The left vertex of the hyperbola is a, the right focus F, P is a point on the hyperbola, and PFA is an isosceles right triangle Please e. to process, thank you!

Let the hyperbolic equation be x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, P (x, y) on the hyperbola. 1) if PA, PF, because a (- A, 0), f (C, 0), then (x + a) (x-C) + y ^ 2 = 0, and (x + a) ^ 2 + y ^ 2 = (x-C) ^ 2 + y ^ 2, the solution is x = (C-A) / 2, y = ± (c + a) / 2, (or draw a picture directly to see that there is no solution

What is the limit of (1-cos 3x) / 2x when x approaches zero

0
Equivalent Infinitesimal Substitution, the above is k * x ^ 2, which is the higher order infinitesimal of X, so the last is 9
You can also use the law of Robida once and get the limit of 0

The limit of X -- > 0, x + x ^ 2 * cos (1 / x)?
The square of X is multiplied by cos (1 / x),

0, because cos (1 / x) is bounded

Find the limit: (1 / x) * cos (1 / X & # 178;) when x tends to zero
Such as the title

This function has no limit
f（x）=f(1/y)
Then f (y) = y cos (y ^ 2) the limit of Y tending to infinity. Obviously, when y tends to positive infinity, f (y) tends to positive infinity, and when y tends to negative infinity, f (y) tends to negative infinity, so this function has no limit

The limit of COS ^ 2 (1 / x) when x tends to zero
X = 0 is the discontinuity of y = cos ^ 2 (1 / x)
So we can find the limit
Request the limit and explain the breakpoint type. If the breakpoint is removable, change the function definition to make it continuous

[1 / 2] + [1 / 2] * cos (2 / x), X & raquo; 0, cos (2 / x) oscillates between - 1 and 1, does not tend to a certain number, has no limit, is a bounded variable, and belongs to the second type of discontinuity (not the first type of discontinuity)
The oscillation discontinuity function oscillates back and forth at this point between some two values, such as - 1 and + 1

It is known that the function f (x) is a decreasing function over the domain of definition (0, positive infinity) and satisfies f (XY) = f (x) + F (y), f (1 / 3) = 1, such as f (2-x)

Because the function f (x) is a decreasing function over the domain of (0, positive infinity)
f(2-x)

F (x) definition field x > 0 is a decreasing function, f (radical 3 / 3) = 1, f (XY) = f (x) + F (y), find the value range of X of F (x) - f (1 / (X-2) > = - 2
F (x) domain x > 0 is a decreasing function, f (radical 3 / 3) = 1, f (XY) = f (x) + F (y). Find the value range of X satisfying f (x) - f (1 / (X-2) > = - 2?

From the definition field of F (x), where x > 0,1 / (X-2) > 0, we can get x > 2F (1) = f (1) + F (1), f (1) = 0f (1) = f (√ 3 / 3) + F (√ 3), f (√ 3) = - 1F (3) = f (√ 3) + F (√ 3) = - 2. The original equation can be transformed into f (1 / (X-2) * x (X-2)) - f (1 / (X-2) > = - 2, that is, f (x (X-2)) > = - 2

It is known that y = f (x) is a decreasing function of the domain over (0, + infinity), and if f (XY) = f (x) + F (y), f (1 / 3) = 1,
If f (2-x)

Let x = 1 be substituted by F (XY) = f (x) + F (y), then f (1) = 0
f(2-x)

FX is an increasing function defined on (0, + ∞), f (XY) = f (x) + F (y). If f (x) + F (X-8) ≤ 2, find the value range of X

Let x = y = 1, f (1) = 2F (1), f (1) = 0,
Let f (a) = 1, then f (a ^) = 2,
From F (x) + F (X-8) ≤ 2
{x>0,
{x-8>0,
{x(x-8)

It is known that y = f (x) is an increasing function defined on r positive integers, and f (XY) = f (x) + F (y). This paper proves that f (x / y) = f (x) - f (y)

Let X / y replace y, that is, f (x) = f [y * (x / y)] = f (x / y) + F (y), so f (x / y) = f (x) - f (y)

The left vertex of the hyperbola is a, the right focus F, P is a point on the hyperbola, and PFA is an isosceles right triangle Please e. to process, thank you!