Thursday January 31 Solutions Functions of several variables Limits 1 For each of the following functions f R2 R draw a sketch of the graph together with pictures of 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 p x 2 y 2 For a the result is one of the many quadric surfaces What is the name for this type Is the graph in b also a quadric surface Solution a The graph of the function f x y x y is 2 0 2 5 0 2 0 5 2 Figure 1 Graph of f x y x y The graph of the level sets f x y 2 1 0 1 2 is 3 0 2 1 0 2 2 1 1 0 0 1 1 2 2 1 2 3 3 2 1 0 1 2 3 Figure 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 hyperbolas and the vertical traces are parabolas b The graph of the function f x x is 2 0 2 3 2 1 0 2 0 2 Figure 3 Graph of f x x The graph of the level sets f x y 0 1 2 3 is 3 2 1 0 0 1 1 2 2 3 3 3 2 0 1 1 2 3 Figure 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 Ax 2 B y 2 C z 2 D x y E y z F xz G x H y I z J 0 It is the top half of a cone which is a quadric surface 2 Consider the function f R2 R given by f x y 2x 3 y x6 y 2 In this problem you ll consider lim x y 0 f x y for x y 6 0 a Look at the values of f on the x and y axes What do these values show the limit lim x y 0 f x y must be if it exists 0 0 x 0 x 6 Solution Along y 0 lim f x y lim f x 0 lim x 0 x y 0 0 0 y 0 y 0 y 2 x y 0 Thus should it exist we must have lim f x y 0 Along x 0 lim f x y lim f 0 y lim x y 0 b Show that along each line in R2 through 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 2mx 4 2mx 2 Along y mx lim f x y lim f x mx lim 6 lim 0 x 0 x 0 x m 2 x 2 x 0 x 4 m 2 x y 0 c Despite this show that the limit lim x y 0 f x y does not exist by finding a curve over which f takes on the constant value 1 2x 6 1 x 0 x 6 x 6 Solution Along y x 3 lim f x y lim f x x 3 lim x 0 x y 0 3 Consider the function f R2 R given by f x y p x y2 x2 y 2 for x y 6 0 In this problem you ll show limh 0 f h 0 a For 1 2 find some 0 so that when 0 h we have f h Hint As with the example in class the key is to relate x and y with h p Solution Note that x y h For 1 2 let 1 2 Then 0 h implies h 3 1 f h h 2 2 h 2 b Repeat with 1 10 p Solution For 1 10 let 1 10 Then 0 h implies f h h 2 2 1 10 c Now show that limh 0 f h 0 That is given an arbitrary 0 find a 0 so that that when 0 h we have f h Solution Given 0 let p Then 0 h implies f h h 2 2 d Explain why the limit laws that you learned in class on Wednesday aren t enough to compute this particular limit Solution f x y cannot be written as f x y g x y h x y so that lim g x and lim h x x 0 both exist and are easier to compute than lim f x x 0 x 0
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