A few questions about higher mathematics, a few questions about higher mathematics

I'd like to ask what the t in the first question is ......

The second problem first is the partial derivative of x and y, and then let it be equal to 0, solve a few points, and then find the second-order partial derivative of a=f to x, the partial derivative of b=f to x and then the partial derivative of y, and the second-order partial derivative of c=f to y. Look at the positive or negative of a to determine whether the maximum or the minimum value.

The third problem is solved using an infinite series. f(x)=(x^(2n))/(2n)!, find the second derivative, you can get the differential equation, solve it, you can get the formula f(x), so that x=1, you get.

In the fourth question, move (-1) into x, become -(-x) n, then find a derivative, and then put the denominator 2 together with x, and you will be familiar with this direction.

That's it. High math is a great need for exercise. Unfamiliar places can continue to communicate

s(x)= [(n 2) *x (n-1)] n from 1 to Itemized integral: n from 1 to : n) *x (n)] =x [n) *x (n-1)] = xf(x).

Itemized for [(n) *x (n-1)] yields: x (n) = x (1-x).

f(x)=[x/(1-x)]'=1/(1-x)^2s(x)=∑[(n^2) *x^(n-1)]=[x/(1-x)^2]'=(1+x)/(1-x)^3

x= 1 series divergence, convergence domain (-1,1).

The first question: 1 x in f(x) is an infinitely large quantity, but cos(1 x) is a function transformed in [-1,1], f(x)=0 when cos(1 x)=0, f(x)=1 x when cos(1 x)=1, and a larger quantity when x approaches 0, so f(x) is a function that constantly changes between positive and negative infinity, and constantly crosses the 0 point;

The second question: in the broad definition of limit, the limit can be infinite, but in the narrow definition, when the limit is infinite, the limit is said to not exist, and it is generally defined in the narrow sense in high school;

The third problem: infinitesimal quantities approach negative infinity;

The fourth problem: when x is not equal to 0, multiply (1+bx) +1, the numerator becomes bx, remove x, f(x)=b ((1+bx) +1), and substitute x=0, b=6;

The fifth problem: the definition of the limit determines that the limit at a certain point is determined by the function near that point, and has nothing to do with the value of the function at that point. When f(x) is a continuous function, the point limit is equal to the point limit, which is also the definition of a continuous function.

As the name suggests, a continuous function is a continuous function, which is connected together in the image

The first cos may be 0, so it's not infinity.

At the second x->1, the left and right limits are unequal.

The third should be the infinitesimal reciprocal that is constant and not 0 is infinite, and 0 has no reciprocal, but it is an infinitesimal quantity.

The product of the fourth 0 and infinity is obviously not infinity.

The fifth x tends to 0 and has nothing to do with x=0, so a can be arbitrary.

These two equations hold only in the case of limits. The principles are all infinitesimal approximations. Specifically, there are the following equivalents.

1+a）^n ~ 1+na

The condition for the above equation to be true is that if x tends to zero, lima=0, that is, a is an infinitesimal quantity that tends to zero, then lim(1+a) n=lim(1+na) sorry, in order to save time, I will not mark x->0 below my lim.

In the original problem, let a=x 2, n=1 3, bring in the above equation to get your first equation, when a=sinx, n=1 2, bring in to get your second equation.

Note that it is not the two equations that are equal, but the equivalence at the limit.

Look at example 1 in the section "Infinitesimal Comparisons", which is also a commonly used equivalent infinitesimal size.

At x >0, (1+u) 1 n is equivalent to 1+1 n u is an infinitesimal where u is a function of x and at x > 0, u > 0. In the process of finding the limit, these two equations can be substituted with each other. So, in your problem, ( 1+x 2 ) 1 3 is replaced by 1+1 3x 2, and (1+sin x) 1 2 is replaced by 1+1 2 sin x, thus giving the latter equation.

You misunderstood the equation in the question, not ( 1+x 2 -1) 3, but ( 1+x 2 -1) 1 3.

For details, please refer to the Tongji Sixth Edition of Advanced Mathematics, the equivalent infinitesimal in the infinitesimal, the part of the equation is specially derived.

The following is the equivalent infinitesimal substitution, see page 58 of the sixth edition of Tongji, and this topic is the promotion of page 58. Extract sinx above.

That's the equivalent infinitesimal At x >0, (1+u) 1 n and 1+1 n u are equivalent infinitesimal , and you get used to it when you use it a lot.

The answer is d

Analysis] Apparently i3=0

Both i2 and i4 are positive.

So, the i3 is the smallest.

In d, |x|+|y|≤1

x|≤1(|x|+|y|)²x|+|y|)^4x²≥x^4

i2＞i4

First solve the differential equation, then determine the arbitrary constant c, find f(x) and find the derivative to get f'(x)。