This is true if and only if |x-2|< 8 = 5=53₁ 1b) We have In our case, domf = R\ {0}. if and only if, for any RE R, there exists > 0, such that, for any x € domf with 1 0< x-1| <8, we have that 1 lim x1 (x-1)2 This is equivalent to 1 lim x+1(x-1)2 we fix R> 0 and we consider the condition that is > R. (x - 1)² To verify that = +∞ f(x) = = +∞ 1 (x - 1)² (x - 1)² < 1/1/2 > R. 1 |x-1|< 8 = √√R
This is true if and only if |x-2|< 8 = 5=53₁ 1b) We have In our case, domf = R\ {0}. if and only if, for any RE R, there exists > 0, such that, for any x € domf with 1 0< x-1| <8, we have that 1 lim x1 (x-1)2 This is equivalent to 1 lim x+1(x-1)2 we fix R> 0 and we consider the condition that is > R. (x - 1)² To verify that = +∞ f(x) = = +∞ 1 (x - 1)² (x - 1)² < 1/1/2 > R. 1 |x-1|< 8 = √√R
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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Exercise 1b) I have attached the solution below
The part I don’t understand is why did they write dom f= R /{0} shouldn’t it be R /{1} correct me if I’m wrong
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