In finding the limit, what can 1 cos3x be replaced with

。The "limit" in mathematics refers to the process of a variable in a certain function, which gradually approaches a certain definite value a in the process of becoming larger (or smaller) forever and "can never coincide to a" ("can never be equal to a, but taking equal to a' is enough to obtain high-precision calculation results"), and the change of this variable is artificially defined as "always approaching without stopping", and it has a "tendency to constantly get extremely close to point a".

Limit is a description of a "state of change". The value a that this variable is always approaching is called the "limit value" (which can also be represented by other symbols).

Criteria for the existence of limits:

Pinch theorem. Monotonous bounded criterion.

Cauchy Criterion. In this problem, when x tends to 0, 1-cosx is equivalent to, then it is obvious that 1-cos3x is equivalent to, i.e.

Keep in mind that when x is approaching 0, 1-cosx is equivalent to.

Well here, apparently, it can be obtained.

1-cos3x equivalent.

<> for the Lu Kao, please smile.

With the formula of double angle:

cos2a=1-2sin²a

1-cos2a=2sin²a

So: 1-cosx=2sin (x 2) 2 (x 2) x 2

The slag circle is as follows: 1-cosx equivalent infinitesimal is x 2

Equivalent infinitesimals are a relationship between infinitesimals, which means that if the limit of the ratio of two infinitesimals is 1 in the process of the same independent variable tendency, the two infinitesimals are said to be equivalent. The infinitesimal equivalence relationship depicts that the velocity of the two middle beams moving infinitesimals towards zero is equal to the speed of selling beams.

Equivalent infinitesimal substitution is a common method for calculating the unshaped limit, which can simplify the problem of finding the limit.

The denominator can be replaced directly.

Because there is addition and subtraction of the molecule, it cannot be replaced at will, but Wu Yingzhou can try.

For example, in this problem, (1-cos3x)sin2x is of the same order as x 3, i.e., it is of the same order as the denominator.

Therefore, the formula can be split into two parts.

lim[（1-cos3x)sin2x]/x^3 - lim[x^4sin(x^-1)]/x^3

The results are easy to get.

But if the denominator is 4th, i.e. the equivalent infinitesimal.

Below the order of the denominator, it cannot be split, because the post-split limit does not exist.

In this case, i.e., the cavity mask makes (1-cos3x) sin2x multiply, and cannot be replaced.

That's where the Lopida rule comes in.

Taylor or physical and chemical methods to deal with.

Answer: Use the formula of double angle of shirt bush.

cos2a=1-2sin²a

1-cos2a=2sin²a

So: 1-cosx = 2sin (x 2) 2 (x 2) x 2 So: 1-cosx is equivalent to infinite or too small.

Quarrel for the X 2

lim√(1-cosx)/tanx

lim-√2sin(x/2)/tanx

lim-√2/2x/x

lim√(1-cosx)/tanx

lim√2sin(x/2)/tanx

lim√2/2x/x

Because lim (1-cosx) tanx≠lim= (1-cosx) tanx

So the limit does not exist, 5, the tea of the heart report.

If the root number is greater than 0, for example, if this question is changed to x, you don't need to use it If the root number happens to be 0, you have to consider it, because the first drawback is that the game cannot be replaced.

It is only in multiplication and division operations that the equivalent infinite substitution can be exchanged for addition and subtraction is not available in textbooks.

For example, :(1-cosx) x

If you substitute 2sinx 2 2*x 2=x lim=1

If we use Robida =lim sinx 1=0,2, lim (1-cosx) tanx is substituted with equivalent infinitesimal substitution to find the limit.

I replaced the numerator (1-cosx) with (2) sinx 2 (2) x 2 denominator tanx with xIn the end, the limit is (2) 2, but the correct result is that the limit does not exist, which is why, I made a mistake in solving the problem.

x tends to infinity and is too big.

2sin (5 x 2) sin (x 2) does not have a sin vertical shirt (oscillating type) fiber hall.

Original tangerine = lim(x->0) or x 2

lim(x->0)(9/2x^2)/x^2

9 Absolute 2

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.

The limit of cosx+2 is equal to cos x is early mountain drain because it tends to zero then x 2 is 0, then limcos(x+a land rot 2) is not left only limcosa, variable x

It doesn't exist anymore, and the restriction that x tends to 0 is useless, and it's cosa.