Under what circumstances can the limits be found separately?

If you can ensure that the limits of separation exist, you can open it and ask for it, note that it is only the prophet whose existence it is, not that it can be found after disassembly. If you encounter something that has been disassembled during the exam, don't be fooled, this kind of limit can definitely not be taken apart, and it can be solved by methods such as rationalization and chemistry.

The premise is that the limit of part A exists, and the limit of part B also exists, and the limit cannot be infinity. The first graph cannot be split because (1-cosx) x 4 is at infinity when x approaches 0.

When infinity approaches this number from the left side of the point, the whole function tends to a particular number; The right limit is the limit when the infinite approach from the right side of this point occurs, and the sufficient and necessary conditions for the existence of the limit are that the left and right limits exist and are equal.

Limit": In the process of a certain variable in a 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"), 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).

With four operations to split, as long as one of them is a definite non-zero constant (5, 6 or something, it can't be ), it can be split, and the other limit exists or not.

1) lim(f(x)+g(x)) = limf(x)+limg(x), provided that limf(x) and limg(x) both exist.

It can be obtained that if lim(f(x)+g(x)) exists, and it is also assumed that limf(x) also exists, then limg(x) must exist. That is, as long as two of the limits exist, the other exists.

2) The situation in lim(f(x)*g(x)) and (1) is the same.

In the above two cases, if part of the limits is a non-zero constant, then you can calculate directly, because the addition, subtraction, multiplication and division of the other part of the limits and this non-zero constant will not affect the limit of the whole, and in addition: a(x)b(x)=c (non-zero constant), if a(x) is an infinitesimal quantity, then b(x) must be an infinitesimal quantity. If a(x) is an infinitesimal quantity, then b(x) ratio is an infinitesimal quantity.

Of course, you can also write it as a divider, in a similar way.

1. For continuous functions, there is no need to divide the left and right limits.

2. For discontinuous (piecewise functions), it is necessary to find the left limit and the existing limit, and if the two are equal, the function limit exists.

is set as a series of infinite real numbers. If there is a real number a, for any positive number, no matter how small, n>0 makes the inequality |xn-a|< is constant on n (n,+ then the constant a is said to be the limit of the series.

This is the four-rule algorithm for using limits. Then determine if you can partially substitute it.

This is basically the same principle when substituting with an equivalent infinitesimal later.

The substitution of the limit bai is the three operations du that follow the limit

Legal. Over here.

zhilim f(x) lim g(x) is already a dao0 0 form, and the third clause of the limit algorithm requires that the denominator limit is not equal to zero, so the limit cannot be substituted. Therefore, it is impossible to put sinx x or cosx into it.

The problem requires you to find the limit, and generally you don't need to consider the left and right limits, that is, the usual limit questions are often not considered.

However, the problem of proving the existence of the limit or verifying the existence of the limit needs to be considered, and if it is a piecewise function, the relationship between the left and right limits and the value of the function at the point must be considered at the breakpoint, and if the limit in the problem tends to 0- or 0+ or something, you need to find the limit, generally speaking, the limit at the point does not exist, but the left or right limit exists.

When you need to find the left and right limits, the value that is the x trend is not in the x definition, or it has an absolute value sign, or here it is a first-class discontinuity, right?

I wonder how to determine whether it is a positive sign or a negative sign when you need to find the left and right limits, and whether all non-elementary functions have to add negative absolute signs?

For example, sinx x 0+ is +1 0- is -1 How is this determined? And the power of x.

The owner of the top floor asks for the same thing.

I don't understand the two values?

The limit is to see if the reciprocal on both sides of the defined number is equal, if it is equal, it is not necessary, and if the reciprocal is not equal, the limit is required, I hope it will help you.

Because when x 0-, 1 x - e (1 x) 0, can be swapped directly.

However, when x 0+, 1 x + e (1 x) + this is divided by e (1 x) to get the answer.

The limit of heavy spike rise is divided into left and right limits. The left and right limits of the limit cannot be brought in directly, and these two questions should be found according to the law of Lobida. The limits of these two problems cannot be directly brought into x, because none of the functions of the limits are 0s.

Limit Solver Note:

If yn is the key to finding y and z or h and g on both sides.

Type 1 is an undefined limit type that cannot be directly substituted.

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Because the left limit and the right limit may not be equal.

1. When x 0+0, 1 x + 1 x - the first half, the numerator and denominator are divided by e (4 x), then lime (-4 x) = 0, lime (-3 x) = 0, and the limit of the first half is 0

The limit of the second half is 1 , (using equivalence, sinx x, ln(1+x) x).

So, when limf(x)=1 (at x 0+0)2 and x 0-0, 1 x - then lime (4 x) = 0, lime (1 x) = 0, and the limit of the first half is 2

The limit of the second half is 1 , (using equivalence, sinx x, ln(1+x) x).

So, limf(x)=1 (when x 0-0) is the original equation=1