Multivariable calculus - Limits

[cargar]

Show that the function f(x,y)=y/(x-y) for x→0, y→0, can take any limit.
Construct the sequences { f(x_n, y_n } with (x_n,y_n)→(0, 0) in such way that the lim _n→∞ f(x_n,y_n) is 3,2,1,0,−2.
Hint: y_n=kx_n.

I am not sure if I am right, but I did the following:

f(x,y) = kx_n/(x_n−kx_n) = kx_n/(x_n(1−k)) = k/(1−k)

k/(1−k) = A

After a few steps I have obtained:

k = A(1+A), A≠−1

So, the function can take any limit except -1.

Now we have: A{3,2,1,0,−2}

If A=3:

k=3/4, y_n=3/4x_n

So lim _n→∞ (3/4x_n)/(1/4x_n)=3.

For A=2, 1, 0 and −2 I did the same.

I am not sure if this is the end of the exercise. I don't understand what exactly in this context is meant by the sequence.

I am also not sure if I should write lim _{(xn, yn)→(0,0)} (3/4x_n)/(1/4x_n)=3 instead of lim _n→∞ (3/4x_n)/(1/4x_n)=3 or just x_n→0, because I have just x_n.

Any help is appreciated.

Thanks in advance.

 

[HallsofIvy]

Well, from k/(1- k)= A, you have k= A- Ak so k+ Ak= k(1+ A)= A and k= A/(1+ A) NOT "k= A(1+A)" but you later use A/(1+ A) so that must be a typo. The point of the exercise is that approaching the point (0, 0) along different lines, y= kx, gives different limits so taking the limits of sequences of point along those different lines gives those different limits. Yes, If you want the limit to be A=3 then k=3/4 so approaching (0, 0) along the line y= (3/4)x (better not to write "y= 3/4x"- many people would interpret that as "y= 3/(4x)") the limit of the function is 3. To get a sequence, take x to be a sequence approaching 0, say {1/n}. Then y= 3/(4n) and f(x, y)= y/(x- y)= [3/(4n)]/[1/n- 3/(4n)]= [3/(4n)]/[1/(4n)]= 3 for all n and so has limit 3.