319HW8Solutions

319HW8Solutions - MAT 319/320 Solutions for HW8 Exercise 1....

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MAT 319/320 Solutions for HW8 Exercise 1. Section 4.3, #5a. Proof. We have that lim x 1+ 1 x - 1 = + . Indeed, for any given α > 0 if we take x (1 , 1 + 1 α ) , then we have that 1 x - 1 > α . Now since lim x 1+ x = 1 , we know that by the comparison the- orem, lim x 1+ x x - 1 = + if and only if lim x 1+ 1 x - 1 = + , which is the case. In conclusion, lim x 1+ x x - 1 = + . s Exercise 2. Section 4.3, #5c. Proof. lim x 0+ x + 2 = 2 and lim x 0+ 1 x = + : indeed for any given α > 0 if we take x (0 , 1 α 2 ) , then we have that 1 x .Thus again by the comparison theorem we know that lim x 0+ x + 2 x = + . s Exercise 3. Section 4.3, #8. Proof. lim x + f ( x ) = L means the following: for any given ε > 0 there exists an α > 0 such that x > α f ( x ) ( L - ε, L + ε ) . If one writes δ = 1/ α the condition x > α is equivalent to y = 1/ x < δ , so the condition is now equivalent to: for any given ε > 0
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This note was uploaded on 01/31/2011 for the course AMS 319 taught by Professor Mark during the Fall '10 term at SUNY Stony Brook.

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319HW8Solutions - MAT 319/320 Solutions for HW8 Exercise 1....

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