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Tutorial02-solutions(1)

# Tutorial02-solutions(1) - Problem Set 2 MTH211 Solutions 1...

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Problem Set 2 MTH211 Solutions 1 Questions on understanding the lecture E What are the domain, the codomain, and the component functions of the following maps? (a) ( ) = + - sin ; (b) ( ) = 1 - 2 - 2 ; (c) ( ) = 3 sin( ) ln( ) S (a) The component functions are 1 ( ) = + 2 ( ) = - sin and the codomain is R 2 . Now we need to find the domain, that is, the set where the map is defined. It’s easy to see that the map is defined for all values of , , , so the domain is R 3 . (b) The component function is just 1 ( ) = 1 - 2 - 2 and the codomain is R . Now we need to find the domain, that is, the set where the map is defined. Expression under the square root must be nonnegative, that is, 1 - 2 - 2 0 , or the domain is 2 + 2 1 . It is geometrically the unit disc centered at (0 0) in R 2 . (c) The component functions are 1 ( ) = 3 2 ( ) = sin( ) 3 ( ) = ln( ) and the codomain is R 3 . Now we need to find the domain, that is, the set where the map is defined. Here, 3 and sin( ) are always defined, but for ln( ) we need > 0 . In other words, we either must have > 0 or < 0 . Thus the domain of this map is the union of the open I and III quadrants. :) 1

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E Given a map : R R , prove that lim x a ( x ) = l holds if and only if lim x a ( x ) - l = 0 does. S We first prove the “only if" part. For this, we assume lim x a ( x ) = l holds and we want to show this implies that lim x a ( x ) - l = 0 . Now it follows from the definition that lim x a ( x ) = l means that for any given > 0 , there exists a δ > 0 such that for any 0 < x - a < δ , we have ( x ) - l < . Note that we have ( x ) - l 0 so that ( x ) - l = | ( x ) - l - 0 | . Therefore, we see that for any > 0 , there exists a δ > 0 such that for any 0 < x - a < δ , we have | ( x ) - l - 0 | = ( x ) - l < , which implies that lim x a ( x ) - l = 0 by definition. We now prove the “if" part. For this, we assume lim x a ( x ) - l = 0 holds and we want to show this implies that lim x a ( x ) = l . Now it follows from the definition that lim x a ( x ) - l = 0 means that for any given > 0 , there exists a δ > 0 such that for any 0 < x - a < δ , we have | ( x ) - l - 0 | < . Note that we have ( x ) - l 0 so that | ( x ) - l - 0 | = ( x ) - l .
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