Limits and Continuity/Partial Derivatives Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011 Christopher Croke Calculus 115 Limits For (x0; y0) an interior or a boundary point of the domain of a function f (x; y). Definition: lim f (x; y) = L (x;y)!(x0;y0) if for every > 0 there is a δ > 0 such that: for all (x; y) in the domain of f if q 2 2 0 < (x − x0) + (y − y0) < δ then jf (x; y) − Lj < . Christopher Croke Calculus 115 Limits For (x0; y0) an interior or a boundary point of the domain of a function f (x; y). Definition: lim f (x; y) = L (x;y)!(x0;y0) if for every > 0 there is a δ > 0 such that: for all (x; y) in the domain of f if q 2 2 0 < (x − x0) + (y − y0) < δ then jf (x; y) − Lj < . Christopher Croke Calculus 115 For example: if lim f (x; y) = L and lim g(x; y) = K (x;y)!(x0;y0) (x;y)!(x0;y0) then lim f (x; y) + g(x; y) = L + K (x;y)!(x0;y0) x2 + 2y + 3 lim =? (x;y)!(1;0) x + y x2 − y 2 lim =? (x;y)!(0;0) x − y This definition is really the same as in one dimension and so satisfies the same rules with respect to +; −; ×; ÷. See for example page 775 for a list. Christopher Croke Calculus 115 then lim f (x; y) + g(x; y) = L + K (x;y)!(x0;y0) x2 + 2y + 3 lim =? (x;y)!(1;0) x + y x2 − y 2 lim =? (x;y)!(0;0) x − y This definition is really the same as in one dimension and so satisfies the same rules with respect to +; −; ×; ÷. See for example page 775 for a list. For example: if lim f (x; y) = L and lim g(x; y) = K (x;y)!(x0;y0) (x;y)!(x0;y0) Christopher Croke Calculus 115 x2 − y 2 lim =? (x;y)!(0;0) x − y This definition is really the same as in one dimension and so satisfies the same rules with respect to +; −; ×; ÷. See for example page 775 for a list. For example: if lim f (x; y) = L and lim g(x; y) = K (x;y)!(x0;y0) (x;y)!(x0;y0) then lim f (x; y) + g(x; y) = L + K (x;y)!(x0;y0) x2 + 2y + 3 lim =? (x;y)!(1;0) x + y Christopher Croke Calculus 115 This definition is really the same as in one dimension and so satisfies the same rules with respect to +; −; ×; ÷. See for example page 775 for a list. For example: if lim f (x; y) = L and lim g(x; y) = K (x;y)!(x0;y0) (x;y)!(x0;y0) then lim f (x; y) + g(x; y) = L + K (x;y)!(x0;y0) x2 + 2y + 3 lim =? (x;y)!(1;0) x + y x2 − y 2 lim =? (x;y)!(0;0) x − y Christopher Croke Calculus 115 f is defined at (x0; y0). lim(x;y)!(x0;y0) f (x; y) exists. They are equal. There is an added complication here. Consider y if (x; y) 6= (0; 0) f (x; y) = x2+y 2 0 if (x; y) = (0; 0) Continuity(The definition is really the same.) f is continuous at (x0; y0) if lim f (x; y) = f (x0; y0): (x;y)!(x0;y0) This means three things: Christopher Croke Calculus 115 lim(x;y)!(x0;y0) f (x; y) exists. They are equal. There is an added complication here. Consider y if (x; y) 6= (0; 0) f (x; y) = x2+y 2 0 if (x; y) = (0; 0) Continuity(The definition is really the same.) f is continuous at (x0; y0) if lim f (x; y) = f (x0; y0): (x;y)!(x0;y0) This means three things: f is defined at (x0; y0). Christopher Croke Calculus 115 They are equal. There is an added complication here. Consider y if (x; y) 6= (0; 0) f (x; y) = x2+y 2 0 if (x; y) = (0; 0) Continuity(The definition is really the same.) f is continuous at (x0; y0) if lim f (x; y) = f (x0; y0): (x;y)!(x0;y0) This means three things: f is defined at (x0; y0). lim(x;y)!(x0;y0) f (x; y) exists. Christopher Croke Calculus 115 There is an added complication here. Consider y if (x; y) 6= (0; 0) f (x; y) = x2+y 2 0 if (x; y) = (0; 0) Continuity(The definition is really the same.) f is continuous at (x0; y0) if lim f (x; y) = f (x0; y0): (x;y)!(x0;y0) This means three things: f is defined at (x0; y0). lim(x;y)!(x0;y0) f (x; y) exists. They are equal. Christopher Croke Calculus 115 Continuity(The definition is really the same.) f is continuous at (x0; y0) if lim f (x; y) = f (x0; y0): (x;y)!(x0;y0) This means three things: f is defined at (x0; y0). lim(x;y)!(x0;y0) f (x; y) exists. They are equal. There is an added complication here. Consider y if (x; y) 6= (0; 0) f (x; y) = x2+y 2 0 if (x; y) = (0; 0) Christopher Croke Calculus 115 It is not enough to check only along straight lines! See the example in the text. Continuity acts nicely under compositions: If f is continuous at (x0; y0) and g (a function of a single variable) is continuous at f (x0; y0) then g ◦ f is continuous at (x0; y0). Example: y 2 sin( ) x2 − 1 If f (x; y) has different limits along two different paths in the domain of f as (x; y) approaches (x0; y0) then lim(x;y)!(x0;y0) f (x; y) does not exist. Christopher Croke Calculus 115 Continuity acts nicely under compositions: If f is continuous at (x0; y0) and g (a function of a single variable) is continuous at f (x0; y0) then g ◦ f is continuous at (x0; y0). Example: y 2 sin( ) x2 − 1 If f (x; y) has different limits along two different paths in the domain of f as (x; y) approaches (x0; y0) then lim(x;y)!(x0;y0) f (x; y) does not exist. It is not enough to check only along straight lines! See the example in the text. Christopher Croke Calculus 115 Example: y 2 sin( ) x2 − 1 If f (x; y) has different limits along two different paths in the domain of f as (x; y) approaches (x0; y0) then lim(x;y)!(x0;y0) f (x; y) does not exist. It is not enough to check only along straight lines! See the example in the text. Continuity acts nicely under compositions: If f is continuous at (x0; y0) and g (a function of a single variable) is continuous at f (x0; y0) then g ◦ f is continuous at (x0; y0). Christopher Croke Calculus 115 If f (x; y) has different limits along two different paths in the domain of f as (x; y) approaches (x0; y0) then lim(x;y)!(x0;y0) f (x; y) does not exist. It is not enough to check only along straight lines! See the example in the text. Continuity acts nicely under compositions: If f is continuous at (x0; y0) and g (a function of a single variable) is continuous at f (x0; y0) then g ◦ f is continuous at (x0; y0). Example: y 2 sin( ) x2 − 1 Christopher Croke Calculus 115 Easy to calculate: just take the derivative of f w.r.t. x thinking of y as a constant. @f @y "partial derivative of f with respect to y" Partial Derivatives of f (x; y) @f @x "partial derivative of f with respect to x" Christopher Croke Calculus 115 Partial Derivatives of f (x; y) @f @x "partial derivative of f with respect to x" Easy to calculate: just take the derivative of f w.r.t. x thinking of y as a constant. @f @y "partial derivative of f with respect to y" Christopher Croke Calculus 115 (xy 3 + xy)3 sin(xy + y) x3y y + x2 @f @f Examples: Compute @x and @y for f (x; y) = 2x2 + 4xy Christopher Croke Calculus 115 sin(xy + y) x3y y + x2 @f @f Examples: Compute @x and @y for f (x; y) = 2x2 + 4xy (xy 3 + xy)3 Christopher Croke Calculus 115 x3y y + x2 @f @f Examples: Compute @x and @y for f (x; y) = 2x2 + 4xy (xy 3 + xy)3 sin(xy + y) Christopher Croke Calculus 115 @f @f Examples: Compute @x and @y for f (x; y) = 2x2 + 4xy (xy 3 + xy)3 sin(xy + y) x3y y + x2 Christopher Croke Calculus 115 @f (2; 0) @y This notion shows up in many fields and many notations have been developed for the same thing. @f @z f z @ f @x x x @x x Example: For f (x; y) = 5 + 2xy − 3x2y 2 find @f (1; 2) @x Christopher Croke Calculus 115 This notion shows up in many fields and many notations have been developed for the same thing. @f @z f z @ f @x x x @x x Example: For f (x; y) = 5 + 2xy − 3x2y 2 find @f (1; 2) @x @f (2; 0) @y Christopher Croke Calculus 115 Example: For f (x; y) = 5 + 2xy − 3x2y 2 find @f (1; 2) @x @f (2; 0) @y This notion shows up in many fields and many notations have been developed for the same thing.