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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 =+ . square 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 > α .Thus again by the comparison theorem we know that lim x 0+ x +2 x =+ . square 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 there exists a δ > 0 such that 0< y < δ f (1/ y ) ( L - ε, L + ε ) . But this exactly means that lim x 0+ f (1/ y )= L . square Exercise 4. Section 5.1, #7. Proof. See the solutions for the Practice midterm II, where this is proved... square Exercise 5. Section 5.1, #12.

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

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