Please detail as much as possible 2. Use the definition of the limit to prove that lim(r+2y – 1) = 1.(1,9)+(0,1)using the examle belowProblem 0.1. By using the definition of the limit,provethatlim(x + y) = 2.(x,y)→(1,1)* Scratch Work: The first step is to let e > 0 be arbitrary and to find a "response" 8 > 0which will satisfy the definition of the limit. Applied to the given function f(x, y) = x + y,L = 2, and (a, b) = (1, 1), the definition gives |(x+y) –- 2| < e when 0 < ||(x, y) – (1, 1)|| < 8,where || - || denotes distance in two dimensions, so this is equivalent to |x + y – 2| < ɛwhenever 0 < v(r – 1)2 + (y – 1)² < 8. From this, note that |r - 1| = V(x – 1)² <V(x – 1)2 + (y – 1)² < 8, and similarly |y – 1| < 8. From the estimate,\f (x, y) – L| = |(x + y) – 2|= |(x – 1) + (y – 1)|< |r – 1|+ |y – 1|< 8 + 8(1)= 28it follows that we should take ɛ = 28 to get the desired inequality. The formal proof is...Proof. Let e > 0 be given and take 8 = €/2. If 0 < /(r – 1)² + (y – 1)² < 8, so that both|x – 1| < 8 and |y – 1| < 8, it follows that\f(x, y) – L| = |(x + y) – 2|= |(x – 1) + (y – 1)|< |r – 1|+ ly – 1|< 28= 2(s/2)(2)Hence,lim(프,y)→(1,1)(x + y) = 2.

Answered: Use the definition of the limit to…

Math

Advanced Math

Use the definition of the limit to prove that lim (r+2y– 1) = 1. (7,4) (0,1)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please detail as much as possible

2. Use the definition of the limit to prove that lim
(r+2y – 1) = 1.
(1,9)+(0,1)
using the examle below
Problem 0.1. By using the definition of the limit,
prove
that
lim
(x + y) = 2.
(x,y)→(1,1)
* Scratch Work: The first step is to let e > 0 be arbitrary and to find a "response" 8 > 0
which will satisfy the definition of the limit. Applied to the given function f(x, y) = x + y,
L = 2, and (a, b) = (1, 1), the definition gives |(x+y) –- 2| < e when 0 < ||(x, y) – (1, 1)|| < 8,
where || - || denotes distance in two dimensions, so this is equivalent to |x + y – 2| < ɛ
whenever 0 < v(r – 1)2 + (y – 1)² < 8. From this, note that |r - 1| = V(x – 1)² <
V(x – 1)2 + (y – 1)² < 8, and similarly |y – 1| < 8. From the estimate,
\f (x, y) – L| = |(x + y) – 2|
= |(x – 1) + (y – 1)|
< |r – 1|+ |y – 1|
< 8 + 8
(1)
= 28
it follows that we should take ɛ = 28 to get the desired inequality. The formal proof is...
Proof. Let e > 0 be given and take 8 = €/2. If 0 < /(r – 1)² + (y – 1)² < 8, so that both
|x – 1| < 8 and |y – 1| < 8, it follows that
\f(x, y) – L| = |(x + y) – 2|
= |(x – 1) + (y – 1)|
< |r – 1|+ ly – 1|
< 28
= 2(s/2)
(2)
Hence,
lim
(프,y)→(1,1)
(x + y) = 2.

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