The law of high numbers Lopita Lopita seeks the limit

Solution The third step of the following process uses equivalent infinitesimal substitutions, the fifth step uses Robida's rule, and the others are identity transformations.

lim(x→0）[（1+x）^(1/x)-e]/x=lim(x→0）/x

Using Lopida's rule, the numerator and denominator are derived at the same time, lim(x 0) x

lim(x→0）exp[ln（1+x）/x]*[ln(1+x)/x]'

Because by using the Lopida rule, we can get lim(x 0)ln(1+x) x=1

So lim(x 0)exp[ln(1+x) x]*[ln(1+x) x].'=e*lim(x→0）[ln(1+x)/x]'

e*lim(x→0）[1-1/(1+x)-ln(1+x)]/x^2

Again, using the Lobida rule, e*lim(x 0)[1-1 (1+x)-ln(1+x)] x 2=e*lim(x 0)[[1 (1+x) 2-1 (1+x)] 2x

e 2 so lim(x 0)[(1+x) (1 x)-e] x=-e 2

Equal to -e 2 mobile phone typing trouble, can't write all the steps for you, prompt you, is to power (1 x) x to e 1 x ln (1+x) and then there is an e? These two items put forward an e at the same time, it is revealed, the form of e u-1, this u drive is 0, e u-1 is equivalent to u, and then use Luobida, it is easy to figure it out!

The solution is as follows: grasp the seepage diagram of the god worship section of the blind spine.

Therefore, it is enough to find the limit of the point x 0-, lim = positive infinity. x 0+, lim=negative, no day, he is poor. The point limit does not exist. I hope we can work together to solve the problem.

A situation where the limit exists but cannot be determined by the law of Lopida. 4. An in-depth understanding of the law of Lopida (insights and answers to the above questions). 5 The blind use of the Lopida rule leads to cumbersome calculations. 6 A variety of solutions to a simple problem.

This series of articles explains the basic content of advanced mathematics, pays attention to the cultivation of learning methods, and often does not hesitate to explain the problems that are not easy for beginners to understand, and connects with high school mathematics as much as possible (advanced mathematics courses need to use some less important content in high school mathematics, such as polar coordinates, which we will supplement when we use them). And appropriately discarded some of the more difficult or advanced mathematics courses (such as the proof of the limit in -δ language, and the proof of some theorems in the textbook).

Those who know me say that I am worried, and I want to ask.

Of course, you can shout and tell me. It's infinite, it's infinite. The derivative of the numerator is -cotx*cscx=-cosx (sinx) 2, and the derivative of the denominator is 1 x, so after slowing down with Lopida's rule, =lim -x*cosx (sinx) 2 =-lim cosx sinx =-infinite Zheng Ming.

I don't understand the first question, why do you use Lopida.

The first question uses one of the two important limits, and it comes out at once (as shown in the upper left corner of the picture). Or it can be replaced with an equivalent infinitesimal hail bridge, which can also be very simple (as shown in the lower left corner of the picture). If you have to use Lopida.

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Hello friends! The bureau letter is detailed and clear about the process, and I hope to help you solve the problem.

1.Overview. Finding the limit is one of the most important contents in the advanced mathematics course (it is a must to learn advanced mathematics along with the derivative, the integral slow bridge, and the dispersion of the judgment series), so it is also a compulsory content for all kinds of advanced mathematics exams.

Lopida's law plays an important role in finding the limit, and it can be said that, except for a few difficult limits that need to be mastered later by Taylor's formula, which is to be learned, most of them are required to be mastered in advanced mathematics.

2.Limit calculation with roots.

3.Commentary to example 1.

This series of articles explains the basic content of advanced mathematics and learning, pays attention to the cultivation of learning methods, and often does not hesitate to explain the problems that are not easy for beginners to understand, and connects them with high school mathematics as much as possible (advanced mathematics courses need to use some less important content in high school mathematics, such as polar coordinates, which we will supplement when we use them). It also appropriately omits some content that is more difficult or does not require too much in advanced mathematics courses (e.g., proving limits in -δ language and proving some theorems in textbooks).

This series of articles is suitable as a reference material for synchronous tutoring in the classroom for beginners in advanced mathematics, for the final review of advanced mathematics and for the first round of review for graduate school entrance examinations. Most of the example problems involved in the sedan rock are routine questions with a solid foundation and conceptual analysis questions that help deepen understanding, with moderate difficulty, and some classic topics in postgraduate mathematics are selected.

Studying High Numbers: Finding the Limit of Lopida's Law Studying for Graduate School of Extreme Numbers: Finding the Limit of Lopida's Law Earlier, the four algorithms of finding the limit were introduced in three aspects: function decomposition, grasping the big head, and discussion of the convergence of the limit.

Question 1. Apply the basic positive acceptance limit formula, the original formula = e a.

Question 2. It belongs to the "0 0" type, and it should be useless to reach the law quietly. Original = lim(x 1)(3x +1 x) e x=4 e.

The solution process is as follows, the first one is to take the correct reserve and then replace it with 1 x=t (you can try this if you encounter a multiplication form without a denominator), and then you can use the equivalent infinitesimal (ln(1+x) x).

The problem is clear with two 0 0 type, Luo Bi is not destroyed after the simplification, the simplified equation is continuous on x=1, and the limit value is equal to the value of the point function, so just bring it in.

Of course, you can shout and tell me. It's infinite, it's infinite. The derivative of the numerator is -cotx*cscx=-cosx (sinx) 2, and the derivative of the denominator is 1 x, so after slowing down with Lopida's rule, =lim -x*cosx (sinx) 2 =-lim cosx sinx =-infinite Zheng Ming.

Except for the point, the function is continuous at all other points, so the limit of any point is the value of the function, however, the endpoint is a one-sided limit, pay attention to yourself.

Therefore, it is only necessary to ask for the limit of the point.

x 0-, lim = positive infinity.

x 0+, lim = negative infinity.

So, (the point limit does not exist.

I hope we can work together to solve the problem.

x→0lim (cotx-1/x)

lim 1/tanx-1/x

lim (x-tanx) xtanx) the limit is 0 0 type, according to l'hospital rule = lim (x-tanx).' xtanx)'

lim (1-1 cos 2x) tanx+x cos 2) = lim (cos 2x-1) sinxcosx+x) this limit is 0 0 type, according to l'hospital rule = lim (cos 2x-1).' sinxcosx+x)'

lim -2cosxsinx / cos^2x-sin^2x+1)=-lim 2sinx / 2cosx

If you don't understand, please ask.

cotx=cosx sinx, and 1 x pass to find lim(xcosx-sinx) xsinx0 0 type infinitesimal.

Use the Lobida rule twice in a row.

The final available limit is -lim(x 0)(sinx+xcosx) (2cosx-xsinx)=0

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