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Unformatted text preview: 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 welldefined and nonzero 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 → 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....
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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.
 Fall '10
 Mark

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