On Thu, Apr 14, 2011 at 3:53 AM, Tom Bachmann <ness...@googlemail.com>wrote:
> On 13 Apr., 21:04, Hector <hector1...@gmail.com> wrote: > > On Wed, Apr 13, 2011 at 11:37 PM, Tom Bachmann <ness...@googlemail.com > >wrote: > > > > > How do you plan on implementing limits of bivariate functions? > > > Computing them is a *very* nontrivial extension over univariate limits > > > (as far as I can tell) ... > > > > To find limit at (x,y) = (0,0) replace "y" by "mx" and check whether the > > given limit is independent of "m" or not. If it is independent, than > limit > > exists and otherwise not. > > No, this is *not* the definition. > > limit f(x,y) as (x,y)->(0,0) = a iff for all e > 0 there exists d > 0 > s.t. x^2+y^2 < d ==> |f(x,y) - a| < e. > > As a counterexample, let f(x,y) = 0 if y <= 0 or x < 0, or x > 0 and y > > exp(-1/x), and f(x,y) = 1 otherwise [so that f(x,y) is zero unless > (x,y) is in the first quadrant und the graph of exp(-1/x)]. Then for > all m, limit f(x,mx) = 0. But limit f(x,y) does not exist. > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to sympy@googlegroups.com. > To unsubscribe from this group, send email to > sympy+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > > Whatever I said was the necessary but is not sufficient condition. I need to work on this before commenting anything more. -- -Regards Hector Whenever you think you can or you can't, in either way you are right. -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sympy@googlegroups.com. To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
