I have 6 questions and for each question, there's an example with a solution:

First question:

Given the function f(x, y) = x2 + y2 10y - 2xy + 10x - 50 , the domain of f(x, y) is {(x, y) ( R2, x / A, y / B} Find A = and B=ac 2 + y2 Given the function f(x, y) = 30y - 10xy + 100x - 300 , the domain of f(x, y) is {(x, y) ER2, x / A, y # B) . Find A = 3 and B = 10 Feedback 2 2 + y2 For the domain of the function f(x, y) = 30y - 10xy + 100x - 300 , We need the denominator to be nonzero. 30y - 10xy + 100x - 300 0 = (3 -x) (10y - 100) / 0 = x / 3, y / 10Evaluate the following statement: the function 3 3 2 2 + y2 does not have a finite limit at the point (0, 0). O True O FalseEvaluate the following statement: the function 2 T 2xy + 22 y2 2-2 2/2 x 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+6123+87y3 (a,y) -(0,0) 15x2+15y2+43x-3 +39g3Let sin 126x2 + 126y2+ 12622) + 44x3 + 40y + 8923 f (x, y, 2) : 18x2 + 18y2 + 1822 Find lim f ( x, y, z ) = 7 (x,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(0))5 + 89(p cos($))3 lim P-+0 18p2 2 (126p2) = lim + lim 44 P-+0 18p2 P-+0 18 p (sin($) cos(8)) + 9p (sin($) sin(0)) + 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 sin (126p2 126 . 2p . cos ( 126p2 lim = lim =7. p-+0 18p2 P-0 18 . 2pEvaluate the following statement: the function 2 y 2xy + /2 2 + y2 Va2 + y2 2 2 + y2 does not have a finite limit at the point (0, 0) . True FalseLet cos ( 220x2 + 220g2) - e200x2+200y + 11x53 f(x, y, 2) := + z cos(x). 20x2 + 20y2 Find lim f(x, y, z ) = -3 (x,y,z)-(0,0,7) Feedback We switch to the cylindrical coordinates {r, 0, z}. Then the limit we need to compute becomes cos (22072 ) - 200r2 + 11(r cos(0))5(rsin(0)) 3 lim (r, z) -(0, 7) 2072 + z cos(r cos(0)) = cos ( 22072 ) - 200r2 = lim + lim 11(r cos(0))(r sin(0)) 3 + lim (r, z) - (0, 7) 20r2 (z cos(r cos(0))) . (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.Let cos ( 144x2 + 14412 ) - 342x2+342y + 63267 f (x, y, 2 ) := + z cos(c). 18x2 + 18y2 Find lim f(x, y, z ) = (x,y,z) ->(0,0,18)Evaluate the following statement: the function 3 y |=| + lyl does not have a nite limit at the point (0, 0}. True 9 False Feedback Let's try to use the two paths test. First we choose the straight line .1 = y. On this straight line the numerator vanishes while the denominator becomes 2 |=| . Theretore the limit at the origin along this straight line is 0. On the other hand we can choose the straight line 2 = y. Then the numerator becomes 2 2. while the denominator is 2 lzl. 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}. \fCompute the following limit 17x2+17y2+93xy2+45x2y+70x3+14y3 lim (z,y) -(0,0) 17x2+17y2+60x3+33y3 Feedback Switching to polar coordinates we obtain the following limit to evaluate: 17 + r (93 cos(0) sin (0) + 45 cos2(0) sin(0) + 70 cos(0) + 14 sin (0) lim 1. 17 + r (60 cos3 (0) + 33 sin3 (0) )