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MIT18_014F10_pset10sols

# MIT18_014F10_pset10sols - 18.014 Problem Set 10 Solutions...

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18.014 Problem Set 10 Solutions Total: 12 points Problem 1: Evaluate log( a + be x ) a. lim b. lim log( x ) log(1 x ). x →∞ a + bx 2 x 1 Solution (4 points) For part (a), we first evaluate the limits lim x →∞ log( a + be x ) and x lim x →∞ a + x bx 2 . The second limit can be computed as follows: x x 2 1 1 lim = lim = lim = . a + bx 2 a + bx 2 t 0 + at 2 + b b x →∞ x →∞ For the first limit, we use L’Hopital’s rule to get log( a + be x ) be x / ( a + be x ) b b lim = lim = lim = lim = 1 . x →∞ x x →∞ 1 x →∞ ae x + b t 0 + ae t + b Now, we know lim f ( x ) g ( x ) = lim f ( x ) lim g ( x ) x →∞ x →∞ x →∞ whenever the two limits on the right hand side exist. Putting it all together, we have log( a + be x ) 1 lim = . x →∞ a + bx 2 b For part (b), note lim x 1 log( 1 x ) = and lim x 1 log(1 x ) = . Using L’Hopital’s rule, we see log(1 x ) 1 / (1 x ) x log( x ) 2 lim = lim = lim . x 1 1 / log( x ) x 1 1 / log( x ) 2 x x 1 (1 x ) Applying L’Hopital’s rule again, we get

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