# (MATH100)worksheet-ch3-sol - (MATH100)worksheet-ch3-sol.pdf...

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MATH 100 (L1)Introduction to Multivariable CalculusSpring 2006-07¥Class WorksheetChapter 3Partial Derivatives1.(Limits)Evaluate the limits (if they exist).(a)lim(x,y)(1,-3)e2x-y2.(b)lim(x,y)(1/2)xy2sinxy.(c)lim(x,y)(0,0)x3+y3x+y.(d)lim(x,y)(0,0)x4-16y4x2+ 4y2.(e)lim(x,y)(0,0)xyx2+y2.(f)lim(x,y)(0,0)x-yx2+y2.(g)lim(x,y)(0,0)cosxyx+y.(h)lim(x,y)(0,0)2x2yx4+y2.(i)lim(x,y)(0,0)xln(|x|+|y|).Solution(a) By direct observation,lim(x,y)(1,-3)e2x-y2=e2(1)-(-3)2=e2-9=e-7.The limit exists.(b)f(x, y) =xy2sinxyis continuous (everywhere), it follows thatlim(x,y)(1/2)f(x, y) =f(12, π),orlim(x,y)(1/2)xy2sinxy= (12)(π)2sin(π2) =π22.The limit exists.(c) Note thatf(x, y) = (x3+y3)/(x+y)is defined at all points of thexy-plane except the origin(0,0).We can still ask whetherlim(x,y)(0,0)f(x, y)exists or not.The reason behind iswhetherf(0,0)is defined or nothas nothing to do with the existence of the limit. By existence of limitswe take care of the function valuesf(x, y)when the points(x, y)areclose to(0,0)but notexactly equal to(0,0).By this problem, since(x, y)6= (0,0),we can do cancellation,lim(x,y)(0,0)x3+y3x+y=lim(x,y)(0,0)(x+y)(x2-xy+y2)x+y=lim(x,y)(0,0)(x2-xy+y2)=(0)2-(0)(0) + (0)2=0.The limit exists.(d) We can do cancellation,lim(x,y)(0,0)x4-16y4x2+ 4y2=lim(x,y)(0,0)(x2-4y2)(x2+ 4y2)x2+ 4y2=lim(x,y)(0,0)(x2-4y2)=(0)2-4(0)2=0.The limit exists.(MATH100)worksheet-ch3-sol.pdf downloaded by twuac from at 2014-02-26 20:42:36. Academic use within HKUST only.
(e) Letf(x, y) =xy/(x2+y2).If we let(x, y)approach(0,0)along thex-axis(y= 0),thenf(x, y) =f(x,0)0(becausef(x,0) = 0identically). Thus,lim(x,y)(0,0)f(x, y)must be zeroif it exists at all. Similarly, at all points of they-axis we havef(x, y) =f(0, y) = 0.However,at points of the liney=x,fhas a different constant value,f(x, x) = 1/2.Since the limit off(x, y)is1/2as(x, y) approaches (0,0) along this line, it follows thatf(x, y)cannot have aunique limit at the origin. That is,lim(x,y)(0,0)xyx2+y2does not exist.(f) The limit does not exist because we can find a path (yes, one path is good enough!) along whichthe limit indeed does not exist.Let(x, y)(0,0)along thex-axis(y= 0),lim(x,y)(0,0)x-yx2+y2=lim(x,0)(0,0)x-0x2+ 0= limx01xwhich does not exist because both one-sided (left and right) limits do not exist:limx0-1x=-∞andlimx0+1x=.(g) The limit does not exist because we can find a path along which the limit indeed does not exist.Let(x, y)(0,0)along thex-axis(y= 0),lim(x,y)(0,0)cosxyx+y= limx0cos 0x+ 0= limx01xwhich does not exist by the same reasoning in the previous part.(h) As in part (e),f(x, y)=2x2y/(x4+y2)vanishes identically on the coordinate axes, solim(x,y)(0,0)f(x, y)must be0if it exists at all.If we examinef(x, y)at points along theoblique straight linesy=kx,we obtainf(x, kx) =2kx3x4+k2x2=2kxx2+k20,asx0(k6= 0).

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