319HW12Solutions

# 319HW12Solutions - MAT 319 Solutions for HW12 Exercise 1...

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MAT 319 Solutions for HW12 Exercise 1. Section 6.3, #7c. Proof. Just write x 3 ln x = ln x 1/ x 3 = f ( x ) g ( x ) and apply L’Hospital’s rule on (0 , + ) (notice that g ( x )= - 3 x 4 is well-defined and non-zero on that interval and that both functions are differen- tiable). Thus one gets f ( x ) g ( x ) = 1/ x 3 x 4 =( - 1/3) .x 3 0 when x 0+ , so the limit is zero. square Exercise 2. Section 6.3, #9c. Proof. As usual write (1 + 3/ x ) x = e x. ln (1+ 3 x ) = exp ( ln (1+3/ x ) 1/ x ) . On the interval (0 , + ) both functions f ( x ) = ln (1 + 3/ x ) , g ( x ) = 1/ x are differentiable and d dx (1/ x ) is nonzero on that interval therefore we can apply L’Hospital’s rule: f ( x ) g ( x ) =( 3 x 2 . 1 1+3/ x ) . ( 1 1/ x 2 )= 3 1+3/ x 3 when x + . So the limit of our function is e 3 . square Exercise 3. Section 6.4, #4. Proof. Write Taylor’s formula at the order one and two for f ( x )= 1+ x between 0 and x : f ( x ) =1 + x.f (0)+ x 2 2 .f ′′ ( c ) = 1 + 1 2 x + x 2 2 . ( 1 2 ) . ( 1 2 ) . (1 + c ) 3/2 , and since the remainder is lessorequalslant 0 , one gets 1+ x lessorequalslant 1+ 1 2 x .

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