Find the following limits LIM (x, y) → (0,1) (2-xy) / (x ^ 2 + 2Y)

Find the following limits LIM (x, y) → (0,1) (2-xy) / (x ^ 2 + 2Y)

F (x, y) = (2-xy) / (X & # 178; + 2Y), which is an elementary function. The elementary function is continuous in the domain of definition, and (0,1) is obviously a point in the domain of definition, so it is continuous. Therefore, the function value can be calculated directly
lim(x,y)→(0,1) (2-xy)/(x^2+2y)=f(0,1)=2/2=1

The limit of LIM x tending to infinity sin π X and the limit of LIM x tending to positive infinity sin π x

Let X1 = 2n, X2 = 2n + 1 / 2, when n tends to infinity, X1 and X2 tend to infinity
But the limit of sin π X1 is 0 and sin π x2 = 1
The limit of sin π X does not exist when x tends to infinity
Note: to prove that the limit of a function does not exist, we only need to show that the limits of its two subsequences are not equal

Limx tends to 0, ln (1-2x) / SiNx, find the extreme value

When x tends to 0, ln (1-2x) and SiNx tend to 0, which is a 0 / 0 type limit
From the law of lobita, we can get
lim ln(1-2x)/sinx=lim -2/(1-2x)cosx
When x tends to zero, lim - 2 / (1 - 2x) cosx = - 2
So limx tends to zero, ln (1-2x) / SiNx = - 2
This question mainly investigates the application of the law of lobita