Математика. Материалы курса

Calculation of limits. Remarkable limits.

Today is a wonderful day! Why all this? Because we will consider remarkable limits. Unfortunately, there are not so many of them, two only, but there are enough consequences, so you will have something to do and study.

First, let’s look at the limit theorem for function composition, which is important when calculating limits. If the function F(x) = g(f (x )). f is an internal function, g is an external one. Lim f(x0) = u0, and lim_(x→0) g (u)=A, then lim_(x→0) F(x0)=A. The easiest way to understand this theorem is to consider the following example:

Sin(5x) / x at the point 0, the uncertainty is 0/0, everything would be fine if it were sin(x)/x, it would be pure. The first remarkable limit, which is 1. But 5 hinders us. Let’s make a substitution. We introduce a new function, t=5x, and express x=t/5. We got a new argument function t → 0, and we see that 5*sin(t)/t is the first remarkable limit that is equal to 1. The answer is 5.

Most often, we will not introduce a new variable, but we will perform transformations, based on this theorem in the following way. Let’s take into consideration some examples.

The first limit, substituting instead of x = 0, remember, sin, arctg, They are all equal to 0 at the point 0, we get the uncertainty – 0/0, it is trigonometry, so this is the first remarkable limit and its consequence, sin and arctg are involved here, the formulas of the first remarkable limit and consequence (lim(x->0) sin(X)/X = 1; lim (x->0) arctg(x) / x =1.)

How do we use them? After all, in the example, we see not sin and not arctg, but something else. Look at what we are doing. What do we do with the numerator? We form a unit using the first remarkable limit. Sin(2x)/2x, but since this is the third power, we multiplied 2x by the third power, further in order: with arctg, we also formed a unit by dividing it by 3X, since this is squared, multiplied by (3X)^2.

And the last operation – sin(5x), we did exactly the same: we divided and multiplied it by 5x. All the time we follow the identity of the transformations, we multiply by what we divided by. We have circled the units. We may not notice them. Consequently, we do not notice the units, we do everything that is left: we raise them to the third power and square them, multiply them, we need to reduce them by x^3 and write down the answer. The result is obtained.

The second example is with Cos.

Trigonometry is the first remarkable limit and its consequences. We remember the formula that we need, look, to get 1, you need to divide by the square of the variable divided in half. To form 1 – in the denominator of the fraction, write ((3x)^2)/2 and in the numerator of the fraction – follow the identity of the transformations. We circle the unit, and then perform multiplication, reduction, and write down the answer.

Another example:

Tg and arcsin are involved here. For tg, we divide by 5x and multiply by 5x. With arcsin, we form a unit under the root. And we multiply by the root of (4x)^2. We circle the units, perform transformations, and write down the answer.

The most important thing is that the uncertainty is 0/0, the variable tends to 0 in all the examples, and the trigonometric functions that participate in the first remarkable limit are transformed according to the desired formula.

Let’s consider the second remarkable limit. We should remember that it deals with a new uncertainty that we haven’t worked with yet, which is 1^∞. If we substitute x for 0, 1-2*0 = 1. And in the exponent 1/0 – infinity. It is necessary to keep in mind that a unit in any degree is a unit, but not to an infinitely large degree. This is an uncertainty that needs to be disclosed.

When the variable tends to 0, we will use the formula (lim_(x→0) (1+x)^(1/x) = e). We have two forms of writing the second remarkable limit: for x → 0 and for x → ∞. What are the important points here? What should you pay attention to? Based on the power of 1+, in our example 1-. Therefore, the first step we took was to write down 1+, which value (-2x), we left the exponent unchanged.

We pay attention to the formula. The exponent must contain the equation: divide 1 by a term that is added to 1. What have we done? We divided by (-2x), and for the transformation to be identical, multiplied by (-2x). And now we can circle what our formula gives the number of e. The base is e, the exponent can be reduced to x, and we get (-2/3), the answer is e^(-2/3).

Second example:

We will calculate: 1-(1/ ∞)=1, this is the base. Indicator: 3x+1 → ∞ this is an uncertainty of 1^∞. We will use the formula when x → ∞. What does it have to be? (1+), we have (1 -), so the first action is to replace the base with (1+), a fraction with the sign ” –“, and we saved the exponent. If we add a value in the formula 1+ 1/x, what is x? This is 1 divided by the second term, so the numerator must be (-2x).

In order for the transformation to be identical, if we have multiplied, we must also divide. We circled e and got: the remaining equation is not reduced. Therefore, you receive another entry. We have to calculate the limit of the ratio of polynomials, here we cannot perform identical transformations, since the limit of the ratio of polynomials on ∞ is equal to the ratio of coefficients for the highest degree of the variable, here 3 and (-2). So we immediately write down the answer: e^(-3/2).

Let’s see how the consequences work. The situation is as follows, we always adjust to the right formula. First, you need to know what the formulas are, and second, how to use them correctly. First, we calculate by substituting 0 instead of x, ln(1)=0, with 0 in the denominator. Uncertainty – (0/0). Everything that is connected with logarithms, exponents, power-exponential functions – this is all the second remarkable limit and its consequences. We recall the necessary formula (lim_(x→0)(ln(1+x))/x =1). Note that in the formula we have ln – (1+). We change it, look how we should write it down: first, 2 was taken out of the limit sign and in the numerator 1+(- x), and in the denominator identical transformations were performed: the “-“ sign appeared both under the limit sign and before the limit. And we receive the specified formula. The circled formula gives 1. The answer is (-1 / 2).

Let’s consider another limit: