I have six-question for each question there's an exact example with a solution.

Question 1:

x2 +y2 Given the function f(x, y) = 10y - 2xy + 10x - 50 , the domain of f(x, y) is {(x, y) ER2, x / A, y / B} Find A = and B =:32 +y2 Given the function it: y) = m {(z,y)eR2,xaA,yaB}. FindA=|3 sin. andB='1U on. , the domain of its, 21;) is Feedback For the domain of the function n: ) $2 + .92 we need the denominator to be '3' _ 3019 1039+ 100a: 300 ' nonzero. 30y102y+1002:300#0 => (3m)(10y100)#0 => 2:73,y#10 Evaluate the following statement: the function 2 2 + y2 does not have a finite limit at the point (0, 0). True O FalseEvaluate the following statement: the function 2 2xy + 22 + y2 a2 + 7/2 2 2 + y 2 does not have a finite limit at the point (0, 0). True False Feedback After simplifying we find that this is just the constant function 1. Hence it does have a finite limit at the origin.Compute the following limit lim 105x2+105y2+36xy2+91x2y+61x3+87y3 (z, y) -(0,0) 15x2+15y2+43x3+39g3Compute the following limit lim 17x2+17y2+93ry2+45x2y+70x3 +14y3 1 (x,y) -(0,0) 17x2+17y2+6013+33y3 Feedback Switching to polar coordinates we obtain the following limit to evaluate: 17 + r (93 cos(0) sin2(0) + 45 cos2 (0) sin(0) + 70 cos3 (0) + 14 sin3(@) ) lim = 1. 7-+0 17 + r (60 cos3 (0) + 33 sin3 (0))Evaluate the following statement: the function 2 T y 2xy 2 2 + y2 V x2 + 12 does not have a finite limit at the point (0, 0). True FalseEvaluate the following statement: the function I gr III + lyl does not have a nite limit at the point (0, 0). True False 0 Feedback Let's try to use the two paths test. First we choose the straight line a: = y. On this straight line the numerator vanishes while the denominator becomes 2 III . Therefore the limit at the origin along this straight line is 0. 0n the other hand we can choose the straight line a = y. Then the numerator becomes 2 2, while the denominator is 2 |z| _ We find that along this straight line the limit is 1 from one side and 1 from the other. Altogether we conclude that the function does not have a limit at (0, 0}. \fLet cos ( 220x2 + 22032) - 200x2+200y + 11253 f(x, y, 2) := + z cos(I). 20x2 + 20y2 Find lim f(, y, z) = -3 (z,y,z) ->(0,0,7) Feedback We switch to the cylindrical coordinates {r, 0, >}. Then the limit we need to compute becomes cos (22072) - e200r + 11(r cos(0))5 (r sin(0)) 3 lim + z cos(r cos(0)) (r, z) -(0, 7) 2072 cos ( 22072 ) - 200r2 lim + lim 11(r cos(0))(r sin(0)) 3 + lim (z cos(r cos(0))) . (r, z) - (0, 7) 2072 (r, z) -(0, 7) 20r2 (r,z) - (0, 7) Using L'Hopital's rule we find that the first summand equals 10. The second summand has to be 0 because sine and cosine are bounded functions. For the same reason the third summand equals 7. Altogther the answer is 7 - 10 = -3.\fLet sin 126x2 + 126y2 + 12622) + 44x3 + 40y5 + 8923 f(x, y, z) := 18x2 + 18y2 + 1822 Find lim f(I, y, z ) = 7 (z,y,z) -+(0,0,0) Feedback We switch to the spherical coordinates {p, 0, }. Then the limit we need to compute becomes sin (126p2 ) + 44(psin($) cos(0))3 + 40(psin($) sin(@))5 + 89(p cos($))3 lim p-+0 18p2 sin (126p2) = lim + lim 44 P-+0 18p2 P-+0 18 p (sin($) cos(0))+ p (sin($) sin(0)) 5 + 89 18 p ( cos ($)) 3 ). Because sine and cosine functions are bounded the second summand has to be 0. Applying L'Hopital's rule to the first summand we find 2(126p2) 126 . 2p . cos ( 126p2) lim lim = 7. p-+0 18p2 p-0 18 . 2p