How many high number limits are asked, high numbers?

1.-1/2.Multiply [root(x2-1) +x] by both up and down, and then divide by x at the same time

0 , sinx approximation is x, so the original approximation is 6 8, and the result is 3 43Lobida's law, once, get 2

Original = x->0, sinx x, special limit, the result is 15Lobida's law, get 1 2

Special limits. The limit is 1, don't think about it, get 1 2

1.The molecule is physicochemical -1 2

2.Equivalence Factor Method 3 4

3.Same as above (split into two parts)2

15.The numerator and denominator are divided by x 1 2

6.There's a formula, which is that the limit of e is normalizing, and I don't remember exactly. 1/2

lim(x-> 0+) x [1 (1+lnx)]lim(x->0+) e [lnx (1+lnx)].

lim(x-> answer 0+) e [(1 x) (1 x)]e

If not, you and I will be fine.

The most talked.

Why can't we just bring in x equals infinity, and the infinite form of 1 equals e? This idea is incorrect, because 1 (+ is an infinitive form, so its limit value should be obtained by simplification.

This limit can be transformed by x=e ln(x) to obtain the (0 0) type, the value of which is calculated using the Lopida's rule.

The detailed calculation process is as follows:

The reason why the diagram on the right is wrong is that the added 1 x is not 0, so it is not an identity deformation, and it has to be infinite.

When x approaches infinity, 1 x approaches 0, but in the intermediate process of finding the limit, it is not yet 0.

Otherwise, the important extremum lim(1+1 x) x becomes lim(1+0) x = 1, which is obviously wrong.

None of the people who know how to calculate the correct answer tell you a single thing about what you really care about.

It is the principle of limit splitting: only the formula of addition, subtraction, multiplication and division can be "possible" to be split to calculate the limit.

The stem of this problem is a function with x in both the exponential base and the power, which does not meet the basic conditions of limit splitting. So you solve for natural errors by substituting the limit in the base part.

This is why the correct method is to convert the formula with the natural logarithm as the base first, and then use the principle of limit difference to solve it.

It cannot be directly substituted, because this is a structural form similar to a power finger function, and it needs to be deformed logarithmically before calculation.

If the f(x) limit is 1, then there is a neighborhood such that |f(x)-1|<1 2 At this time|f(x)|=f(x)-1+1|>1-|f(x)-1|>1/2

Ropita Law lim f''(x) (x-x0)=a then there is f in the neighborhood of x0''(x)/(x-x0) >a/2>0

i.e. left f''(x)<0 f(x) convex right f''(x)>0 f(x) concave.

The derivative of the numerator x is 1 and the derivative of the denominator e (x e) is e (x e) (x e).'=e^(x/e) (1/e)

At x->0+ limxlnx =lim lnx (1 x)=lim (1 x) (1 x 2)=lim(-x)=0

limx(lnx)^k=lim (lnx)^k /(1/x)=lim k(lnx)^(k-1)(1/x)/(1/x^2)=-klimx(lnx)^(k-1)

Mathematical induction can be used to prove that the positive integer k limx(lnx) k=0

k is not a positive integer, and can be pinched with two integers.

The latter is similar.

Here's how, please refer to:

Bring in x 0, where the limit of cosx is 1 and x is 0

x tends to be 0 positive, cosx tends to 1 positive, does not affect the formula, and the remaining x square is divided by sinx, and x square is the high-order infinitesimal of sinx, which is naturally 0

When x tends to 2, the numerator and denominator tend to search forward to zero, and Lopida's rule applies.

After using Lopida's rule, the numerator is cosx, the denominator is 1, and it is flame sensitive.

So the limit is 0

For reference, please smile.

The result of this limit is 0, and perhaps the teacher said that the reciprocal of this limit is infinity.

The first derivative is equal to zero, and the standing point is obtained.

If the second derivative is greater than zero, the resulting function image is a canopy pin function, i.e., the left subtracts and the right increases.

Therefore, there is a unique minimum chain carrying value point, and the hidden skin is the function with a minimum value.

For reference, please smile.

<> the value of k can be found using the second important limit formula.

If this limit is proved by definition (-definition), then the proof process is quite cumbersome.

In general, it can be proved by using the method of finding the limits using the two important limits or the Lopida's rule.

Take the Lopida Law as an example:

The derivative of the numerator is -1 to the power of (1+x) and the derivative of the denominator is 1, so this limit is equal to -1 power of (1+0).

For reference, please smile.

Apply the Robita rule to solve, because the 0 0 is indefinite, you can use the Lobida rule.

lim(x->0) /x

lim(x->0) [u(1+x)^(u-1)]/1u。

Hello dear. "Limit" is a fundamental concept of calculus, a branch of mathematics, and "limit" in the broad sense means "infinitely close and never reachable". "Limit" in mathematics refers to:

In the process of a certain variable in a function, which gradually approaches a certain definite value a 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).

You can send us the specific problem and I will help you solve it

Since it is 0 0 amorphous, you can use the Lopida rule.

lim(x->0) /x

lim(x->0) [u(1+x)^(u-1)]/1u

Just find the relation between a and b.

That's the good way.