University of Toronto ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2010/2011Term Test Solutions1.(a) From the lecture notes we haveLetf:U⊂Rn→Rkbe a given function. We say thatfisdifferentiable ata∈Uif the partial derivatives offexist ataand iflimx→abardblf(x)−f(a)−Df(a) (x−a)bardblbardblx−abardbl= 0,whereDf(a) is thek×nmatrixparenleftbigg∂fi∂xjparenrightbiggevaluated ata.Df(a) is called thederivative offata.(b)(i) LetA⊂Rn. A pointa∈Ais called aninterior point ofAifBr(a)⊂A, forsomer >0.(Br(a) is an open ball of radiusrcentered ata.)(ii) Supposef:U⊂Rn→R. A pointa∈Uis called alocal (relative) minimumoffif there is an open ballBr(a) such thatf(x)≥f(a) for allx∈Br(a).2.(a)lim(x,y)→(0,0)x y−y2√x+√y=lim(x,y)→(0,0)y(x−y)√x+√y=lim(x,y)→(0,0)y(√x−√y)(√x+√y)√x+√y=lim(x,y)→(0,0)y(√x−√y)= 0.(b) If (x, y)negationslash= (0,0),f(x, y) =x3+ 2x2+ 2xy2+ 4y2x2+ 2y2. Since rational functions arecontinuous on their domains,f(x, y) is continuous for (x, y)negationslash= (0,0).Forfto be continuous at (0,0), we needlim(x,y)→(0,0)f(x, y) =−2 =f(0,0). Nowlim(x,y)→(0,0)f(x, y) =lim(x,y)→(0,0)x3+ 2x2+ 2xy2+ 4y2x2+ 2y2divide=lim(x,y)→(0,0)(x+ 2) =2negationslash=−2 =f(0,0). Hence we conclude thatf(x, y) is not continuous at (0,0).