Wikipedia:Reference desk/Archives/Mathematics/2016 January 15

= January 15 =

Alzebra
Solve it                                                                              (a+b)^4  — Preceding unsigned comment added by 49.126.0.41 (talk) 00:39, 15 January 2016 (UTC)
 * There's nothing to solve, but you can expand it to a^4 + 4(a^3)b + 6(a^2)(b^2) + 4a(b^3) + b^4 using the Binomial theorem. -- ℕ   ℱ  02:25, 15 January 2016 (UTC)
 * From the title, I assumed that this was a strip(p)ed-down problem →86.139.120.76 (talk) 12:36, 18 January 2016 (UTC)

Multivariate Limit
Let $$f:\R^n\to\R, x\in\R^n, I\subset\R^n$$. Assume that f is continuous at $$I-\{x\}$$ (the "minus" sign denotes set difference), and that $$\forall v\in\R^n $$ the limit $$lim_{a\to 0} f(x+av)$$ exists (and finite). Then, the multivariate limit $$lim_{y\to x} f(y)$$ exists? 213.8.204.40 (talk) 08:48, 15 January 2016 (UTC)
 * I believe the classical counterexample is
 * $$f(x,y) = \frac{x y}{x^2+y^2}.$$
 * around the origin .Dmcq (talk) 10:41, 15 January 2016 (UTC)
 * Another counterexample is $$f(x) = \begin{cases}2y/x^2-y^2/x^4&x>0, 00, y=x^2\\0&\textrm{Otherwise}\end{cases}$$ and modified it to be continuous everywhere except the origin. I hope I didn't mess it up.) -- Meni Rosenfeld (talk) 11:24, 15 January 2016 (UTC)