Answer:
Step-by-step explanation:
Given that,
f(x, y)=7x−yx+y
We want to show that the limit doesn't exist as (x, y)→(0,0).
Limits typically fail to exist for one of four reasons:
1. The one-sided limits are not equal
2. The function doesn't approach a finite value
3. The function doesn't approach a particular value
4. The x - value is approaching the endpoint of a closed interval
a. Considering the case that y=3x
lim(x,y)→(0,0) 7x−yx+y
Since y=3x
lim(x,3x)→(0,0) 7x−3x(x)+3x
lim(x,3x)→(0,0) 7x−3x(x)+3x
lim(x,3x)→(0,0) 10x−3x²
Therefore,
lim(x,3x)→(0,0) 10x−3x² = 0-0=0
b. Let also consider at y=4x
lim(x,y)→(0,0) 7x−yx+y
Since y=4x
lim(x,4x)→(0,0) 7x−4x(x)+4x
lim(x,4x)→(0,0) 7x−4x(x)+4x
lim(x,4x)→(0,0) 11x−4x²
Therefore,
lim(x,4x)→(0,0) 11x−4x² = 0-0=0
c. Let also consider it generally at y=mx
lim(x,y)→(0,0) 7x−yx+y
Since y=mx
lim(x,mx)→(0,0) 7x−mx(x)+mx
lim(x,mx)→(0,0) 7x−mx(x)+mx
lim(x, mx)→(0,0) (7+m)x−mx²
Therefore,
lim(x, mx)→(0,0) (7+m)x−mx² = 0-0=0
But the limit of the given function exist.
So let me assume the function is wrong and the question meant.
f(x, y)= (7x−y) / (x+y)
So, let analyze again
a. Considering the case that y=3x
lim(x,y)→(0,0) (7x−y)/(x+y)
Since y=3x
lim(x,3x)→(0,0) (7x−3x)/(x+3x)
lim(x,3x)→(0,0) 4x/4x
lim(x,3x)→(0,0) 1
Therefore,
lim(x,3x)→(0,0) 1= 1
So the limit is 1
b. Let also consider at y=4x
lim(x,y)→(0,0) (7x−y)/(x+y)
Since y=4x
lim(x,4x)→(0,0) (7x−4x)/(x+4x)
lim(x,4x)→(0,0) 3x/5x
lim(x,4x)→(0,0) 3/5
Therefore,
lim(x,4x)→(0,0) 3/5 = 3/5
So the limit is 3/5
This show that the limit does not exit.
Since one of the condition given above is met, then the limit does not exist. i.e. The function doesn't approach a particular value
c. Let also consider it generally at y=mx
lim(x,y)→(0,0) (7x−y)/(x+y)
Since y=mx
lim(x,mx)→(0,0) (7x−mx)/(x+mx)
lim(x,mx)→(0,0) (7-m)x/(1+m)x
lim(x, mx)→(0,0) (7-m)/(1+m)
Therefore,
lim(x, mx)→(0,0) (7-m)/(1+m) = (7m)/(1+m)
Then, the limit is (7-m)/(1+m)
So the limit doesn't not have a specific value, it depends on the value of m, so the limit doesn't exist.
