Thursday, February 6 ∗ Solutions ∗ Functions of several variables; Limits.1. For each of the following functions f : R2→R, draw a sketch of the graph together with picturesof some level sets.(a) f (x, y) =x y(b) f (x) = |x|. Please note here that x is a vector. In coordinates, this function is f (x, y) =px2+y2.For (a), the result is one of the many quadric surfaces. What is the name for this type? Is thegraph in (b) also a quadric surface?Solution.(a) The graph of the function f (x, y) = x y is-202-202-505Figure 1: Graph of f (x, y) = x y.The graph of the level sets f (x, y) =−2,−1,0, 1,2 is210-1-2-2-10120-3-2-10123-3-2-10123Figure 2: Graph of Level Sets of f (x, y) = x y.The graph of f (x, y) = x y is a hyperbolic paraboloid since the horizontal traces are hyper-bolas and the vertical traces are parabolas.(b) The graph of the function f (x) =|x| is-202-2020123Figure 3: Graph of f (x) =|x|.The graph of the level sets f (x, y) =0,1,2, 3 is0123-3-2-10123-3-2-10123Figure 4: Graph of Level Sets of f (x) =|x|.The graph of f (x) =|x| is not a quadric surface because it cannot be written as Ax2+B y2+C z2+Dx y +E yz +F xz +Gx +H y +I z +J =0. It is the top half of a cone, which is a quadricsurface.2. Consider the function f : R2→R given byf (x, y) =2x3yx6+y2for (x, y) 6=0In this problem, you’ll consider lim(x,y)→0f (x, y).(a) Look at the values of f on the x- and y-axes. What do these values show the limit lim(x,y)→0f (x, y)must be if it exists?Solution. Along y =0, lim(x,y)→0f (x, y) = limx→0f (x,0) = limx→00x6=0.Along x =0, lim(x,y)→0f (x, y) = limy→0f (0, y) = limy→00y2=0.Thus, should it exist, we must have lim(x,y)→0f (x, y) =0.(b) Show that along each line in R2through the origin, the limit of f exists and is 0.Solution. Any line through the origin besides x =0 or y =0 can be written as y =mx, m 6=0.Along y =mx, lim(x,y)→0f (x, y) = limx→0f (x,mx) = limx→02mx4x6+m2x2= limx→02mx2x4+m2=0.(c) Despite this, show that the limit lim(x,y)→0f (x, y) does not exist by finding a curve overwhich f takes on the constant value 1.Solution. Along y =x3, lim(x,y)→0f (x, y) = limx→0f (x, x3) = limx→02x6x6+x6=1.3. Consider the function f : R2→R given byf (x, y) =x y2px2+y2for (x, y) 6=0In this problem, you’ll show limh→0f (h) =0.(a) For ² = 1/2, find some δ >0 so that when 0 < |h| <δ we have |f (h)| < ². Hint: As with theexample in class, the key is to relate |x| and |y| with |h|.Solution. Note that | x|,|y|≤|h|. For ² =1/2, let δ =1/p2. Then 0 <|h|<δ implies|f (h)|≤|h|3|h|=|h|2<δ2=12.(b) Repeat with ² =1/10.Solution. For ² =1/10, let δ =1/p10. Then 0 <|h|<δ implies|f (h)|≤|h|2<δ2=110.(c) Now show that limh→0f (h) = 0. That is, given an arbitrary ² > 0, find a δ > 0 so that thatwhen 0 <|h|<δ we have |f (h)|<².Solution. Given ² >0, let δ =p². Then 0 <|h|<δ implies|f (h)|≤|h|2<δ2=².(d) Explain why the limit laws that you learned in class on Wednesday aren’t enough to com-pute this particular limit.Solution. f (x, y) cannot be written as f (x, y) = g (x, y)h(x, y) so that lim|x|→0g (x) and lim|x|→0h(x)both exist and are easier to compute than lim|x|→0f
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