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Unformatted text preview: Problem Set 3 Calculus 1 Abdullah Khalid, Hassan Bukhari December 3, 2010 Thomas’ Calculus, Exercise 2.6 1. The function is not continuous at x = 2 because the function is not defined at x = 2. Hence the function is not continuous on the internal [1,3]. 2. The function is not continuous because the function is leftdiscontinuous at x = 3 (i.e. the lim x → 3+ g ( x ) 6 = g (3)). Note the kink at x = 2 does not allow us to differentiate g at that point, but g is still continuous at x = 2. 5. a. f ( 1) = 0 and hence it exists b. lim x → 1 + f ( x ) = 0 c. lim x → 1 + f ( x ) = f(1) = 0 d. The function is only rightcontinuous at x = 1. It is not continuous at x = 1 29. lim x → π sin( x sin x ) = 0 because sin( x sin x ) = g ◦ f ( x ) where f ( x ) = x sin x and g ( x ) = sin x . Since f and g are continuous everywhere on the real line g ◦ f is continuous everywhere on the real line by the theorem of composition of continuous functions. Hence lim x → π g ◦ f ( x ) = g ◦ f ( π ) = 0. 35. Consider lim x → 3 x 2 9 x 3 = lim x → 3 ( x 3)( x + 3) ( x 3) = lim x → 3 ( x + 3) = 6 Note that the function g ( x ) is not defined at x = 3 but the limit exists as x approaches 3. To make it continuous we have no choice but to define a new function f(x) as f ( x ) = ( ( x 2 9) ( x 3) for x 6 = 3 6 for x = 3 Note: The book wants you to still call this new function...
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 Spring '10
 RazaSuleman

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