problemset4s - MAT 137Y 2008-2009 Winter Session Solutions...

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MAT 137Y 2008-2009 Winter Session, Solutions to Problem Set 4 1 (a) We are given f ( x ) M and lim x a f ( x ) = L exists. Suppose L > M . Since lim x a f ( x ) = L , then for any ε > 0, there exists δ > 0 such that 0 < | x - a | < δ implies | f ( x ) - L | < ε . In particular, let ε = L - M > 0. Then there exists δ > 0 such that 0 < | x - a | < δ implies | f ( x ) - L | < L - M , which implies L - ( L - M ) < f ( x ) < L +( L - M ) = M < f ( x ) < 2 L - M = f ( x ) > M , which is a contradiction, since f ( x ) M for all x . Therefore it follows that L M . (b) The statement is false. Consider the function f ( x ) = ( - x 2 , x 6 = 0 , - 1 , x = 0 . Then f ( x ) < 0 for all x , but lim x 0 f ( x ) = 0, so L = M . 2 (SHE 2.3) 44. Consider the functions f ( x ) = ( 0 , x < c , 1 , x c , g ( x ) = ( 1 , x < c , 0 , x c . Both limits do not exist at x = c , but lim x c f ( x ) g ( x ) = 0. 46. The statement is false. Since lim x c [ f ( x ) + g ( x )] and lim x c f ( x ) exists, then by the properties of limits, lim x c g ( x ) = lim x c [ g ( x )+ f ( x ) - f ( x )] = lim x c [ f ( x )+ g ( x )] - lim x c f ( x ) , so lim x c g ( x ) must exist. 48. The statement is false. Let f ( x ) = - 1. Then lim x c f ( x ) = - 1 for any c , but p f ( x ) is not defined at any x = c , so lim x c p f ( x ) does not exist. 50. The statement is false since it need not be true that lim x c f ( x ) or lim x c g ( x ) exists. So, for example, let f ( x ) = 0 and g ( x ) be the Dirichlet function (as defined in SHE 2.2 #47). Then for any c , lim x c g ( x ) does not exist. However, if we stipulate that the limits must exist, then the statement is true.
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problemset4s - MAT 137Y 2008-2009 Winter Session Solutions...

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