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assig0_soln

# assig0_soln - Math 237 Review Assignment Not to be handed...

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Math 237 Review Assignment Not to be handed in 1: Evaluate the following limits. a) lim x 0 x - sin x x 2 arctan x lim x 0 x - sin x x 2 arctan x = lim x 0 1 - cos x x 2 1+ x 2 + 2 x arctan x by L’HR = lim x 0 sin x 2 x (2+ x 2 ) (1+ x 2 ) 2 + 2 arctan x by L’HR = lim x 0 cos x 2(2 - 3 x 2 - x 4 ) (1+ x 2 ) 3 + 2 1+ x 2 by L’HR = 1 6 b) lim x →∞ x ( e 1 /x - 1) lim x →∞ x ( e 1 /x - 1) = lim x →∞ e 1 /x - 1 1 x = lim x →∞ - 1 x 2 e 1 /x - 1 x 2 = lim x →∞ e 1 /x = 1 c) lim x 0 x 4 sin(1 /x ) We observe that - 1 sin(1 /x ) 1 for x 6 = 0 and hence - x 4 x 4 sin(1 /x ) x 4 for x 6 = 0. Since lim x 0 - x 4 = 0 = lim x 0 x 4 we have lim x 0 x 4 sin(1 /x ) = 0 by the squeeze theorem. d) lim x →∞ ln x x lim x →∞ ln x x = lim x →∞ 1 /x 1 = 0 , by L’Hosptial’s rule. 1

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2 2: Let g ( x ) = 2 sin x - cos x if x 0 be cx if x > 0 . a) Determine the value of b that makes g ( x ) continuous at x = 0. To be continuous at x = 0 we must have lim x 0 - g ( x ) = lim x 0 + g ( x ). lim x 0 - g ( x ) = lim x 0 - 2 sin x - cos x = - 1 , and lim x 0 + g ( x ) = lim x 0 + be cx = b. Therefore, we must have b = - 1 to get lim x 0 g ( x ) = - 1 = g (0). b) Find all b and c that make g ( x ) diﬀerentiable at x = 0, giving reasons for you answer. In order for g to be diﬀerentiable at x = 0 we must have that g is continuous at x = 0, hence b = - 1, and we must have lim x 0 - g 0 ( x ) = lim x 0 + g 0 ( x ) . We have
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assig0_soln - Math 237 Review Assignment Not to be handed...

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