continuous Determine the set of points at which the function is continuous. G x, y sx y sx y 31. f x, y x

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1 SECTION 11. LIMITS AND CONTINUITY LIMITS AND CONTINUITY A Click hee fo answes. S Click hee fo solutions. Copyight 013, Cengage Leaning. All ights eseved. 1 Find the it, it exists, o show that the it does not exist. 1.. x y 3 x 3 y x sin x y 6. x, y l, 4 x y x y x y x y x x y x y 5 3y x, y l, 3 x 3 3x y 5y 3 1 x, y l 3, 4 x, y l, 0 sx y 1 1 x y x, y l 0, 1 x, y l 1, 1 xz y z x, y, z l 1,, 3 z 1 x, y, z l, 3, 0 xez ln x y x y z x, y, z l 0, 0, 0 x y z yz zx x, y, z l 0, 0, 0 x y z x, y, z l 0, 0, 0 x 3 y x y y x y 4x 4 x x y y 1 x y x y x y x y x y z x y z x y x, y l, 1 x y sx y e x, y l 1, 4 x y 8x y x 4 y 4 1 x y Find h x, y t f x, y and the set on which h is continuous. 3. t t e t cos t, f x, y x 4 x y y 4 4. t z sin z, 5 38 Detemine the set of points at which the function is continuous F x, y x 6 x 3 y 3 y 6 F x, y x y 1 x y 1 x 3 y 3 7. F x, y tan x 4 y 4 8. G x, y e sin x y 9. F x, y 1 x y 30. F x, y ln x 3y G x, y sx y sx y z f x, y, z x y z 33. G x, y x tan y 34. f x, y, z x ln yz 35. f x, y, z x ysx z x y f x, y x 0 y x y 3 f x, y x 0 y f x, y y ln x f x, y x 0 y 39. Pove, using Definition 1, that a x a b x, y l a, b c c c x, y l a, b 40. Use pola coodinates to find sin x y x y x, y l a, b y b [If, ae pola coodinates of the point x, y with o, note that l 0 + as.]

2 SECTION 11. LIMITS AND CONTINUITY 11. ANSWERS E Click hee fo execises. S Click hee fo solutions x, y y x} x, y x 4 y 4 n +1,nan intege} e 3 7. Does not exist 8. Does not exist 8. R } 9. x, y y x 9. Does not exist 10. Does not exist 30. x, y x +3y>0} Does not exist Does not exist 17. Does not exist Does not exist 1. Does not exist h x, y e x4 +x y +y 4 cos x 4 + x y + y 4, R 4. h x, y sinyln x, x, y x>0} 5. x, y x + y 1 0 } 31. x, y y x} 3. x, y, z z x + y } 33. x, y y n +1,nan intege} 34. x, y, z yz > 0} 35. x, y, z x + z 0} 36. x, y x, y 0, 0} 37. R 38. x, y x, y 0, 0} Copyight 013, Cengage Leaning. All ights eseved.

3 SECTION 11. LIMITS AND CONTINUITY SOLUTIONS E Click hee fo execises. Copyight 013, Cengage Leaning. All ights eseved. 1. The function is a polynomial, so the it equals The function is a polynomial, so the it equals Since this is a ational function defined at 0, 0, the it equals / This is a ational function defined at, 1, sotheit equals 4 +1/ The poduct of two functions continuous at,, sothe it equals sin The composition of two continuous functions, so the it equals e 1+8 e Let f x, y x y. Fist appoach 0, 0 along the x + y x-axis. Then f x, 0 x x 1 and f x, 0 doesn t x x 0 exist. Thus f x, y doesn t exist. x,y 0,0 8. Let f x, y. As x, y 0, 0 along the x +y x-axis, f x, y 0. Butasx, y 0, 0 along the line y x, f x, x x 3x,sof x, y as x, y 0, 0 3 along this line. So the it doesn t exist. x + y 9. f x, y.asx, y 0, 0 along the x-axis, x + y f x, y 1. Butasx, y 0, 0 along the line y x, f x, x 4x fo x 0,sof x, y. Thus, the x it does not exist. 10. f x, y 8x y. Appoaching 0, 0 along the x-axis x 4 + y4 gives f x, y 0. Appoaching 0, 0 along the line y x, f x, x 8x4 4fo x 0, so along this line x4 f x, y 4 as x, y 0, 0. Thus the it doesn t exist. x x,y 0,0 x + y x 0 x,y 0, Since is a ational function defined at 0, 0 the x + y it is f x, y x3 y. We use the Squeeze Theoem: x + y x 3 y 0 x + y x 3 since y x + y,and x 3 0 as x, y 0, 0. So f x, y 0. x,y 0,0 y y x 14. f x, y x + y 4x +4 y +x.then f x, 0 0 fo x,sof x, y 0 as x, y, 0 along the x-axis. But x x x f x, x x +x x 1 fo x,sof x, y 1 as x, y, 0 along the line y x x. Thus, the it doesn t exist. 15. We have x y x + y x y ationalize x + y x + y +1+1 x y x + y x [since y x + y ] But x,y 0,0 x 0, so, by the Squeeze Theoem, x y x,y 0,0 x + y x 16. Let f x, y.thenf0,y0fo x + y y +1 y 1,sof x, y 0 as x, y 0, 1 along the y-axis. x x +1 1 But f x, x +1 x +x fo x 0so f x, y 1 as x, y 0, 1 along the line y x +1. Thus, the it doesn t exist. 17. Let f x, y x + y x y x + y x +y + x 1 +y 1 x 1 +y +1 Then f 1,y y 1 y +1. Thus, asx, y 1, 1 along the line x 1, the it of f x, y doesn t exist and so the it doesn t exist. xz y z 18. x,y,z 1,,3 z since the 5 function is continuous at 1,, x,y,z,3,0 [xez +lnx y] e 0 +ln4 3 since the function is continuous at, 3, Let f x, y, z x y z. Then f x, 0, 0 1 fo x + y + z x 0and f 0,y,0 1 fo y 0,soas x, y, z 0, 0, 0 along the x-axis, f x, y, z 1 but as x, y, z 0, 0, 0 along the y-axis, f x, y, z 1. Thus the it doesn t exist.

4 4 SECTION 11. LIMITS AND CONTINUITY Copyight 013, Cengage Leaning. All ights eseved. + yz + zx 1. Let f x, y, z. Thenf x, 0, 0 0 fo x + y + z x 0,soasx, y, z 0, 0, 0 along the x-axis, f x, y, z 0. But f x, x, 0 x / x 1 fo x 0,soasx, y, z 0, 0, 0 along the line y x, z 0,fx, y, z 1. Thus the it doesn t exist.. We can show that the it along any line though the oigin is 0 and thus suspect that this it exists and equals 0. Let ε>0be given. We need to find δ>0such that x y z x + y + z 0 <ε wheneve 0 < x + y + z <δo equivalently x y z x + y + z <εwheneve 0 < x + y + z <δ. But x x + y + z and similaly fo y and z,so x y z x x + y + z + y + z 3 x + y + z fo x + y + z x + y + z 0. Thus choose δ ε 1/4 and let 0 < x + y + z <δ. Then x y z x + y + z 0 x + y + z Hence by definition x + y + z 4 <δ 4 x,y,z 0,0,0 O: Use the Squeeze Theoem: 0 since z x + y + z,andx y 0 as x, y, z 0, 0, 0. ε 1/4 4 ε x y z x + y + z h x, y g f x, y g x 4 + x y + y 4 x y z x + y + z x y e x4 +x y +y 4 cos x 4 + x y + y 4 Since f is a polynomial it is continuous thoughout R and g is the poduct of two functions, both of which ae continuous on R,his continuous on R. 4. h x, y g f x, y sin y ln x. Since f x, y y ln x it is continuous on its domain x, y x>0} and g is continuous thoughout R. Thush is continuous on its domain D x, y x>0}, the ight half-plane excluding the y-axis. 5. F x, y is a ational function and thus is continuous on its domain D x, y x + y 1 0 },thatis,f is continuous except on the cicle x + y F x, y is a ational function and thus is continuous on its domain D x, y x 3 + y 3 0 } x, y y x}, R except the line y x. 7. F x, y g f x, y whee f x, y x 4 y 4, a polynomial so continuous on R and g t tant, continuous on its domain t t n +1,nan intege}. Thus F is continuous on its domain D x, y x 4 y 4 n +1,nan intege}. 8. G x, y g x, y f x, y whee g x, y e and f x, y sinx + y both of which ae continuous on R. Thus G is continuous on R F x, y is a ational function x y and thus is continuous on its domain x, y x y 0 } x, y y x },sof is continuous on R except the paabola y x. 30. F x, y lnx +3y g f x, y whee f x, y x +3y, continuous on R and g t lnt, continuous on its domain t t>0}. ThusF is continuous on its domain D x, y x +3y>0}. 31. G x, y g 1 f 1 x, y g f x, y whee f 1 x, y x + y and f x, y x y, both of which ae polynomials so continuous on R,andg 1 t t, g s s, both of which ae continuous on thei espective domains t t 0} and s s 0}. Thusg 1 f 1 is continuous on its domain D 1 x, y x + y 0} x, y y x} and g f is continuous on its domain D x, y x y 0} x, y y x}. ThenG, being the dfeence of these two composite functions, is continuous on its domain D D 1 D x, y x y x} x, y y x} z 3. f x, y, z is a ational function x + y z and thus is continuous on its domain x, y, z x + y z 0 } x, y, z z x + y }, so f is continuous on R 3 except on the cicula paaboloid z x + y. 33. G x, y g f x, y whee f x, y x tan y which is continuous on its domain x, y y n +1,nan intege} and g t t which is continuous on R. ThusG x, y is continuous on its domain D x, y y n +1,nan intege}. 34. f x, y, z xg f y,z whee f y, z yz, continuous on R and g t lnt, continuous on its domain t t>0}. Since h x x is continuous on R,fx, y, z is continuous on its domain D x, y, z yz > 0}.

5 SECTION 11. LIMITS AND CONTINUITY f x, y, z h x+k y g f x, z whee h x x and k y y, both continuous on R and f x, z x + z, continuous on R,gt t cont inuous on its domain D t t 0}. Thus f is continuous on its domain D x, y, z x + z 0}. In Poblems 36 38, each f is a piecewise defined function whose fist piece is a ational function defined eveywhee except at the oigin. Thus each f is continuous on R except possibly at the oigin. So fo each we need only check f x, y. x,y 0,0 36. Letting z x, x y x,y 0,0 x + y z,y 0,0 z y which doesn t z + y exist by Example 1. Thus f is not continuous at 0, 0 and the lagest set on which f is continuous is x, y x, y 0, 0}. 37. Since x x + y,wehave x y 3 x + y y 3. We know that y 3 0 as x, y 0, 0. So, by the Squeeze Theoem, x,y 0,0 f x, y x,y 0,0 x y 3 x + y 0. Also f 0, 0 0, so f is continuous at 0, 0. Fo x, y 0, 0,fx, y is equal to a ational function and is theefoe continuous. Thus f is continuous thoughout R. 38. Let g x, y x + + y. Theng x, 0 0/x 0 fo x 0,sog x, y 0 as x, y 0, 0 along the x-axis. But g x, x x 3x 1 fo x 0,sog x, y as x, y 0, 0 along the line y x. Thus x,y 0,0 doesn t exist, so f is not x + + y continuous at 0, 0 and the lagest set on which f is continuous is x, y x, y 0, 0}. 39. a Let ε>0be given. We need to find δ>0such that x a <εwheneve 0 < x a +y b <δ. But x a x a x a +y b. Thus setting δ ε and letting 0 < x a +y b <δ,wehave x a x a +y b <δ ε. Hence, by Definition 1, x a. x,y a,b b The agument is the same as in a with the oles of x and y intechanged. c Let ε>0 be given and set δ ε. Then f x, y L c c 0 x a +y b <δ ε wheneve 0 < x a +y b <δ. Thus, by Definition 1, c c. x,y a,b sin x + y sin 40., which is an x,y 0,0 x + y 0 + indeteminate fom of type 0. Using l Hospital s Rule, we 0 get sin cos cos 1 sin θ O: Use the fact that 1. θ 0 θ Copyight 013, Cengage Leaning. All ights eseved.