Calculus Calculus Math Forum

 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
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,696
Thanks: 2681

Math Focus: Mainly analysis and algebra
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
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,696
Thanks: 2681

Math Focus: Mainly analysis and algebra
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. Tags calculus, limits, multivariable, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Leonardox Calculus 1 October 1st, 2018 08:49 PM akansel Calculus 6 October 16th, 2017 09:44 AM maluita659 Calculus 7 February 15th, 2014 02:15 PM summerleei Real Analysis 1 December 5th, 2013 06:49 AM Ecomat Math Books 1 July 2nd, 2012 04:07 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      