Why is it that when x tends to infinity, the limit of fraction is 1 and the limit of denominator is 0, then the molecular limit is 0

Why is it that when x tends to infinity, the limit of fraction is 1 and the limit of denominator is 0, then the molecular limit is 0

When the limit of fraction and denominator exists, Lim molecule = Lim fraction × Lim denominator = 0

Find the limit of a fraction when n tends to infinity. The denominator is 1 / N power of n! And the denominator is n

This problem is difficult and can be divided into three steps: 1. Let xn = [n! ^ (1 / N)] / N, then ㏑ xn = ㏑ {[n! ^ (1 / N)] / N} = (1 / N) ㏑ [n! / N ^ n] = (1 / N) [㏑ 1 / N + ㏑ 2 / N + +㏑ n / N] = (1 / N) ∑ (k = 1, n) ㏑ K / N (it can be understood as integral sum) 2. Transform to definite integral: = ∫ (0,1) lnxdx = [xlnx -

Is the derivative of binary function the existence of partial derivative?

It means that there are partial derivatives in X and Y directions

If the intersection of the line y = 2x-3 with the X and Y axes is a and B, then the area of the triangle AOB is?

When x = 0, y = - 3
When y = 0, x = 3 / 2
So a (3 / 2,0), B (0, - 3)
S△ABC=1/2*3/2*3=9/4

Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) (n belongs to n positive) are all on the image of function y = 3x-2
Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) (n belongs to n positive) are all on the image of function y = 3x-2
1. Find the general term formula of sequence {an}
2. Let BN = 3 / ana (n + 1), tn be the sum of the first n terms of the sequence {BN},
Let tn be

I don't know how to get the last "1 / a (n-1) - an" in his step (1 / 2) * (1-1 / 7 + 1 / 7-1 / 13 +. + 1 / a (n-1) - an). I get this solution: BN = 3 / ana (n + 1) = 3 / (6n-5) (6 (n + 1) - 5) = 3 / (6n-5) (6N + 1) = (1 / 2) * (1 / (6n-5) - 1 / (6N + 1)) TN = (1 / 2) * (1-1 / 7

The volume of a cuboid is 60cm, and the length, width and height are three consecutive natural numbers. What is the surface area of the cuboid in square centimeter?
If the height of a 5cm cylinder is reduced by 2cm, the surface area will be reduced by 37.68cm?
Put two identical cubes together into a cuboid. The total length of the cuboid's edges is 64cm. What's the cuboid's volume in cubic centimeter

The length, width and height are 3, 4 and 5 respectively
Surface area = (3 × 4 + 4 × 5 + 3 × 5) × 2 = 94 square cm
2, perimeter = 37.68 △ 2 = 18.84
Radius = 18.84 △ 3.14 △ 2 = 3cm
The original volume is 3.14 × 3 × 3 × 5 = 141.3 cubic centimeter
3. Make a cuboid and reduce the length of 8 edges
So the edge length of the cube is 64 (24-8) = 4cm
Then the length of the cuboid = width = 4cm
Height = 4 × 2 = 8 cm
The cuboid volume = 4 × 4 × 8 = 128 cubic centimeters

The equation LXL = ax + 1 has a positive root and a negative root

When x > 0
x=ax+1
x=1/(1-a)>0
Then 1-A > 0
a

How much smaller is the sum of - 4, - 5 and + 7 than the sum of the absolute values of these three numbers?

According to the meaning of the question: | - 4 | + | - 5 | + | + 7 | - (- 4-5 + 7) = 4 + 5 + 7 + 4 + 5-7 = 18, then the sum of - 4, - 5, + 7 is 18 less than the sum of the absolute values of the three numbers

The function FX = the maximum of 2sinx (SiNx + cosx), as long as the maximum

fx=2sinx(sinx+cosx)=2sinxsinx+2sinxcosx)
=1-cos2x+sin2x
=sin2x-cos2x+1
=√2sin(2x-π/4)+1
So the maximum is √ 2 + 1

Given the line L1: x-2y-1 = 0, line L2: ax by + 1 = 0, a, B ∈ {123456}. Find the probability of line L1 ∩ L2 = 0

1/12
L1 ∩ L2 = 0, that is, L1 ∩ L2, that is, a: B = 1:2
There are three kinds: a = 1, B = 2; a = 2, B = 4; a = 3, B = 6
There are: 6 × 6 = 36 kinds (a has 6 choices, B has 6 choices, multiplication principle)
The probability is 1 / 12