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asst8-f10-solns - Math 137 1(a lim x0 Assignment No 8...

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Math 137 Assignment No. 8 - Solutions Fall 2010 1. (a) lim x 0 parenleftbigg cot x - 1 x parenrightbigg . This is an indeterminate form of the type ∞ - ∞ . If we put the two functions over a common denominator, i.e., lim x 0 parenleftbigg cot x - 1 x parenrightbigg = lim x 0 x cos x - sin x x sin x = lim x 0 f ( x ) g ( x ) : a 0 0 indeterminate form. If we now differentiate numerator and denominator, we obtain lim x 0 f ( x ) g ( x ) = lim x 0 - x sin x sin x + x cos x , another 0 0 indeterminate form. Differentiating again (with faith), lim x 0 f ′′ ( x ) g ′′ ( x ) = lim x 0 - sin x - x cos x cos x + cos x - x sin x = 0. We can now work backward and conclude that lim x 0 parenleftbigg cot x - 1 x parenrightbigg = lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) = lim x 0 f ′′ ( x ) g ′′ ( x ) = 0. (b) lim x →∞ ( xe 1 /x - x ). This is an indeterminate form of the type ∞ - ∞ . If we factor the expression and then rewrite it as a product, i.e., ( xe 1 /x - x ) = x ( e 1 /x - 1) = e 1 /x - 1 1 /x , we have produced a 0 0 indeterminate form since e 1 /x e 0 = 1 as x → ∞ . It is convenient to make the change of variable y = 1 x so that lim x →∞ e 1 /x - 1 1 /x = lim y 0 e y - 1 y . Now differentiate numerator and denominator to find that lim y 0 e y - 1 y = lim y 0 e y 1 = 1. Therefore lim x →∞ ( xe 1 /x - x ) = 1. 2. Required to find a such that lim x →∞ parenleftbigg x + a x - a parenrightbigg x = e. (1) Method No. 1: Using the logarithm. If we take the natural logarithm of the LHS of (1), ln parenleftbigg x + a x - a parenrightbigg x = x ln parenleftbigg x + a x - a parenrightbigg = ln bracketleftbig ( x + a )( x - a ) 1 bracketrightbig x 1 . From (1), we would like the limit as x → ∞ of this expression to be ln e = 1. In the limit x → ∞ , the rightmost term is a 0 0 indeterminate form. Rewrite this term as follows, f ( x ) g ( x ) = ln( x + a ) - ln( x - a ) x 1 , 1
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which implies that f ( x ) g ( x ) = ( x + a ) 1 - ( x - a ) 1 - x 2 = · · · = 2 ax 2 ( x + a )( x - a ) . Then lim x →∞ f ( x ) g ( x ) = lim x →∞ 2 ax 2 ( x + a )( x - a ) = lim x →∞ 2 a (1 + ax 1 )(1 - ax 1 ) = · · · = 2 a. Recalling that this limit must be equal to ln e = 1, we have 2 a = 1, implying that a = 1 2 . Method No. 2: Using the limit definition for e . The appearance of x both inside the brackets as well as in the exponent suggests that the limit definition for e may possibly be involved, i.e., lim n →∞ parenleftbigg 1 + 1 n parenrightbigg n = e (2) We’ll return to this equation. In the original expression, let x - a = y which implies that x = y + a , so that x
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