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 March 20th, 2019, 04:11 AM #1 Newbie   Joined: Mar 2019 From: - Posts: 3 Thanks: 0 Multivariable calculus - Limits Show that the function f(x,y)=y/(x-y) for x→0, y→0, can take any limit. Construct the sequences { f(xn, yn } with (xn,yn)→(0, 0) in such way that the lim n→∞ f(xn,yn) is 3,2,1,0,−2. Hint: yn=kxn. I am not sure whether I am right, but I did the following: f(x,y) = kxn/(xn−kxn) = kxn/(xn(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, yn=3/4xn So lim n→∞ (3/4xn)/(1/4xn)=3. For A = 2, 1, 0 and −2, I did the same. I am not sure whether 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 whether I should write lim (xn, yn)→(0,0) (3/4xn)/(1/4xn)=3 instead of lim n→∞ (3/4xn)/(1/4xn)=3 or just xn→0, because I have just xn. Any help is appreciated. Thanks in advance. Last edited by skipjack; April 8th, 2019 at 04:47 PM.
 March 21st, 2019, 12:20 PM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 820 Thanks: 113 Math Focus: Elementary Math If you change the order of limits you get $\displaystyle -1=\lim_{x\rightarrow 0} \frac{0}{-x}$ which is not true . Last edited by idontknow; March 21st, 2019 at 12:25 PM.
March 21st, 2019, 12:52 PM   #3
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Quote:
 Originally Posted by cargar So, the function can take any limit except -1.
You were asked to show that it can take any limit. How are you going to get -1? Hint: approach along the $y$ axis.
Quote:
 Originally Posted by cargar I am also not sure if I should write lim (xn, yn)→(0,0) (3/4xn)/(1/4xn)=3 instead of lim n→∞ (3/4xn)/(1/4xn)=3 or just xn→0, because I have just xn.
Having shown, for example, that with $y_n=\frac34 x_n$ you get $\displaystyle \lim_{n \to \infty} f(x_n,y_n) = 3$, you should interpret what $y_n=\frac34x_n$ means in terms of the approach to $(0,0)$. In this case you are approaching the origin along the straight line $y = \frac34x$. This is the answer that I consider most appropriate, although the talk of sequences may mean that your teacher has something else in mind.

March 21st, 2019, 12:54 PM   #4
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Quote:
 Originally Posted by idontknow If you change the order of limits you get $\displaystyle -1=\lim_{x\rightarrow 0} \frac{0}{-x}$ which is not true .
There is no "order of limits" here. The idea is simply to determine a path of approach to $(0,0)$ that delivers the appropriate limit.

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