Four algorithms of function limit: If there is a function, if LIMf(x) = A and LIMg(x) = B are in the same changing process of independent variables f(x) and g(x), then

lim[f(x)g(x)]= limf(x)limg(x)= A B

lim[f(x)？ g(x)]=limf(x)？ limg(x)=A？ B

lim==(B≠0)

(Similar to four algorithms with order restrictions) Now take the discussion function as an example.

For the function limit in the form of sum, difference, product and quotient, we will naturally think of four algorithms of limit, but to use these algorithms, we often need to do some same deformation or simplification of the function first, and then use the four algorithms of limit. These methods are:

1. Direct substitution

For the limit f(x) of elementary function f(x), if the function value f(x) of f(x) at point x exists, then f(x)=f(x).

The essence of direct substitution method is that as long as x=x is substituted into the function expression, if it makes sense, its limit is the function value.

Example 1: Find the limit (x+3).

Solution: (x+3)=2+3=7.

2. Conversion method between infinity and infinitesimal

In the same change process, if the variable does not take zero value, then the variable is infinite? Its reciprocal is infinitely small. For some special limits, we can use the reciprocal relationship between infinity and infinitesimal to solve them.

(1) When the limit of denominator is "0" and the limit of numerator is not "0", the arithmetic of limit quotient cannot be directly used. Instead, the limit of f(x) is obtained by using the reciprocal relationship between infinity and infinitesimal.

Example 2: Seek.

Solution: ∫= = 0

∴=∞。

(2) When the limit of denominator is ∞ and the numerator is constant, the limit of f(x) is 0.

Example 3: Seek.

Solution: =0.

3. Divide by an appropriate infinite method

For the limit is ""type, the limit quotient algorithm cannot be used directly, and the denominator and numerator must be divided by an appropriate infinite quantity x first.

Example 4: Calculation.

Solution: = = 3.

The general situation has the following conclusions:

Let a≠0, b≠0, m and n be positive integers, then

=0, when n > m, when n=m ∞, when n < m.

4. Physical and chemical methods

Applies to rooted limits.

Example 5: Calculate (-).

Answer: (-) =

==0。

Second, use pinch criterion to find the limit.

Pinch theorem of function limit: let functions f(x), g(x) and h(x) be defined in the centripetal neighborhood of x (or | x |x|>N), if ① f (x) ≤ g (x) ≤ h (x); ②f(x)=h(x)=A (or f(x)=h(x)=A), then g(x) (or g(x)) exists, and g(x)=A (or g(x)=A). (similar to the pinch theorem that can get the limit of sequence)

The key to using pinch criterion is to choose appropriate inequalities.

Example 6: Calculate x [].

Solution: when x > 0, there is 1-x < x [] ≤ 1. With the pinch criterion, there is (1-x)= 1, so there is x [] = 1.

Third, use monotone bounded criterion to find the limit.

Monotone Bounded Criterion: Monotone Bounded Sequence Must Have Limit.

Firstly, the monotonicity and boundedness of sequence are discussed by mathematical induction, and then the limit is obtained by solving the equation.

Example 7: Prove that the sequence … has a limit, and find its limit.

It is proved that (1) proves that the sequence is bounded, and it is easy to know that {x} is increasing, and x≥

Prove that x≤2 by mathematical induction, obviously x = < 2,

If x≤2, then x =≤2.

(2) Prove that the sequence is monotonically increasing, and x-x=-x==.

Use (1) 0 < x < 2? … x-x-x>0。

(3) Using monotone bounded convergence criterion, x = a.

(4) x=，x = 2+x。

Take the limit at both ends of the equation to get a=2+a, and get a=2 or a=- 1 (obviously unsatisfactory, omitted).

So x=2.

Fourth, find the limit by equivalent infinitesimal substitution

Common examples of equivalent infinitesimal are: when x→0, sinx ~ x；; tanx ~ x； 1-cosx ~ x； e- 1 ~ x； ln( 1+x)~ x； arcs inx ~ x； arctanx ~ x； ( 1+x)- 1~x .

Substitution theorem of equivalent infinitesimal: Let α(x), α′ (x), β(x) and β′ (x) all be infinitesimal of independent variable X in the same changing process, and α (x) ~ α′ (x), β (x) ~ β′ (x) and lim exist, then lim=lim.

Example 8: Calculation.

Solution: Using equivalent infinitesimal substitution,

Yes = = =.

Note: When the denominator or numerator is the subtraction of two equivalent infinitesimals, they cannot be simply replaced by their respective equivalent infinitesimals, otherwise it will lead to wrong results. From another point of view, equivalent infinitesimal substitution is suitable for product sum quotient, but not for simple substitution in addition and subtraction.

For example, when x→0, tanx ~ x and sinx ~ x have ==0.

The above formula is wrong because when x→0, although tanx ~ x and sinx ~ x, the speed at which tanx and sinx(x→0) tend to zero can only be approximately equal, but not completely equal.

Five, using the nature of infinitesimal to find the limit

In the nature of infinitesimal, especially the property that the product of infinitesimal and bounded variable is still infinitesimal is used to find the limit.

Example 9: Calculate xsin.

Solution: When x→0, x is infinitesimal, from |sin|≤ 1, that is, sin is bounded, so xsin is infinitesimal, so xsin=0.

Six, using two important limits to find the limit

When using two important limits = 1 and (1+)=e to find the limit, the key is to properly deform the given function or sequence to make it have the corresponding form, and sometimes the problem can be simplified by variable replacement.

Example 10: Calculation.

Solution: = = 2.

Example 1 1: Calculation ().

Solution: () = [(1+)] = e.

Seven, using L'H?pital's law to find the limit.

If the functions f(x) and g(x) both tend to zero or infinity at x→a (or x→∞), they may or may not exist. Usually, such a limit is called "type" or "type" indefinite formula respectively. Generally speaking, the limit algorithm cannot be applied to this kind of limit, but the limit can be found by using the law of Lobida.

L'H?pital's law:

Let (1) limit be of type or indefinite type;

(2)f(x) and g(x) are differentiable in eccentric neighborhood (x) or | x | > x, and g' (x) ≠ 0;

(3) Existence or infinitesimal, then =.

Other infinitives, such as "0? Type ∞, Type ∞ -∞, type 1, type 0, type∞, L'H?pital rule cannot be used directly, and L'H?pital rule can only be used if it is changed to type or type.

Example 12: Calculation. (type)

Solution: ==2.

Example 18: calculation (sinx). (Type 0)

Solution: (sinx) = e = e = e = e = e = e = e = e = e = e = e = e = e =1.

Eight, use Taylor formula to find the limit.

If the function f(x) has a derivative up to order n in the open interval (a, b) containing x, then when x is in (a, b), there is always f (x) = f (x)+f' (x)+(x-x)+…+(x-x).

Where o [(x-x)] is called piano remainder, and when x=0, the above equation is called McLaughlin formula.

For some complicated limit problems, McLaughlin formula can be used to solve them.

Example 19: Calculation.

Answer: =

==。

When using Taylor formula to find the limit, we should use it flexibly to distinguish which items need to be expanded and which items can be kept. Taylor formula is a powerful and effective tool to find the limit of complex variable function.

Nine, using the definition of definite integral to find the limit

If you encounter some summation limit problems, it can be expressed as the integral sum of an integrable function, and you can use definite integral to find the limit. The key is to determine the integrable function and integral interval according to the given summation formula.

Example 15: calculate sin+sin+…+sinπ.

Solution: the original formula =sin+sin+…+sinπ+sinπ=? Sin π xdx = [cos π x] =