The horizontal and vertical asymptotes of the curve y = e ^ (x ^ (- 2)) arctan ((x + 1) / (x-1)) are?

The horizontal and vertical asymptotes of the curve y = e ^ (x ^ (- 2)) arctan ((x + 1) / (x-1)) are?

The answer of finaifi is wrong, it should be vertical asymptote: x = 0, horizontal asymptote: y = π / 4, but x = 1 is the first kind of jumping discontinuity of curve, not asymptote

The integral domain of mathematical integral calculation is - pi / 2 to pi / 2, and the integrand is arctan (e ^ x) + arctan (e ^ (- x)),

The original function of arctan (e ^ x) + arctan (e ^ (- x)) is (e ^ x-1) / (e ^ x + 1), which is an odd function. So the definite integral of symmetric interval is 0

We all know that when we reach the speed of light, our time will slow down. When we exceed the speed of light, it is equivalent to going back to the future through time and space. Can we go back to the past
It's based on Einstein's special theory of relativity But history can't change, so does it create a space of different dimensions

It's undeniable that it's a great thing to have a time machine. Who doesn't want to go back to the past, change history, and become a superstar? Scientists have also listed some ways to realize time travel. But wait a minute, they seem to forget to send you a code of time travel. Otherwise, you may have a life to cross, but no life to come back

A problem about the balance of two forces in Physics
When an object with a weight of 10N is placed on a horizontal desktop, the force of the object pressing the desktop is 10N, and the force of the desktop supporting the object is 10N. Are these two forces balanced? Why? Why is the object in equilibrium?
By the way, is pressure and supporting force balanced or is gravity and supporting force balanced?

"The force of the object pressing on the table top" is the table top, while "the force of the table top supporting the object" is the object. These two forces are two forces acting on different objects, which are mutual forces rather than equilibrium forces. The so-called equilibrium forces must be two forces acting on the same object. In this problem, the gravity of the object and the supporting force of the table top on the object constitute a pair of equilibrium forces

If the side length of a square is increased by 3 cm, the area will be increased by 45 square cm. The original area of a square is 3 cm______ ．

(45-3 × 3) / (3 + 3), = 36 △ 6, = 6 (CM); the area of the original square: 6 × 6 = 36 (cm 2); answer: the area of the original square is 36 cm 2. So the answer is: 36 cm 2

Please use the 24 point algorithm to calculate 1 12 11

12*(12-11+1)=24

The range X of function y = 1 / 2x-1 is exponential

The domain of function is (- ∞, 0) ∪ (0, + ∞). When x ∈ (- ∞, 0) ∪ (0, + ∞), 2 ^ X-1 ∈ (- 1,0) ∪ (0, + ∞). So the domain of original function is (- ∞, - 1) ∪ (0, + ∞)

In a rectangular grassland, there is a winding road with a width of 1 meter. As shown in the picture, how many square meters is the area of the lawn?
Length 18, width 12. (CM)

18*12-(18+12)*1=186
If your unit is right, that's the answer
Otherwise, how can there be a one meter road within a 12 * 18 range

Choose any number in 1.2.3.4.5.6.7.8.9, multiply this number by 7, and then multiply by 15873. What rule do you find?

The six digits are the same
Because 7 × 15873 = 111111

In the sequence an, A1 = 2, an = 2A (n-1) + 3 (n belongs to n *, and is greater than or equal to 2), find an

a(n+1)=2a(n) + 3^(n+1),
a(n+1)/3^(n+1) = (2/3)[a(n)/3^n] + 1,
a(n+1)/3^(n+1) - 3 = (2/3)[a(n)/3^n] - 2 = (2/3)[a(n)/3^n - 3]
{a (n) / 3 ^ n - 3} is an equal ratio sequence whose first term is a (1) / 3 - 3 = 2 / 3 - 3 = - 7 / 3 and the common ratio is 2 / 3
a(n)/3^n - 3 = (-7/3)(2/3)^(n-1),
a(n) - 3^(n+1) = -7*2^(n-1),
a(n) = 3^(n+1) - 7*2^(n-1)