This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 2513, Fall 2007 Simon Fraser University Assignment 5: Solutions Additional Question Due: 4:30pm, Wednesday 24 October 2007 1. (a) Find the limit, if it exists, or show carefully that the limit does not exist: lim ( x,y ) → (0 , 0) 3 xy (2 cos( x )) x 2 + 2 y 2 Solution: We can show that the limit does not exist by taking the limit lim ( x,y ) → (0 , 0) f ( x, y ) = lim ( x,y ) → (0 , 0) 3 xy (2 cos( x )) x 2 + 2 y 2 along two different paths, and showing that they are different. For instance, along the xaxis y = 0 , we have f ( x, 0) = 3 x · 0(2 cos( x )) x 2 + 0 = 0 = ⇒ lim ( x,y ) → (0 , 0) f ( x, y ) = 0 along y = 0 (similarly, the limit along the yaxis x = 0 is 0). However, along the line y = x , f ( x, x ) = 3 x · x (2 cos( x )) x 2 + 2 x 2 = 2 cos( x ) , so the limit along this line is lim ( x,y ) → (0 , 0) f ( x, y ) = lim x → (2 cos( x )) = 2 1 = 1 6 = 0 along y = x. Since the limits along different paths are different, the limit lim ( x,y ) → (0 , 0) f ( x, y ) does not exist. [In general, along the straight line y = mx we have f ( x, mx ) = 3 x · mx (2 cos( x )) x 2 + 2( mx ) 2 = 3 m 1 + 2 m 2 (2 cos( x )) = ⇒ lim ( x,y ) → (0 , 0) f ( x, y ) = 3 m 1 + 2 m 2 along y = mx....
View
Full
Document
This note was uploaded on 10/09/2009 for the course MATH macm 101 taught by Professor Jcliu during the Spring '09 term at Simon Fraser.
 Spring '09
 jcliu

Click to edit the document details