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Math 412 Week #4 Practice Quiz Solutions for Limits - Prof. Scott Annin, Quizzes of Mathematics

California State University (CSU) - FullertonMathematics
Prof. Scott Annin

The solutions to problem 1 and problem 2 from the week #4 practice quiz of math 412 course. The problems involve evaluating limits using algebraic manipulations and justifications, as well as using an (ฮต, ฮด) argument.

Typology: Quizzes

Pre 2010

Uploaded on 08/16/2009

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February 16-20, 2009 Week #4 Practice Quiz Name:
Math 412
SOLUTIONS
Problem 1. (6 points each) Evaluate each limit below. If the limit does not exist,
supply a justification.
(a):
lim
zโ†’2โˆ’i=zโˆ’2 + i
z3โˆ’3z2+z+ 5.
SOLUTION: We have
zโˆ’2 + i
z3โˆ’3z2+z+ 5 =zโˆ’2 + i
(z+ 1)(zโˆ’(2 โˆ’i))(zโˆ’(2 + i)) =1
(z+ 1)(zโˆ’(2 + i)),
provided z6= 2 โˆ’i. Therefore,
lim
zโ†’2โˆ’i=zโˆ’2 + i
z3โˆ’3z2+z+ 5 = lim
zโ†’2โˆ’i
1
(z+ 1)(zโˆ’(2 + i)) =1
(3 โˆ’i)(โˆ’2i)=โˆ’1+3i
20 .
In the last steps, we have applied the fact that rational functions are continuous wherever the
denominator is nonzero, and hence, we can evaluate the last limit by plugging in z= 2 โˆ’ito the
final fraction. ๎˜ƒ
(b):
lim
zโ†’3โˆ’2ix2yโˆ’2xi + 3y.
SOLUTION: In this case, xโ†’3 and yโ†’ โˆ’2, so
lim
zโ†’3โˆ’2ix2yโˆ’2xi + 3y= 32ยท(โˆ’2) โˆ’6iโˆ’6 = โˆ’24 โˆ’6i.
๎˜ƒ
Problem 2. (8 points) Use an (๎˜, ฮด)argument to show that
lim
zโ†’โˆ’1+5i(2zโˆ’3) = โˆ’5 + 10i.
SOLUTION: Let ๎˜ > 0 be given. Note that |f(z)โˆ’(โˆ’5 + 10i)|=|2zโˆ’3+5โˆ’10i|= 2|z+ 1 โˆ’5i|=
2|zโˆ’(โˆ’1+5i)|. To bound this by ๎˜, we will choose ฮด=๎˜
2.
Assume that 0 <|zโˆ’(โˆ’1+5i)|< ฮด =๎˜
2. Then
|(2zโˆ’3) โˆ’(โˆ’5 + 10i)|=|2z+ 2 โˆ’10i|= 2|zโˆ’(โˆ’1+5i)|<2ฮด=๎˜,
as required. ๎˜ƒ

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Download Math 412 Week #4 Practice Quiz Solutions for Limits - Prof. Scott Annin and more Quizzes Mathematics in PDF only on Docsity!

February 16-20, 2009 Week #4 Practice Quiz Name:

Math 412

SOLUTIONS

Problem 1. (6 points each) Evaluate each limit below. If the limit does not exist, supply a justification.

(a):

zโ†’^ lim 2 โˆ’i =^

z โˆ’ 2 + i z^3 โˆ’ 3 z^2 + z + 5

SOLUTION: We have

z โˆ’ 2 + i z^3 โˆ’ 3 z^2 + z + 5

z โˆ’ 2 + i (z + 1)(z โˆ’ (2 โˆ’ i))(z โˆ’ (2 + i))

(z + 1)(z โˆ’ (2 + i))

provided z 6 = 2 โˆ’ i. Therefore,

zโ†’lim 2 โˆ’i =^

z โˆ’ 2 + i z^3 โˆ’ 3 z^2 + z + 5 =^ zโ†’lim 2 โˆ’i

(z + 1)(z โˆ’ (2 + i)) =^

(3 โˆ’ i)(โˆ’ 2 i) =^

โˆ’1 + 3i

In the last steps, we have applied the fact that rational functions are continuous wherever the denominator is nonzero, and hence, we can evaluate the last limit by plugging in z = 2 โˆ’ i to the final fraction. 

(b):

zโ†’^ lim 3 โˆ’ 2 i x^2 y^ โˆ’^2 xi^ + 3y.

SOLUTION: In this case, x โ†’ 3 and y โ†’ โˆ’2, so

zโ†’lim 3 โˆ’ 2 i x^2 y^ โˆ’^2 xi^ + 3y^ = 3^2 ยท^ (โˆ’2)^ โˆ’^6 i^ โˆ’^ 6 =^ โˆ’^24 โˆ’^6 i. 

Problem 2. (8 points) Use an (, ฮด) argument to show that

zโ†’โˆ’^ lim1+5i(2z^ โˆ’^ 3) =^ โˆ’5 + 10i.

SOLUTION: Let  > 0 be given. Note that |f (z) โˆ’ (โˆ’5 + 10i)| = | 2 z โˆ’ 3 + 5 โˆ’ 10 i| = 2|z + 1 โˆ’ 5 i| = 2 |z โˆ’ (โˆ’1 + 5i)|. To bound this by , we will choose ฮด =  2

Assume that 0 < |z โˆ’ (โˆ’1 + 5i)| < ฮด =  2

. Then

|(2z โˆ’ 3) โˆ’ (โˆ’5 + 10i)| = | 2 z + 2 โˆ’ 10 i| = 2|z โˆ’ (โˆ’1 + 5i)| < 2 ฮด = ,

as required.