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S2x + 3 for x 2 | Hw6) 2et f(x) = 24x for x

S2x + 3 for x 2 | Hw6) 2et f(x) = 24x for x

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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Please solve homework question in the following format like in two columns: 

 

•statment          |             •justifications 

 

Note: My previous homework was cancelled due to not following format.

[I have attached the required format for guidance]

i want to give you thumbs up. So please follow attached format

The required format is: 1) Given e > 0, Choose N. = !. 1) Since e > 0, it is possible to choose N, = !. %3D 2) If k 2 Ne, then k 2) Since N, = ! 3) No switch when the inequality multiplied by e > 0. 3) 1 < ke 4) 1 < (2k)e 4) Since k < 2k for every natural number k. 5) 1 < (2k + 1)e 5) Since 2k < 2k + 1, for every real number. 6) No switch when the inequality divided by 6) 2k+1 くE 2k+1 7) 2k+1 7) By the definition of absolute value. くE 8) < e 8) |-1| = 1 9) |1 + - 1| < e - 1| <e 9) Algebra 2k+1 10) 2k+1 -1 10) Sincol 2k+1
Transcribed Image Text:The required format is: 1) Given e > 0, Choose N. = !. 1) Since e > 0, it is possible to choose N, = !. %3D 2) If k 2 Ne, then k 2) Since N, = ! 3) No switch when the inequality multiplied by e > 0. 3) 1 < ke 4) 1 < (2k)e 4) Since k < 2k for every natural number k. 5) 1 < (2k + 1)e 5) Since 2k < 2k + 1, for every real number. 6) No switch when the inequality divided by 6) 2k+1 くE 2k+1 7) 2k+1 7) By the definition of absolute value. くE 8) < e 8) |-1| = 1 9) |1 + - 1| < e - 1| <e 9) Algebra 2k+1 10) 2k+1 -1 10) Sincol 2k+1
S2x+3 for x?| Hw6| Let flx)= { 4x f(x) = 2 4 x for x<l Prove lim f(x) = 5 メ→」+ lim f(x)=4
Transcribed Image Text:S2x+3 for x?| Hw6| Let flx)= { 4x f(x) = 2 4 x for x<l Prove lim f(x) = 5 メ→」+ lim f(x)=4
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