Mathematical calculus is solved one problem, and one calculus problem is solved.

arcsiny)'=(1-y 2) (1 2) This is easy, and then the integration gives the formula arcsiny.

arctany)'=1 1+x 2 is easier, refer to x n=1 1-x, and then integrate to get arctany's formula.

Then the numerator and denominator can be done, and it is not too cumbersome. The software is as follows.

series expansion sintanx-tansinx=-x^7/30-29x^9/756+o(x^9)

series expansion arcsinarctanx-arctanarcsinx=-x^7/30+13x^9/756+o(x^9)

Limit 1 hope helps

The limit is 0 because in both cases x->0+ and x->0-, the numerator of the equation approaches 0 and the denominator is a non-zero number. All you need to do is prove that the limit exists at that point, and then use the entrapment criterion to prove that the limit is 0Why use the pinch criterion?

Since sin (infinity) cannot determine its value, it needs to be scaled a bit. At this time, you have to use the coercing criterion, otherwise it will not be easy to do.

e x(1+bx+cx 2)-1=ax+ (x 2), e x(1+bx+cx 2)=1+ax+ (x 2), x->0.

e x(1+bx+cx 2)-1] x=a+ (x),x->0 for the above equation to find the limit.

limx->0[e^x(1+bx+cx^2)+e^x(b+2cx)]=a

limx->0[e^x(b+1)+e^x(bx+cx^2+2cx)]=a

So b+1=a, and c is any number.

Taylor forgot it.

e^x = 1 + x + x^2 / 2 + x^3 / 3! +

1 + x + x^2 / 2 + o(x^2))(1+bx+cx^2)=1+ax+o(x^2)

1+bx+cx^2) +x+bx^2+o(x^2)) x^2 / 2 + o(x^2)) o(x^2))=1+ax+o(x^2)

1 + 1+b)x+(c+b+1/2)x^2 + o(x^2) = 1 + ax + o(x^2)

1 + 1+b)x+(c+b+1/2)x^2 + o(x^2) = 1 + ax + o(x^2)

I forgot that the o(x2) package doesn't include x2.

Well, according to the upstairs, not included.

a=1+b.

Answer: In this integral problem, as long as the radical is written as (x+1) +1, it becomes the general standard substitution type, so that tanu = 1+x, and then use the rational fractional integration method, it can be accumulated.

For details, see the figure below. Click to enlarge, enlarge the screen and zoom in again:

See the ** description below to understand why d.

The lower limit function of this problem is 0, and it is still 0 after the derivative;

The upper bound function is derivative of 2t, so only d is correct.

The question is not missing a dx

It is known according to the meaning of the definite integral.

f(t) is 1 (t 2+1) for t 2 and 2t (t 4+1) for t

Because here is the derivative of x, and x=t squared.

Hello, this is the integral upper limit function. There are formulas to find it. Trust your teacher to give the formula. The derivative of f[ x)] x)=f(t) is the answer dρ x)=t^2.

It's easier to calculate with a Y shape.

If there is anything you don't understand, you can ask at any time, I will try my best to answer, I wish you academic progress, thank you.

If the issue is resolved, click "Select as satisfactory answer" below

III. 1The words of the index are too small to read.

With the formula: z x = z u* u x + z v* v x, and so on. May I?