \\ to 0) \\ frac (\\ sin \\ alpha) (\\ alpha) \u003d 1 $$
We note that for $ \\ alpha \\ to 0 $ we get $ \\ sin \\ alpha \\ to 0 $, thus, we have zeros in the numerator and in the denominator. Thus, the formula for the first remarkable limit is needed to disclose the uncertainties $ \\ frac (0) (0) $.
To apply the formula, two conditions must be met:
- The expressions in the sine and denominator of the fraction are the same
- Expressions in the sine and denominator of a fraction tend to zero
Attention! $ \\ lim_ (x \\ to 0) \\ frac (\\ sin (2x ^ 2 + 1)) (2x ^ 2 + 1) \\ neq 1 $ Although the expressions under the sine and in the denominator are the same, but $ 2x ^ 2 + 1 \u003d 1 $, for $ x \\ to 0 $. The second condition is not met, therefore, the formula is NOT allowed!
Consequences
Rarely enough in tasks you can see a clean first remarkable limit in which you could immediately write down the answer. In practice, things look a little more complicated, but for such cases it will be useful to know the consequences of the first remarkable limit. Thanks to them, you can quickly calculate the required limits.
$$ \\ lim _ (\\ alpha \\ to 0) \\ frac (\\ alpha) (\\ sin \\ alpha) \u003d 1 $$
$$ \\ lim _ (\\ alpha \\ to 0) \\ frac (\\ sin (a \\ alpha)) (\\ sin (b \\ alpha)) \u003d \\ frac (a) (b) $$
$$ \\ lim _ (\\ alpha \\ to 0) \\ frac (tg \\ alpha) (\\ alpha) \u003d 1 $$
$$ \\ lim _ (\\ alpha \\ to 0) \\ frac (\\ arcsin \\ alpha) (\\ alpha) \u003d 1 $$
$$ \\ lim _ (\\ alpha \\ to 0) \\ frac (arctg \\ alpha) (\\ alpha) \u003d 1 $$
Solution examples
Consider the first remarkable limit, examples of solving which on the calculation of limits containing trigonometric functions and uncertainty $ \\ bigg [\\ frac (0) (0) \\ bigg] $
| Example 1 |
| Evaluate $ \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (4x) $ |
| Decision |
|
Consider the limit and note that there is a sine in it. Next, substitute $ x \u003d 0 $ in the numerator and denominator and get the uncertainty zero divided by zero: $$ \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (4x) \u003d \\ frac (0) (0) $$ Already two signs that you need to apply a wonderful limit, but there is a small nuance: we cannot immediately apply the formula, since the expression under the sine sign differs from the expression in the denominator. And we need them to be equal. Therefore, using elementary transformations of the numerator, we will turn it into $ 2x $. To do this, we take two out of the denominator of the fraction by a separate factor. It looks like this: $$ \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (4x) \u003d \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (2 \\ cdot 2x) \u003d $$ $$ \u003d \\ frac (1) (2) \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (2x) \u003d \\ frac (1) (2) \\ cdot 1 \u003d \\ frac (1) (2) $$ Note that at the end $ \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (2x) \u003d 1 $ is obtained by the formula. If you can't solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the course of the calculation and get information. This will help you get credit from your teacher in a timely manner! |
| Answer |
| $$ \\ lim_ (x \\ to 0) \\ frac (\\ sin2x) (4x) \u003d \\ frac (1) (2) $$ |
| Example 2 |
| Find $ \\ lim_ (x \\ to 0) \\ frac (\\ sin (x ^ 3 + 2x)) (2x-x ^ 4) $ |
| Decision |
|
As always, you first need to know the type of uncertainty. If it is zero divided by zero, then we pay attention to the presence of a sine: $$ \\ lim_ (x \\ to 0) \\ frac (\\ sin (x ^ 3 + 2x)) (2x-x ^ 4) \u003d \\ frac (0) (0) \u003d $$ This uncertainty allows us to use the formula for the first remarkable limit, but the expression from the denominator is not equal to the sine argument? Therefore, the formula cannot be applied "head-on". You need to multiply and divide the fraction by the sine argument: $$ \u003d \\ lim_ (x \\ to 0) \\ frac ((x ^ 3 + 2x) \\ sin (x ^ 3 + 2x)) ((2x-x ^ 4) (x ^ 3 + 2x)) \u003d $$ Now, by the properties of the limits, we write: $$ \u003d \\ lim_ (x \\ to 0) \\ frac ((x ^ 3 + 2x)) (2x-x ^ 4) \\ cdot \\ lim_ (x \\ to 0) \\ frac (\\ sin (x ^ 3 + 2x)) ((x ^ 3 + 2x)) \u003d $$ The second limit just fits the formula and is equal to one: $$ \u003d \\ lim_ (x \\ to 0 ) \\ frac (x ^ 3 + 2x) (2x-x ^ 4) \\ cdot 1 \u003d \\ lim_ (x \\ to 0) \\ frac (x ^ 3 + 2x) (2x-x ^ 4) \u003d $$ Substitute again $ x \u003d 0 $ into a fraction and we get the uncertainty $ \\ frac (0) (0) $. To eliminate it, it is sufficient to exclude $ x $ from parentheses and reduce to it: $$ \u003d \\ lim_ (x \\ to 0) \\ frac (x (x ^ 2 + 2)) (x (2-x ^ 3)) \u003d \\ ) (2) \u003d 1 $$ |
| Answer |
| $$ \\ lim_ (x \\ to 0) \\ frac (\\ sin (x ^ 3 + 2x)) (2x-x ^ 4) \u003d 1 $$ |
| Example 4 |
| Evaluate $ \\ lim_ (x \\ to0) \\ frac (\\ sin2x) (tg3x) $ |
| Decision |
|
Let's start the calculation by substituting $ x \u003d 0 $. As a result, we get the uncertainty $ \\ frac (0) (0) $. The limit contains sine and tangent, which hints at a possible development of the situation using the formula for the first remarkable limit. We transform the numerator and denominator of the fraction under the formula and consequence: $$ \\ lim_ (x \\ to0) \\ frac (\\ sin2x) (tg3x) \u003d \\ frac (0) (0) \u003d \\ lim_ (x \\ to0) \\ frac (\\ frac (\\ sin2x) (2x) \\ cdot 2x ) (\\ frac (tg3x) (3x) \\ cdot 3x) \u003d $$ Now we see expressions suitable for the formula and consequences appeared in the numerator and denominator. The sine argument and tangent argument are the same for the corresponding denominators $$ \u003d \\ lim_ (x \\ to0) \\ frac (1 \\ cdot 2x) (1 \\ cdot 3x) \u003d \\ frac (2) (3) $$ |
| Answer |
| $$ \\ lim_ (x \\ to0) \\ frac (\\ sin2x) (tg2x) \u003d \\ frac (2) (3) $$ |
In the article: "The first remarkable limit, examples of solutions" was told about the cases in which it is advisable to use this formula and its consequences.
Usually, the second remarkable limit is written in this form:
\\ begin (equation) \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (1) (x) \\ right) ^ x \u003d e \\ end (equation)
The number $ e $ indicated on the right-hand side of equality (1) is irrational. The approximate value of this number is as follows: $ e \\ approx (2 (,) 718281828459045) $. If we make the substitution $ t \u003d \\ frac (1) (x) $, then formula (1) can be rewritten as follows:
\\ begin (equation) \\ lim_ (t \\ to (0)) \\ biggl (1 + t \\ biggr) ^ (\\ frac (1) (t)) \u003d e \\ end (equation)
As with the first remarkable limit, it does not matter which expression stands for the variable $ x $ in formula (1) or instead of the variable $ t $ in formula (2). The main thing is the fulfillment of two conditions:
- The base of the degree (i.e., the expression in brackets of formulas (1) and (2)) should tend to one;
- The exponent (i.e. $ x $ in formula (1) or $ \\ frac (1) (t) $ in formula (2)) should tend to infinity.
The second remarkable limit is said to expose the $ 1 ^ \\ infty $ uncertainty. 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.
Note that there are also several useful consequences from the second remarkable limit. Examples on the use of the second remarkable limit, as well as the consequences from it, are very popular with the compilers of standard typical calculations and control works.
Example # 1
Calculate the limit of $ \\ lim_ (x \\ to \\ infty) \\ left (\\ frac (3x + 1) (3x-5) \\ right) ^ (4x + 7) $.
Note right away that the base of the degree (i.e. $ \\ frac (3x + 1) (3x-5) $) tends to unity:
$$ \\ lim_ (x \\ to \\ infty) \\ frac (3x + 1) (3x-5) \u003d \\ left | \\ frac (\\ infty) (\\ infty) \\ right | \u003d \\ lim_ (x \\ to \\ infty) \\ frac (3+ \\ frac (1) (x)) (3- \\ frac (5) (x)) \u003d \\ frac (3 + 0) (3-0) \u003d 1. $$
In this case, the exponent (expression $ 4x + 7 $) tends to infinity, i.e. $ \\ lim_ (x \\ to \\ infty) (4x + 7) \u003d \\ infty $.
The base of the degree tends to one, the exponent tends to infinity, i.e. we are dealing with $ 1 ^ \\ infty $ uncertainty. Let's apply the formula to reveal this uncertainty. At the base of the degree of the formula is the expression $ 1 + \\ frac (1) (x) $, and in the example we are considering, the base of the degree is as follows: $ \\ frac (3x + 1) (3x-5) $. Therefore, the first step is to formally fit the expression $ \\ frac (3x + 1) (3x-5) $ to look like $ 1 + \\ frac (1) (x) $. First, add and subtract one:
$$ \\ lim_ (x \\ to \\ infty) \\ left (\\ frac (3x + 1) (3x-5) \\ right) ^ (4x + 7) \u003d | 1 ^ \\ infty | \u003d \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (3x + 1) (3x-5) -1 \\ right) ^ (4x + 7) $$
It should be noted that you cannot just add a unit. If we are forced to add a unit, then we need to subtract it, so as not to change the value of the entire expression. To continue the solution, take into account that
$$ \\ frac (3x + 1) (3x-5) -1 \u003d \\ frac (3x + 1) (3x-5) - \\ frac (3x-5) (3x-5) \u003d \\ frac (3x + 1- 3x + 5) (3x-5) \u003d \\ frac (6) (3x-5). $$
Since $ \\ frac (3x + 1) (3x-5) -1 \u003d \\ frac (6) (3x-5) $, then:
$$ \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (3x + 1) (3x-5) -1 \\ right) ^ (4x + 7) \u003d \\ lim_ (x \\ to \\ infty) \\ Let's continue the "fit". In the expression $ 1 + \\ frac (1) (x) $ of the formula, the numerator of the fraction contains 1, and in our expression $ 1 + \\ frac (6) (3x-5) $, the numerator contains $ 6 $. To get $ 1 $ in the numerator, omit $ 6 $ in the denominator using the following conversion:
$$ 1+ \\ frac (6) (3x-5) \u003d 1 + \\ frac (1) (\\ frac (3x-5) (6)) $$
Thus,
$$ \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (6) (3x-5) \\ right) ^ (4x + 7) \u003d \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (1) (\\ frac (3x-5) (6)) \\ right) ^ (4x + 7) $$
So, the base of the degree, i.e. $ 1 + \\ frac (1) (\\ frac (3x-5) (6)) $, adjusted to the form $ 1 + \\ frac (1) (x) $, which is required in the formula. Now let's start working with the exponent. Note that in the formula, the expressions in the exponents and in the denominator are the same:
This means that in our example, the exponent and the denominator must be reduced to the same form. To get the exponent $ \\ frac (3x-5) (6) $, simply multiply the exponent by this fraction. Naturally, to compensate for such a multiplication, you will have to immediately multiply by the reciprocal, i.e. by $ \\ frac (6) (3x-5) $. So, we have:
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$$ \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (1) (\\ frac (3x-5) (6)) \\ right) ^ (4x + 7) \u003d \\ lim_ (x \\ to \\ ) \\ cdot (4x + 7)) \u003d \\ lim_ (x \\ to \\ infty) \\ left (\\ left (1+ \\ frac (1) (\\ frac (3x-5) (6)) \\ right) ^ (\\ Consider separately the limit of the fraction $ \\ frac (6 \\ cdot (4x + 7)) (3x-5) $, located in the power:
$$ \\ lim_ (x \\ to \\ infty) \\ frac (6 \\ cdot (4x + 7)) (3x-5) \u003d \\ left | \\ frac (\\ infty) (\\ infty) \\ right | \u003d \\ lim_ (x \\ to \\ infty) \\ frac (6 \\ cdot \\ left (4+ \\ frac (7) (x) \\ right)) (3- \\ frac (5) (x)) \u003d 6 \\ cdot \\ $$
: $ \\ lim_ (x \\ to (0)) \\ biggl (\\ cos (2x) \\ biggr) ^ (\\ frac (1) (\\ sin ^ 2 (3x))) \u003d e ^ (- \\ frac (2) (9)) $.
AnswerExample No. 4
Find the limit $ \\ lim_ (x \\ to + \\ infty) x \\ left (\\ ln (x + 1) - \\ ln (x) \\ right) $.
Since for $ x\u003e 0 $ we have $ \\ ln (x + 1) - \\ ln (x) \u003d \\ ln \\ left (\\ frac (x + 1) (x) \\ right) $, then:
$$ \\ lim_ (x \\ to + \\ infty) x \\ left (\\ ln (x + 1) - \\ ln (x) \\ right) \u003d \\ lim_ (x \\ to + \\ infty) \\ left (x \\ cdot \\ ln \\ Expanding the fraction $ \\ frac (x + 1) (x) $ into the sum of fractions $ \\ frac (x + 1) (x) \u003d 1 + \\ frac (1) (x) $ we get:
$$ \\ lim_ (x \\ to + \\ infty) \\ left (x \\ cdot \\ ln \\ left (\\ frac (x + 1) (x) \\ right) \\ right) \u003d \\ lim_ (x \\ to + \\ infty) \\ left (x \\ cdot \\ ln \\ left (1+ \\ frac (1) (x) \\ right) \\ right) \u003d \\ lim_ (x \\ to + \\ infty) \\ left (\\ ln \\ left (\\ frac (x + 1) (x) \\ right) ^ x \\ right) \u003d \\ ln (e) \u003d 1. $$
: $ \\ lim_ (x \\ to + \\ infty) x \\ left (\\ ln (x + 1) - \\ ln (x) \\ right) \u003d 1 $.
Example No. 5
AnswerFind the limit $ \\ lim_ (x \\ to (2)) \\ biggl (3x-5 \\ biggr) ^ (\\ frac (2x) (x ^ 2-4)) $.
Since $ \\ lim_ (x \\ to (2)) (3x-5) \u003d 6-5 \u003d 1 $ and $ \\ lim_ (x \\ to (2)) \\ frac (2x) (x ^ 2-4) \u003d \\ infty $, then we are dealing with an uncertainty of the form $ 1 ^ \\ infty $. Detailed explanations are given in example # 2, but here we will restrict ourselves to a short solution. Making the substitution $ t \u003d x-2 $, we get:
$$ \\ lim_ (x \\ to (2)) \\ biggl (3x-5 \\ biggr) ^ (\\ frac (2x) (x ^ 2-4)) \u003d \\ left | \\ begin (aligned) & t \u003d x-2 ; \\; x \u003d t + 2 \\\\ & t \\ to (0) \\ end (aligned) \\ right | \u003d \\ lim_ (t \\ to (0)) \\ biggl (1 + 3t \\ biggr) ^ (\\ frac (2t + 4) (t ^ 2 + 4t)) \u003d \\\\ \u003d \\ lim_ (t \\ to (0) ) \\ biggl (1 + 3t \\ biggr) ^ (\\ frac (1) (3t) \\ cdot 3t \\ cdot \\ frac (2t + 4) (t ^ 2 + 4t)) \u003d \\ lim_ (t \\ to (0) ) \\ left (\\ biggl (1 + 3t \\ biggr) ^ (\\ frac (1) (3t)) \\ right) ^ (\\ frac (6 \\ cdot (t + 2)) (t + 4)) \u003d e ^ 3. $$
You can solve this example in a different way, using the replacement: $ t \u003d \\ frac (1) (x-2) $. Of course, the answer is the same:
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$$ \\ lim_ (x \\ to (2)) \\ biggl (3x-5 \\ biggr) ^ (\\ frac (2x) (x ^ 2-4)) \u003d \\ left | \\ begin (aligned) & t \u003d \\ frac ( 1) (x-2); \\; x \u003d \\ frac (2t + 1) (t) \\\\ & t \\ to \\ infty \\ end (aligned) \\ right | \u003d \\ lim_ (t \\ to \\ infty) \\ left (1+ \\ frac (3) (t) \\ right) ^ (t \\ cdot \\ frac (4t + 2) (4t + 1)) \u003d \\\\ \u003d \\ 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)) \u003d \\ lim_ (t \\ to \\ infty) \\ left (\\ left (1+ \\ frac (1) (\\ frac (t) ( 3)) \\ right) ^ (\\ frac (t) (3)) \\ right) ^ (\\ frac (6 \\ cdot (2t + 1)) (4t + 1)) \u003d e ^ 3. $$
Answer: $ \\ lim_ (x \\ to (2)) \\ biggl (3x-5 \\ biggr) ^ (\\ frac (2x) (x ^ 2-4)) \u003d e ^ 3 $.
Example No. 6
Find the limit $ \\ lim_ (x \\ to \\ infty) \\ left (\\ frac (2x ^ 2 + 3) (2x ^ 2-4) \\ right) ^ (3x) $.
Let us find out what the expression $ \\ frac (2x ^ 2 + 3) (2x ^ 2-4) $ tends to under the condition $ x \\ to \\ infty $:
$$ \\ lim_ (x \\ to \\ infty) \\ frac (2x ^ 2 + 3) (2x ^ 2-4) \u003d \\ left | \\ frac (\\ infty) (\\ infty) \\ right | \u003d \\ lim_ (x \\ to \\ infty) \\ frac (2+ \\ frac (3) (x ^ 2)) (2- \\ frac (4) (x ^ 2)) \u003d \\ frac (2 + 0) (2 -0) \u003d 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) \u003d | 1 ^ \\ infty | \u003d \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (2x ^ 2 + 3) (2x ^ 2-4) -1 \\ right) ^ (3x) \u003d \\\\ \u003d \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (7) (2x ^ 2-4) \\ right) ^ (3x) \u003d \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (1) (\\ frac (2x ^ 2-4) (7)) \\ right) ^ (3x) \u003d \\\\ \u003d \\ 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) \u003d \\ 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)) \u003d e ^ 0 \u003d 1. $$
Answer: $ \\ lim_ (x \\ to \\ infty) \\ left (\\ frac (2x ^ 2 + 3) (2x ^ 2-4) \\ right) ^ (3x) \u003d 1 $.
The formula for the second remarkable limit is lim x → ∞ 1 + 1 x x \u003d e. Another notation looks like this: lim x → 0 (1 + x) 1 x \u003d e.
When we talk about the second remarkable limit, we have to deal with an uncertainty of the form 1 ∞, i.e. a unit to an infinite degree.
Consider problems in which the ability to calculate the second remarkable limit is useful.
Example 1
Find the limit lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4.
Decision
Substitute the desired formula and perform the calculations.
lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 \u003d 1 - 2 ∞ 2 + 1 ∞ 2 + 1 4 \u003d 1 - 0 ∞ \u003d 1 ∞
In our answer, we got one to the power of infinity. To determine the solution method, we use the uncertainty table. We choose the second remarkable limit and change the variables.
t \u003d - x 2 + 1 2 ⇔ x 2 + 1 4 \u003d - t 2
If x → ∞, then t → - ∞.
Let's see what we got after the replacement:
lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 \u003d 1 ∞ \u003d lim x → ∞ 1 + 1 t - 1 2 t \u003d lim t → ∞ 1 + 1 t t - 1 2 \u003d e - 1 2
Answer: lim x → ∞ 1 - 2 x 2 + 1 x 2 + 1 4 \u003d e - 1 2.
Example 2
Compute the limit lim x → ∞ x - 1 x + 1 x.
Decision
Substitute infinity and get the following.
lim x → ∞ x - 1 x + 1 x \u003d lim x → ∞ 1 - 1 x 1 + 1 x x \u003d 1 - 0 1 + 0 ∞ \u003d 1 ∞
In the answer, we again got the same thing as in the previous problem, therefore, we can again use the second remarkable limit. Next, we need to select the whole part at the base of the power function:
x - 1 x + 1 \u003d x + 1 - 2 x + 1 \u003d x + 1 x + 1 - 2 x + 1 \u003d 1 - 2 x + 1
After that, the limit takes the following form:
lim x → ∞ x - 1 x + 1 x \u003d 1 ∞ \u003d lim x → ∞ 1 - 2 x + 1 x
We replace the variables. Suppose that t \u003d - x + 1 2 ⇒ 2 t \u003d - x - 1 ⇒ x \u003d - 2 t - 1; if x → ∞, then t → ∞.
After that, we write down what we got in the original limit:
lim x → ∞ x - 1 x + 1 x \u003d 1 ∞ \u003d lim x → ∞ 1 - 2 x + 1 x \u003d lim x → ∞ 1 + 1 t - 2 t - 1 \u003d \u003d lim x → ∞ 1 + 1 t - 2 t 1 + 1 t - 1 \u003d lim x → ∞ 1 + 1 t - 2 t lim x → ∞ 1 + 1 t - 1 \u003d \u003d lim x → ∞ 1 + 1 tt - 2 1 + 1 ∞ \u003d e - 2 (1 + 0) - 1 \u003d e - 2
To perform this transformation, we used the basic properties of limits and degrees.
Answer: lim x → ∞ x - 1 x + 1 x \u003d e - 2.
Example 3
Find the limit lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5.
Decision
lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d lim x → ∞ 1 + 1 x 3 1 + 2 x - 1 x 3 3 2 x - 5 x 4 \u003d \u003d 1 + 0 1 + 0 - 0 3 0 - 0 \u003d 1 ∞
After that, we need to transform the function to apply the second remarkable limit. We got the following:
lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d 1 ∞ \u003d lim x → ∞ x 3 - 2 x 2 - 1 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d \u003d lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5
lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d \u003d lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5
Since now we have the same exponents in the numerator and denominator of the fraction (equal to six), the limit of the fraction at infinity will be equal to the ratio of these coefficients at the highest powers.
lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 2 x 2 + 2 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d \u003d lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 6 2 \u003d lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3
Replacing t \u003d x 2 + 2 x 2 - 1 - 2 x 2 + 2 gives us a second remarkable limit. Means what:
lim x → ∞ 1 + - 2 x 2 + 2 x 3 + 2 x 2 - 1 x 3 + 2 x 2 - 1 - 2 x 2 + 2 - 3 \u003d lim x → ∞ 1 + 1 tt - 3 \u003d e - 3
Answer: lim x → ∞ x 3 + 1 x 3 + 2 x 2 - 1 3 x 4 2 x 3 - 5 \u003d e - 3.
conclusions
Uncertainty 1 ∞, i.e. the unit to an infinite degree is a power uncertainty, therefore, it can be opened using the rules for finding the limits of exponential functions.
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There are several wonderful limits, but the most famous are the first and second wonderful limits. The remarkable thing about these limits is that they have a wide application and can be used to find other limits found 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, it is not necessary to disclose the uncertainties they contain, since the values \u200b\u200bof these limits have long been deduced by great mathematicians.
The first wonderful limit 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 a trigonometric function is under the limit sign, this is almost a sure sign that this expression can be reduced to the first remarkable limit.
Example 1.Find the limit.
Decision. Substitution instead of x zero leads to uncertainty:
.
The denominator is sine, therefore, the expression can be reduced to the first remarkable limit. Let's start transformations:
.
The denominator contains the sine of three x, and the numerator contains only one x, which means that you need to get three x in the numerator. For what? To represent 3 x = a and get an expression.
And we come to a variation on the first wonderful limit:
because it doesn't matter which letter (variable) in this formula stands for the x.
We multiply x by three and then divide:
.
In accordance with the observed first remarkable limit, we replace the fractional expression:
Now we can finally solve this limit:
.
Example 2.Find the limit.
Decision. Direct substitution again leads to the zero-divide-by-zero ambiguity:
.
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 are with the same coefficient. Let this coefficient be equal to 2. To do this, we represent the current coefficient at x as below, performing actions with fractions, we get:
.
Example 3.Find the limit.
Decision. When substituting, we again obtain the "zero divided by zero" uncertainty:
.
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 x squares in the numerator and the sine in the denominator by the same factors, and to get the same coefficients for the x and sine, divide the x in the numerator by 3 and then multiply by 3. We get:
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Example 4.Find the limit.
Decision. Again we get the uncertainty "zero divided by zero":
.
We can get the ratio of the first two remarkable limits. Divide both the numerator and the denominator by x. Then, so that the coefficients for sines and for x 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.
Decision. And again the uncertainty "zero divided by zero":
Remember from trigonometry that tangent is the ratio of sine to cosine, and the cosine of zero is equal to one. We make transformations and get:
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Example 6.Find the limit.
Decision. The trigonometric function under the limit sign again suggests the idea of \u200b\u200busing the first remarkable limit. We represent it as the ratio of sine to cosine.
Collected formulas, properties and theorems used in solving problems that can be solved using the first remarkable limit. Detailed solutions of examples are given using the first remarkable limit of its consequences.
ContentSee also: Proof of the first remarkable limit and its consequences
Applicable formulas, properties and theorems
Here we look at examples of solutions to limit computation problems that use the first remarkable limit and its consequences.
Below are the formulas, properties and theorems that are most often used in this kind of calculations.
- The first remarkable limit and its consequences:
. - Trigonometric formulas for sine, cosine, tangent and cotangent:
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at,;
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Solution examples
Example 1
For this.
1. We calculate the limit.
Since the function is continuous for all x, including at the point, then
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2. Since the function is not defined (and, therefore, is not continuous) for, then we need to make sure that there is a punctured neighborhood of the point at which. In our case, at. Therefore, this condition is fulfilled.
3. We calculate the limit. In our case, it is equal to the first remarkable limit:
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$$ \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (6) (3x-5) \\ right) ^ (4x + 7) \u003d \\ lim_ (x \\ to \\ infty) \\ left (1+ \\ frac (1) (\\ frac (3x-5) (6)) \\ right) ^ (4x + 7) $$
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Similarly, we find the limit of the function in the denominator:
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at;
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Finally, we apply the arithmetic properties of the function limit:
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Let's apply.
When. From the table of equivalent functions we find:
at; at.
Then.
Example 2
Find the limit:
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Solution with the first remarkable limit
For,,. It's an uncertainty of the kind 0/0 .
We transform the function beyond the limit sign:
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Let's change the variable. Since and for, then
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Similarly, we have:
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Since the cosine function is continuous on the entire number axis, then
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We apply the arithmetic properties of the limits:
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Solution with Equivalent Functions
Let us apply the theorem on the replacement of functions by equivalent ones in the quotient limit.
When. From the table of equivalent functions we find:
at; at.
Then.
Example 3
Find the limit:
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Substitute in the numerator and denominator of the fraction:
;
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It's an uncertainty of the kind 0/0
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Let's try to solve this example using the first great limit. Since the value of the variable in it tends to zero, we will make a substitution so that the new variable tends not to, but to zero. To do this, we pass from x to the new variable t, making the substitution,. Then for,.
First, we transform the function beyond the limit sign by multiplying the numerator and denominator of the fraction by:
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Substitute and use the above trigonometric formulas.
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The function is continuous at. We find its limit:
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We transform the second fraction and apply the first remarkable limit:
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We have made a substitution in the numerator of the fraction.
We apply the property of the limit of the product of functions:
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Example 4
Find the limit:
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For,,. We have an uncertainty of the kind 0/0 .
Let us transform the function under the limit sign. Let's apply the formula:
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Let's substitute:
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We transform the denominator:
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Then
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Since and for, we make a substitution, and apply the theorem on the limit of a complex function and the first remarkable limit:
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We apply the arithmetic properties of the function limit:
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Example 5
Find the limit of the function:
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It is easy to see that in this example we have an uncertainty of the form 0/0
... To reveal it, we apply the result of the previous problem, according to which
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Let's introduce the notation:
(A5.1) ... Then
(A5.2) .
From (A5.1) we have:
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Substitute into the original function:
,
where,
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;
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We use (A5.2) and the continuity of the cosine function. We apply the arithmetic properties of the function limit.
,
here m is a nonzero number,;
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Example 6
Find the limit:
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When, the numerator and denominator of the fraction tend to 0
... It's an uncertainty of the kind 0/0
... To expand it, we transform the numerator of the fraction:
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Let's apply the formula:
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Let's substitute:
;
,
where.
Let's apply the formula:
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Let's substitute:
;
,
where.
Fraction numerator:
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The function outside the limit sign becomes:
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Let us find the limit of the last factor, taking into account its continuity at:
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Let's apply the trigonometric formula:
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Substitute,
... Then
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Divide the numerator and denominator by, apply the first remarkable limit and one of its consequences:
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Finally, we have:
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Note 1. You could also apply the formula
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Then.