[RESOLVED] Limit

12012016, 10:43 AM
(This post was last modified: 12082016 02:53 PM by jrozsas.)
Post: #1




[RESOLVED] Limit
I can not solve this equation:
Appears this error: Anybody can help me? Leo 

12012016, 01:51 PM
Post: #2




RE: Limit
limit is a 1 dimensional command. For dimension > 1, you must specify yourself the direction, in your example set X=2+u*h and Y=4+v*h where the direction is (u,v) and take the limit as h > 0 (or 0+ for halfdirection limit).


12012016, 02:47 PM
Post: #3




RE: Limit
Also, the variables you've shown are in upper case, instead of the usual lower case for [CAS] variables.


12022016, 08:50 AM
Post: #4




RE: Limit
(12012016 02:47 PM)DrD Wrote: Also, the variables you've shown are in upper case, instead of the usual lower case for [CAS] variables. Lower case: Result in error : Leo 

12022016, 10:49 AM
Post: #5




RE: Limit
I don't have my Prime with me at present but it looks as though (2,4) is being interpreted as a complex number. Is this what you intend?
Also, you could do the limits separately, one after the other: y>4 on the original expression followed by x>2 on the result. In general, limits involving two or more variables depend on the precise route followed in taking the limit. I think that the Prime's failure to return a result in this situation is a good thing, although the error message should explain the reason for the failure more clearly. Nigel (UK) 

12022016, 11:57 AM
Post: #6




RE: Limit  
12042016, 03:16 PM
Post: #7




RE: Limit
(12022016 10:49 AM)Nigel (UK) Wrote: I don't have my Prime with me at present but it looks as though (2,4) is being interpreted as a complex number. Is this what you intend?The first problem in this attachment is what you would like to solve at HP Prime. I thought that when typing the equation as it is in the attachment, HP Prime would solve or give some other information. Leo 

12052016, 07:21 AM
(This post was last modified: 12052016 07:23 AM by Tim Wessman.)
Post: #8




RE: Limit
Have you tried nesting two limit calls?
lim(lim(...,x,2),y,4) type of thing? limit(limit((y+4)*x/(x^2*yx*y+4*x^24*x),x,2),y,4) TW Although I work for the HP calculator group, the views and opinions I post here are my own. 

12052016, 08:50 AM
Post: #9




RE: Limit
(12052016 07:21 AM)Tim Wessman Wrote: Have you tried nesting two limit calls? Tim: I do not know how to speak English, but in Portuguese we say: "Estou soltando fogos". Expression used when celebrating something really incredible. Thank you very much for your attention! Leo 

12052016, 10:39 AM
(This post was last modified: 12052016 10:48 AM by DrD.)
Post: #10




RE: Limit
Nice solution, Tim! I noticed that this process worked regardless of the order of the variables:
limit(limit((y+4)*x/(x^2*yx*y+4*x^24*x),y,4),x,2); ==> 1 limit(limit((y+4)*x/(x^2*yx*y+4*x^24*x),x,2),y,4); ==> 1 (I tried unsuccessfully, using various approaches, lists, etc.). Wolfram has a special widget that handles multidimensional limits. Something similar to the Wolfram multidimensional limit widget would be a nice enhancement, and with your approach, it seems like it wouldn't be too difficult to implement. http://www.wolframalpha.com/widgets/view...f5ac88eefd Dale 

12052016, 12:17 PM
(This post was last modified: 12052016 12:25 PM by DrD.)
Post: #11




RE: Limit
Thinking about this further, is it fair to say that as long as "regardless of the order" of the variables always converges to the same limit value, the limit exists, but if reordering the variables does NOT converge to the same limit, that a limit does not exist?
This is based on the idea that different paths to the approached value must converge to the same limiting value if a limit exists for multidimensional limits. Not knowing how the command works, this idea may not be correct. 

12052016, 09:38 PM
(This post was last modified: 12052016 09:40 PM by Han.)
Post: #12




RE: Limit
Be careful about assuming that a complex limit is equivalent to a double limit over real variables. The following limit does not exist.
\[ \lim_{(x,y)\to (0,0)} \left( \frac{x^2y^2}{x^2+y^2} \right)^2 \] Reason: \[ \lim_{(x,0)\to (0,0)} \left( \frac{x^2y^2}{x^2+y^2} \right)^2 = \lim_{(x,0)\to (0,0)} \left( \frac{x^20}{x^2+0} \right)^2 = 1\] \[ \lim_{(x,x)\to (0,0)} \left( \frac{x^2y^2}{x^2+y^2} \right)^2 = \lim_{(x,x)\to (0,0)} \left( \frac{x^2x^2}{x^2+x^2} \right)^2 = 0\] (12052016 12:17 PM)DrD Wrote: Thinking about this further, is it fair to say that as long as "regardless of the order" of the variables always converges to the same limit value, the limit exists, but if reordering the variables does NOT converge to the same limit, that a limit does not exist? The example above shows that "swapping order" is not sufficient as the squaring of the fraction produces a limit of 1 even after interchanging x and y. Graph 3D  QPI  SolveSys 

12052016, 09:59 PM
Post: #13




RE: Limit
Thanks, Han. Things are seldom as simple as I might wish. I appreciate your time and information!
Dale 

« Next Oldest  Next Newest »

User(s) browsing this thread: 1 Guest(s)