{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math solution 3 - Homework due 7th February Solutions to...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework due 7th February Solutions to compulsory questions Problem ( § 2.2 Exercise 8b). Compute the limit lim ( x,y ) (0 , 0) sin( xy ) y if it exists. Solution. Using the substitution u = xy we have lim ( x,y ) (0 , 0) sin( xy ) ( xy ) = lim u 0 sin u u = 1 . Therefore, by the properties of limits, lim ( x,y ) (0 , 0) sin( xy ) y = lim ( x,y ) (0 , 0) x × sin( xy ) ( xy ) = lim ( x,y ) (0 , 0) x lim ( x,y ) (0 , 0) sin( xy ) ( xy ) = 0 . Alternatively, by expanding sin( xy ) into a Taylor series we have lim ( x,y ) (0 , 0) sin( xy ) y = lim ( x,y ) (0 , 0) ( xy ) - 1 3! ( xy ) 3 + · · · y = lim ( x,y ) (0 , 0) x 1 - ( xy ) 2 3! + · · · = 0 . Problem ( § 2.2 Exercise 10c). Compute the limit lim ( x,y ) (0 , 0) ( x - y ) 2 x 2 + y 2 if it exists. Solution. Along the line y = x we have f ( x, x ) = ( x - x ) 2 x 2 + x 2 = 0 x 0 -→ 0 , and along the line y = - x we have f ( x, - x ) = ( x + x ) 2 x 2 + x 2 = 2 x 0 -→ 2 . Therefore the limit does not exist. Problem ( § 2.2 Exercise 15). If f : R n R and g : R n R are continuous, show that the function f 2 + g : R n -→ R x -→ [ f ( x )] 2 + g ( x ) is continuous. Solution. If a R n then lim x a ( f 2 + g )( x ) = lim x a ( [ f ( x )] 2 + g ( x ) ) (definition of f 2 + g ) = lim x a f ( x ) 2 + lim x a g ( x ) (using limit properties) = [ f ( a )] 2 + g ( a ) (since f, g continuous) = ( f 2 + g )( a ) (definition of f 2 + g ) . Therefore f 2 + g is continuous at each a R n , and so f 2 + g is continuous on R n . 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Problem ( § 2.3 Exercise 2ac). Evaluate the partial derivatives ∂z/∂x and ∂z/∂y for the functions: (a) z = a 2 - x 2 - y 2 at (0 , 0) and at ( a/ 2 , a/ 2). (c) z = e ax cos( bx + y ) at (2 π/b, 0). Solution. (a) We compute ∂z ∂x = - x a 2 - x 2 - y 2 and ∂z ∂y = - y a 2 - x 2 - y 2 Therefore ∂z ∂x (0 , 0) = ∂z ∂y (0 , 0) = 0 and ∂z ∂x ( a/ 2 , a/ 2) = ∂z ∂y ( a/ 2 , a/ 2) = - 1 2 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}