The method of finding the limit of the following function, the method of finding the limit of the fo

Shouldn't there be an approximation value for the limit?

Directly find the derivative, the first derivative is y=5 3x (2 3)-2 3x (-1 3), and then find the second derivative, we can see that at x=2 5, the first derivative is zero, and the second derivative is not zero, so x=2 5 is the extreme point. (Extremum second sufficient condition).

You proposed 1 3x (-1 3) above, and the result is the same, what intermediate process are you asking?

You're coming up with 1 3x (-1 3). In fact, you don't care about the form, you don't have to transform into this form at all, just make Y'=0, and then get the equation, and the cubic of both sides will solve.

2.[1+(3-x)/(x+2)]^x+2)/(3-x)*(3-x)/(x+2)*(1/(x-3))

e^(-1/5)

tends to infinity, 1 x tends to 0, cos(2 x)-1 1 x) 2 primitive=-1

Summary. Hello dear<>

From what I understand, the question should be like this:1Original ==0(2) The original formula ===3) =, so the original formula == Analysis: (1) is "type, should be deformed: (2) 3) is the type, and it is necessary to remove the "commission factor".

Find the limits of the following functions.

Hello dear<>

According to my reasoning, the question should be as follows:1Original ==0(2) the original formula ===3) =, the pure burned to the original formula == analysis: (1) is the "type, should be deformed: (2) 3) is the type, need to be about peeled to the "factor".

Dear, can you type, the backstage** can't see Oh, you're not in a hurry, I'm waiting for you.

Only ** can be sent.

Dear, you can also type, I can help you this way.

What. Dear, can you type, the backstage** can't see Oh, you're not in a hurry, I'm waiting for you. You can type too, I can help you with that.

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

If there is help from the landslide, please celebrate.

Summary. The numerator denominator is physical and chemical, and the numerator and denominator are multiplied by ( x+1)[x (2 3)+x (1 3)+1)]limx 1 ( x 1) x 1lim[(x-1)( x+1)] lim( x+1) (x 2 3+x 1 3+1)=2 3

Question 2 and Question 5.

The third big question. There are 4 major questions and small questions.

This is the third question.

The numerator denominator is physical and chemical, and the denominator of the band family is multiplied by ( ascending Qi Zhuoling x+1)[x (2 3)+x (1 3)+1)]limx 1 ( x 1) x 1lim[(x-1)( x+1)] lim( x+1) (x 2 3+x 1 3+1)=2 3

The fifth sub-question of the second major question is directly substituted into the calculation, and the limit is 1

lim(x—>2) 4 under the number (x-1)=1

limx 1 ( x 1) (x 1) This is also based on the Lopida rule.

I didn't get it. Hurry up with Lopida.

Directly substitute x=2 into the calculation.

The third problem is that the left limit is equal to the right limit, and then the value of a is calculated.

The result of the fifth question of the fourth question is divergent, and the limit of the sixth question is 0, the highest power method, and the numerator and denominator are divided by x

In the fourth sub-question of the fourth question, the limit is 1, and the equivalent infinitesimal can be used, and e x-1 x becomes x x

Or the Law of Lopida.

Solution: The landlord wants to be a year old, and he must not have learned Robida!

This is solved using three methods:

1° (Limit of Importance).

Original limit. lim(x→a)

ln(x/a)

x-a)lim(x→a)

ln(x/a)^[1/(x-a)]

lim(x→a)

ln(a+x-a)/a]^[1/(x-a)]lim(x→a)

ln1+(x-a)/a]^[a/(x-a)]·x-a)/a][1/(x-a)]

lne^(1/a)

1 a2° (definition of the number of finch sizes).

Check the number of laughs on the ruler: y=lnx, obviously:

lnx)'|x=a

lim(x→a)

lnx-lna)

x-a)(1/x)|x=a

1 a3° (Laguera median theorem).

Examine the function: y=lnx, obviously continuous, derivable.

Then in the interval (x,a) or (a,x).

lnξ)'lnx-lna)

x-a) thus:

lim(x→a)

lnx-lna)

x-a)lim(ξ→a)

lnξ)'1/a

Add some points, three ways!

The first two questions are reached by Luo Bichun, and the last defeat pants are used to point to the important limit, please ask if you don't understand, and you are satisfied.

1. lim(cosx(sinx x)), because when x 0 has lim(sinx x)=1, the original limit is 1,2, lim (1+x3) 2x3 (x numerator and denominator are divided by x to the 3rd power, which is equal to 1 2

3. lim (ex+1) x ( x 2) both make sense at this point, so the limit is (e2+1) 2