Question

(1 point) In this problem we show that the function f(x,y)=7x−yx+y f(x,y)=7x−yx+y does not have a limit as (x,y)→(0,0)(x,y)→(0,0). (a) Suppose that we consider (x,y)→(0,0)(x,y)→(0,0) along the curve y=3xy=3x. Find the limit in this case: lim(x,3x)→(0,0)7x−yx+y=lim(x,3x)→(0,0)7x−yx+y= (b) Now consider (x,y)→(0,0)(x,y)→(0,0) along the curve y=4xy=4x. Find the limit in this case: lim(x,4x)→(0,0)7x−yx+y=lim(x,4x)→(0,0)7x−yx+y= (c) Note that the results from (a) and (b) indicate that ff has no limit as (x,y)→(0,0)(x,y)→(0,0) (be sure you can explain why!). To show this more generally, consider (x,y)→(0,0)(x,y)→(0,0) along the curve y=mxy=mx, for arbitrary mm. Find the limit in this case: lim(x,mx)→(0,0)7x−yx+y=lim(x,mx)→(0,0)7x−yx+y= (Be sure that you can explain how this result also indicates that ff has no limit as (x,y)→(0,0)(x,y)→(0,0).

Answer 1

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.