Math 23B. DodsonWeek 6 Homework:14.2 limits14.3 partial derivatives, 2nd order derivWeek 6 Homework:14.2 limitsProblem 14.2.6:Find the limitlim(x,y)→(6,3)(xycos (x-2y)).Solution:We see that the functionf(x, y) =xycos (x-2y)is continuous. (why?)Sincexyandx-2yare polynomials;cosx-2yis a composite of continuous functions;and thenfis a product of continuous functions(that is, ofxyand cosx-2y). So thelimit isf(6,3) = 6·3 cos 6-6 = 18·1 = 18.Problem 14.2.13:Show that the limitlim(x,y)→(0,0)2x2yx4+y2does not exist.
2.Solution:Along thex-axis (x,0) the limit islim(x,0)→(0,0)2x2yx4+y2=limx→00x4= 0,recalling thatx= 0 whilex→0.But alongthe parabolay=x2, so (x, x2),lim(x,x2)→(0,0)2x2yx4+y2=limx→02x42x4= 1,againusing thatx= 0 whilex→0.Since we havetwo different limits among points with (x, y)→(0,0),the limit does not exist. Note that the limit alonglines (0, y) (they-axis) gives0y2→0and (x, mx) (withy=mx, m= 0) gives2mx3x4+m2x2→2mxx2+m2→0m2= 0,(sincem= 0).So the limits along all lines approach 0,and we needed the parabola to get a non-zero value andshow that the limit doesn’t exist.Problem 14.3.15:Find the partial derivatives ofthe functionz=f(x, y) =xe3y.Findfx(2,1).
3Solution:Forfxwe (temporarily) holdyconstant,soe3yconstant, givingfas constant·x.We then take the derivative inx, with(constant·x)’ = constant, orfx=e3y.Likewise, withxconstant,fy=x∂∂y(e3y)=xddy(e3y) = 3xe3y.Finally,fx(2,1) =e3gives the rate of changeoffat (2,1) with respect tox; which wemay constrast withfy(2,1) = 6e3,the rate ofchange offat (2,1) with respect toy.For example,fis growing six times more rapidlyinythan inx. We also solved #49, 14.3, andin particular verified that the second partialszxy=zyx.