limit of (x,y)--->(1,0) of ln(1+y^2/x^2+xy))
Find the limit, if it exists, or show that the limit does
not exist.
To find the limit of a function as (x, y) approaches a point (a, b), we need to check if the function approaches the same value regardless of how we approach the point (a, b) in the domain of the function.
In this case, we want to find the limit of the function ln(1 + y^2/x^2 + xy) as (x, y) approaches (1, 0).
To start, let's substitute (x, y) = (1, 0) into the function and see what happens:
ln(1 + 0^2/1^2 + 1*0) = ln(1 + 0 + 0) = ln(1) = 0
So, at point (1, 0), the function evaluates to 0.
Now, let's consider different paths towards (1, 0) and check if the function approaches the same value.
1. Approach along the x-axis (y = 0):
ln(1 + 0^2/x^2 + 0*x) = ln(1 + 0 + 0) = ln(1) = 0
As x approaches 1, the function evaluates to 0.
2. Approach along a curve y = mx, where m is a constant:
In this case, the function becomes:
ln(1 + (mx)^2/x^2 + mx) = ln(1 + m^2 + mx)
As x approaches 1, this becomes ln(1 + m^2 + m) where m can take different values.
Since ln(x) is a continuous function, we know that ln(1 + m^2 + m) is also continuous.
Therefore, the function approaches different values depending on the chosen m.
This shows that the limit of the function does not exist as (x, y) approaches (1, 0).
In conclusion, the limit of ln(1 + y^2/x^2 + xy) does not exist as (x, y) approaches (1, 0).