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6.4. Prove that lim (x² + 2y) = 5. y-2 Method 1, using definition of limit.

6.4. Prove that lim (x² + 2y) = 5. y-2 Method 1, using definition of limit.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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6.4) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.

Limits and continuity 6.4. Prove that lim (x² +2y) = 5. y2 Method 1, using definition of limit. We must show that, given any ô > 0, we can find ô > 0 such that |x + 2y – 5| < 8 when 0 < |x– 1 |<8, 0< ly-2| < 8. If 0 < |x – 1| < 8 and 0 <|y – 2| < 8, then 1 – 8 x < 1 + < 8 and 2 – 8 <y< 2+ 8, excluding x = 1, y = 2. Thus, 1– 28 + 8 ² <x² < 1 + 28 + 8 ² and 4 – 28 < 2y < 4 + 28. Adding 5 – 48 + 82<x² + 2y < 5 + 48 + 8 ² or -48 + 8 ² <x² + 2y – 5 < 48 + 8 ²2 Now, if ô < 1, it certainly follows that –58 <x² + 2y – 5<58; i.e., |x² + 2y – 5| < 58 whenever 0 < |x- 1| <8,0< ly-2| <8. Then, choosing 58= e, i.e., 8= e/5 (or d = 1, whichever is smaller), it follows that |+2y-5|| <e when 0< |x-1| <8,0< [y-2|<ô; i.e., lim ( +2y) = 5. y→2 Method 2, using theorems on limits. lim(x + 2y) = lim x² + lim 2y= 1+4= 5
Transcribed Image Text:Limits and continuity 6.4. Prove that lim (x² +2y) = 5. y2 Method 1, using definition of limit. We must show that, given any ô > 0, we can find ô > 0 such that |x + 2y – 5| < 8 when 0 < |x– 1 |<8, 0< ly-2| < 8. If 0 < |x – 1| < 8 and 0 <|y – 2| < 8, then 1 – 8 x < 1 + < 8 and 2 – 8 <y< 2+ 8, excluding x = 1, y = 2. Thus, 1– 28 + 8 ² <x² < 1 + 28 + 8 ² and 4 – 28 < 2y < 4 + 28. Adding 5 – 48 + 82<x² + 2y < 5 + 48 + 8 ² or -48 + 8 ² <x² + 2y – 5 < 48 + 8 ²2 Now, if ô < 1, it certainly follows that –58 <x² + 2y – 5<58; i.e., |x² + 2y – 5| < 58 whenever 0 < |x- 1| <8,0< ly-2| <8. Then, choosing 58= e, i.e., 8= e/5 (or d = 1, whichever is smaller), it follows that |+2y-5|| <e when 0< |x-1| <8,0< [y-2|<ô; i.e., lim ( +2y) = 5. y→2 Method 2, using theorems on limits. lim(x + 2y) = lim x² + lim 2y= 1+4= 5
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