Does the limit exist at theorigin?f (x, y) = e−xycos(x + y)SOLUTION:f is continuous at the origin.The limit is f (0, 0) = 1.Does the limit exist at theorigin?f (x, y) = e−xycos(x + y)SOLUTION:f is continuous at the origin.The limit is f (0, 0) = 1.Does the limit exist at (1, 0)?f (x, y) = ln1 + y2x2+ xySOLUTION:f is continuous at (1, 0). Thelimit is f (1, 0) = ln(1) = 0.Does the limit exist at (1, 0)?f (x, y) = ln1 + y2x2+ xySOLUTION:f is continuous at (1, 0). Thelimit is f (1, 0) = ln(1) = 0.Does the limit exist at (0, 0)?f (x, y) =x2+ sin2(y)2x2+ y2SOLUTION:Limit along x = 0:lim(x ,y )→(0,0)sin(y)y2= 1Limit along y = 0:lim(x ,y )→(0,0)x22x2=12The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =x2+ sin2(y)2x2+ y2SOLUTION:Limit along x = 0:lim(x ,y )→(0,0)sin(y)y2=1Limit along y = 0:lim(x ,y )→(0,0)x22x2=12The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =x2+ sin2(y)2x2+ y2SOLUTION:Limit along x = 0:lim(x ,y )→(0,0)sin(y)y2= 1Limit along y = 0:lim(x ,y )→(0,0)x22x2=12The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =x2+ sin2(y)2x2+ y2SOLUTION:Limit along x = 0:lim(x ,y )→(0,0)sin(y)y2= 1Limit along y = 0:lim(x ,y )→(0,0)x22x2=12The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =x2+ sin2(y)2x2+ y2SOLUTION:Limit along x = 0:lim(x ,y )→(0,0)sin(y)y2= 1Limit along y = 0:lim(x ,y )→(0,0)x22x2=12The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =6x3y2x4+ y4SOLUTION:Limit along y = x:lim(x ,y )→(0,0)6x42x4+ x4= 2Limit along y = −x:lim(x ,y )→(0,0)−6x42x4+ x4= − 2The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =6x3y2x4+ y4SOLUTION:Limit along y = x:lim(x ,y )→(0,0)6x42x4+ x4=2Limit along y = −x:lim(x ,y )→(0,0)−6x42x4+ x4= − 2The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =6x3y2x4+ y4SOLUTION:Limit along y = x:lim(x ,y )→(0,0)6x42x4+ x4= 2Limit along y = −x:lim(x ,y )→(0,0)−6x42x4+ x4= − 2The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =6x3y2x4+ y4SOLUTION:Limit along y = x:lim(x ,y )→(0,0)6x42x4+ x4= 2Limit along y = −x:lim(x ,y )→(0,0)−6x42x4+ x4=− 2The limit DNE at the origin.Does the limit exist at (0, 0)?f (x, y) =6x3y2x4+ y4SOLUTION:Limit along y = x:lim(x ,y )→(0,0)6x42x4+ x4= 2Limit along y = −x:lim(x ,y )→(0,0)−6x42x4+ x4= − 2The limit DNE at the origin.Does the limit exist at theorigin?f (x, y) =x4− y4x2+ y2SOLUTION: The numeratorcan be factored(x2+ y2)(x2− y2)x2+ y2= x2− y2so the limit is zero.Does the limit exist at theorigin?f (x, y) =x4− y4x2+ y2SOLUTION: The numeratorcan be factored(x2+ y2)(x2− y2)x2+ y2= x2− y2so the limit is zero.Does the limit exist at theorigin?f (x, y) =x4− y4x2+ y2SOLUTION: The numeratorcan be factored(x2+ y2)(x2− y2)x2+ y2= x2− y2so the limit is zero.Does the limit exist at theorigin?f (x, y) =x4− y4x2+ y2SOLUTION: The numeratorcan be factored(x2+ y2)(x2− y2)x2+ y2= x2− y2so the limit is zero.Does the limit exist at theorigin?f (x, y) =x2sin2(y)x2+ 2y2SOLUTION: Squeeze it!0 ≤x2sin2(y)x2+ 2y2≤x2x2+ 2y2≤ x2By the Squeeze Theorem, thelimit is zero.Does the limit exist at theorigin?f (x, y) =x2sin2(y)x2+ 2y2SOLUTION:Squeeze it!0 ≤x2sin2(y)x2+ 2y2≤x2x2+ 2y2≤ x2By the Squeeze Theorem, thelimit is zero.Does the limit exist at theorigin?f (x, y) =x2sin2(y)x2+ 2y2SOLUTION: Squeeze it!0 ≤x2sin2(y)x2+ 2y2≤x2x2+ 2y2≤ x2By the Squeeze Theorem, thelimit is