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ie. whenever, 12-201 < € 191 gyo. any 2 with 0212-201 <ó we have we choose 6 = & clearly lal Also for 1 az + b = (azo + b) | 0 and for any (x, y) ER with o0 be arbitary. choose Then we choose 8₁ = &₁ and 5₂ = €//2 clearly an 6₁ 8₂ 20.

ie. whenever, 12-201 < € 191 gyo. any 2 with 0212-201 <ó we have we choose 6 = & clearly lal Also for 1 az + b = (azo + b) | 0 and for any (x, y) ER with o0 be arbitary. choose Then we choose 8₁ = &₁ and 5₂ = €//2 clearly an 6₁ 8₂ 20.

Advanced Engineering Mathematics
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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a> Lim 2→% Complex Constant and sol": Let az + b = Now Now, us wheneven ie. whenever, a zo + b a fo. | az+b la ²₁ + b) | = |az_azo | = 101|2 - 201 choose его arbitary. laz+b = (az₂ +b)/ < E 1a1 1Z-Zol <E E 191 12-201 < we choose 6 = {2₁ E... be Also for any 2 with 1 az + b = (azo + b) | <e This shows that lim Z→ Zo where a, b, to are an clearly & so. کی نے 201 - 0212 az + b = we have azo+b. b> Now, We Now, We NOW, Then We Let 2 = x+iy. => 1-1 => (x,y) → (1, -1) lim (x + i(2x + y)) = lim (u(x,y) + i v(x, y)) 241-1 x 1 y I now Lim x → 1 y →→ Let Then where u(x,y) find Lim X→ 1 Y = -1 |u(x, y) = 1/ Clearly иск, у) be X an исх, у) Then 8 >0 and 50, of The n we for any (x, y) ER with ox/x-11cd ando<ly +11<8 have | u(x, y) = 1 | <E. This shows that = 1. 12-11 E >O arbitary. |u(x, y) — 1| <& when ever 1/ choose 8 = E and and v(x, y) | V(x,y) - 11 = 12x+y=11 an &, 6,8₂ 70. Lim x → 1 y-1 2x+y = V(x, y) i.e. [v(x,y)-1| ≤ 2 |x-1| + ly +1| E >O be arbitary. choose we choose 8₁ = &₁ and 8/₂ = 2/1/2 |x-1| < E 12x = 2 + y +11|
Transcribed Image Text:a> Lim 2→% Complex Constant and sol": Let az + b = Now Now, us wheneven ie. whenever, a zo + b a fo. | az+b la ²₁ + b) | = |az_azo | = 101|2 - 201 choose его arbitary. laz+b = (az₂ +b)/ < E 1a1 1Z-Zol <E E 191 12-201 < we choose 6 = {2₁ E... be Also for any 2 with 1 az + b = (azo + b) | <e This shows that lim Z→ Zo where a, b, to are an clearly & so. کی نے 201 - 0212 az + b = we have azo+b. b> Now, We Now, We NOW, Then We Let 2 = x+iy. => 1-1 => (x,y) → (1, -1) lim (x + i(2x + y)) = lim (u(x,y) + i v(x, y)) 241-1 x 1 y I now Lim x → 1 y →→ Let Then where u(x,y) find Lim X→ 1 Y = -1 |u(x, y) = 1/ Clearly иск, у) be X an исх, у) Then 8 >0 and 50, of The n we for any (x, y) ER with ox/x-11cd ando<ly +11<8 have | u(x, y) = 1 | <E. This shows that = 1. 12-11 E >O arbitary. |u(x, y) — 1| <& when ever 1/ choose 8 = E and and v(x, y) | V(x,y) - 11 = 12x+y=11 an &, 6,8₂ 70. Lim x → 1 y-1 2x+y = V(x, y) i.e. [v(x,y)-1| ≤ 2 |x-1| + ly +1| E >O be arbitary. choose we choose 8₁ = &₁ and 8/₂ = 2/1/2 |x-1| < E 12x = 2 + y +11|
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Step 1: Part(a):

T o space S h o w colon space limit as straight z rightwards arrow straight z subscript 0 of open parentheses a z plus b close parentheses equals az subscript 0 plus straight b space where space straight a comma space straight b comma space straight z subscript 0 space are space complex space constant space and space straight a not equal to 0.

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