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Full Version: [RESOLVED] Limit
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I can not solve this equation:

Appears this error:

Anybody can help me?
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 half-direction limit).
Also, the variables you've shown are in upper case, instead of the usual lower case for [CAS] variables.
(12-01-2016 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 :
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)

https://en.wikipedia.org/wiki/Limit_(mathematics)
(12-02-2016 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?

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)
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.
Have you tried nesting two limit calls?

lim(lim(...,x,2),y,4) type of thing?

limit(limit((y+4)*x/(x^2*y-x*y+4*x^2-4*x),x,2),y,-4)
(12-05-2016 07:21 AM)Tim Wessman Wrote: [ -> ]Have you tried nesting two limit calls?

lim(lim(...,x,2),y,4) type of thing?

limit(limit((y+4)*x/(x^2*y-x*y+4*x^2-4*x),x,2),y,-4)

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!
Nice solution, Tim! I noticed that this process worked regardless of the order of the variables:

limit(limit((y+4)*x/(x^2*y-x*y+4*x^2-4*x),y,-4),x,2); ==> 1

limit(limit((y+4)*x/(x^2*y-x*y+4*x^2-4*x),x,2),y,-4); ==> 1

(I tried unsuccessfully, using various approaches, lists, etc.). Wolfram has a special widget that handles multi-dimensional 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-

[attachment=4252]
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.

[attachment=4253]
Be careful about assuming that a complex limit is equivalent to a double limit over real variables. The following limit does not exist.

$\lim_{(x,y)\to (0,0)} \left( \frac{x^2-y^2}{x^2+y^2} \right)^2$

Reason:

$\lim_{(x,0)\to (0,0)} \left( \frac{x^2-y^2}{x^2+y^2} \right)^2 = \lim_{(x,0)\to (0,0)} \left( \frac{x^2-0}{x^2+0} \right)^2 = 1$

$\lim_{(x,x)\to (0,0)} \left( \frac{x^2-y^2}{x^2+y^2} \right)^2 = \lim_{(x,x)\to (0,0)} \left( \frac{x^2-x^2}{x^2+x^2} \right)^2 = 0$

(12-05-2016 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.
Thanks, Han. Things are seldom as simple as I might wish. I appreciate your time and information!

-Dale-
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