m116m1solution

# m116m1solution - Solution to MATH 116 INTERMEDIATE CALCULUS...

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Unformatted text preview: Solution to MATH 116 INTERMEDIATE CALCULUS III MIDTERM I EXAM Date: June 20, 2005, Time: 9:00-11:00 Problem 1. (a) Use the ε- δ definition of limit to verify that lim ( x,y ) → (1 ,- 1) (2 x 2- 4 y 2 ) =- 2 (b) Find lim ( x,y ) → (0 , 0) sin x sin 3 y 1- cos( x 2 + y 2 ) or explain why it doesn’t exist. Solution . (a) For a given ε > 0 take δ = min { 1 , ε 18 } . Then for any ( x,y ) satisfying p ( x- 1) 2 + ( y + 1) 2 < δ , we have 1) 0 < x < 2,- 2 < y < 0 (since δ ≤ 1) 2) | 2 x 2- 4 y 2- (- 2) | = | (2 x 2- 2)- (4 y 2- 4) | ≤ 2 | x- 1 | ( x +1)+4 | y +1 || y- 1 | ≤ 6 | x- 1 | + 12 | y +1 | ≤ 18 p ( x- 1) 2 + ( y + 1) 2 < 18 δ ≤ ε . It implies that lim ( x,y ) → (1 ,- 1) (2 x 2- 4 y 2 ) =- 2. (b) Since lim ( x,y ) → (0 , 0) ,alongy = mx sin x sin 3 y 1- cos( x 2 + y 2 ) = lim x → sin x sin 3 mx 1- cos( x 2 (1 + m 2 )) = lim x → x 4 m 3 x 4 (1+ m 2 ) 2 2 = 2 m 3 (1 + m 2 ) 2 , then by the Two-Path Test the limit does not exist....
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## This note was uploaded on 10/29/2010 for the course MATH 116 taught by Professor Ahmetguloglu during the Spring '10 term at Bilkent University.

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m116m1solution - Solution to MATH 116 INTERMEDIATE CALCULUS...

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