Find the limit. LIM (x → 0) [√ (XY + 4) - 2] / XY (Y → 0) lim (x→0)(y→0)[√ (xy+4) -2 ]/xy

Find the limit. LIM (x → 0) [√ (XY + 4) - 2] / XY (Y → 0) lim (x→0)(y→0)[√ (xy+4) -2 ]/xy

Molecules can be obtained after rationalization, using the square difference
As a result, I can't type

Find the limit Lim "XY / (x ^ 2 + y ^ 2)" ^ (x ^ 2) (x, y) tends to positive infinity

x^2+y^2>=2xy
——》0limx,y→+∞ 0^(x^2)=0,
limx,y→+∞ (1/2)^(x^2)=0,
According to the pinch theorem:
Original formula = 0

Find the limit: LIM (2 - (XY + 4) ^ 0.5) / (x ^ 2 + y ^ 2) ^ 0.5 (x, y) → (0,0)

Let xy = t
Original formula = [2 - (T + 4) ^ 0.5] / T * t / (x ^ 2 + y ^ 2) ^ 0.5, where t → 0
= -1/[2+(t+4)^0.5]* t/(x^2+y^2)^0.5
= -1/4*xy/(x^2+y^2)^0.5
= -1/4* 1/(1/X^2+1/y^2)^0.5
= 0

LIM (XY / (x ^ 2 + y ^ 2)) ^ x ^ 2 x, y approaches infinity to find the limit

The limit doesn't exist
When x = KY (k is greater than 0), the limit value and
When x = y ^ 2, the limit values are not equal, so the limit does not exist
For multivariate functions, to make the limit exist, the limit value must be the same from all directions

Seek advice on the problem of binary function f (x + y, X-Y) = XY + y ②

s=x+y,t=x-y
x=(s+t)/2,y=(s-t)/2
f(x+y,x-y)=xy+y^2=y(x+y)=[(s-t)/2]*s
f(s,t)=(s^2-st)/2
f(x,y)=(x^2-xy)/2

Binary function image f (x, y) = XY / (x ^ + y ^), x ^ + y ≠ 0, x ^ + y ^ = 0

Z partial derivative of x = y-50 / x ^ 2
Z partial derivative of y = x-20 / y ^ 2
To find the extreme value, make y-50 / x ^ 2 = 0 and x-20 / y ^ 2 = 0
Because x > 0, Y > 0
The solution is x = 5, y = 2
Because when x, Y - > infinity, Z approaches infinity
So it is obvious that (5,2) is the minimum point of Z = XY + 50 / x + 20 / y
Then the minimum value of Z = XY + 50 / x + 20 / y is 10 + 10 + 10 = 30