Question 7. Don t read that wrong. How do you write without Lopida s rule?

Using McLaurin on the molecule, the result is x2 x[ln2+x(ln2) 2 2!+x^2(ln2)^3/3!+x^3(ln2)^4/4!

The whole equation becomes x2 x[ln2+x(ln2) 2 2!+x^2(ln2)^3/3!

x^3(ln2)^4/4!+.sinx, since the limit of x sinx is 1, it is reduced to 2 x[ln2+x(ln2) 2 2!

x^2(ln2)^3/3!+x^3(ln2)^4/4!+.

When x tends to 0, all the terms containing x in parentheses tend to 0, leaving only ln2, and 2 x outside the parentheses tends to 1, so the whole limit is 1*ln2=ln2

You can do it without Robida The above is changed to sinx*(cosx-1) cosx (extract the common factor sinx) below with the equivalent infinitesimal i.e. x sinx ln(1+x) x then it is the following =sinx*x 2 up and down at the same time to get sinx to get the original formula =(cosx-1) cosx x 2 that is=(cosx-1) x 2 (the cosx that is approximated is the result of bringing in x=0) and since 1-cosx x 2 2 then the original formula =-x 2 2 x 2 =- 1 2

Question (1) is because the limit of lim(x)sinx does not exist, so the limit of its molecule does not exist.

So it's not a type, so you can't use Lopida's law.

It should be: lim(x)x+sinx) x=lim(x)1+sinx x]=1

This is because sinx is finite, while lim(x)1 x = 0, infinitesimal multiplied by finite is still infinitesimal.

The reason for question (2) is the same, since the limit of lim(x 0)sin(1 x) does not exist, it is not of type 0 0, so it cannot be used by Lopida's rule, it can only be.

lim(x→0)x^2sin(1/x)/sinx=lim(x→0)[x/sinx]*xsin(1/x)=0

No, the second question is no, when the Lopida law requires that when the numerator and denominator are found to be the limits, they are either 0 or infinite.

The second question denominator is not.

Say I'm here to watch!

The first question, intuitively, the answer is one. sinx is a bounded function, and x tends to infinity. It can be seen that it is equal to 1

But it seems that the conditions of Lobida are met again! Finding 1+cosx with Lopida is an oscillation function!!

The answer is different!!

The second question, substitution with an equivalent infinitesimal is equal to x*sin(1 x).Since sin is bounded, it is equal to 0!

With Lopida, the formula is too complicated to write, and the answer is also an oscillation function!

It's really weird.

The use of equivalent non-fighting staring such as the poor base Qi small substitution. Zeming.

The detailed process of carrying the paragraph is shown by the stool Xun rt, and the argument is very clear.

Yes, the numerator and denominator are approaching 0 at the same time, and you can use Lopida's rule.

Lopida's rule is that the numerator and denominator are derivatives separately.

The numerator is: ((x+1)ln(x+1))).'=ln(x+1)+1 denominator is: (x).'=1

So: lim[x 0]((x+1)ln(x+1)) x=1

If x tends to 0, calculate 1+x first, and then substitute it with equivalence, 1

Yes, it can be used, but it is not easy to use.

Directly use the important limit, sinx and x are equivalent infinitesimal and the numerator and denominator can be directly reduced.

1. The essence of Lobida's law is consistent convergence, in your problem, sin(1 x) is not convergent when x 0, so it does not meet the condition of consistent convergence;

2. Understand the best application of Luobida, don't just memorize those formulas, memorizing formulas is a waste material, useless! 3. This question can only be calculated by using the Cauchy criterion (pinch criterion).

According to the equivalent infinitesimal definition, the original formula is equivalent to:

lim(x→0) xsin(1/x)

Whereas: -x xsin(1 x) x

Original = 0

If the former collapse, the slag shown in the figure below shows the wisdom shed.

<> such as a frank touch of socks let the excitement be noisy.