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Solution_2 - MA 1505 Mathematics I Tutorial 2 Solutions 1(a...

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MA 1505 Mathematics I Tutorial 2 Solutions 1. (a) lim x π/ 2 1 - sin x 1 + cos 2 x = lim x π/ 2 - cos x - 2 sin 2 x = lim x π/ 2 sin x - 4 cos 2 x = 1 4 . (b) lim x 0 ln(cos ax ) ln(cos bx ) = lim x 0 - a sin ax cos ax - b sin bx cos bx = lim x 0 a sin ax cos bx b sin bx cos ax = a 2 b 2 . (c) lim x →∞ x tan 1 x = lim x →∞ tan( x - 1 ) x - 1 = lim x →∞ - x - 2 sec 2 ( x - 1 ) - x - 2 = lim x →∞ cos - 2 ( x - 1 ) = 1. (d) lim x 0+ x a ln x = lim x 0+ ln x x - a = lim x 0+ 1 x - ax - a - 1 = lim x 0+ x a - a = 0 . (e) lim x 1 ln x 1 1 - x = lim x 1 ln x 1 - x = lim x 1 1 x - 1 = - 1. So lim x 1 x 1 1 - x = e - 1 . (f) Using (1d) we have lim x 0+ ln x sin x = lim x 0+ sin x ln x = lim x 0+ sin x x · x ln x = lim x 0+ sin x x lim x 0+ x ln x = 0 . So lim x 0+ x sin x = e 0 = 1. (g) lim x 0 ln " sin x x 1 x 2 # = lim x 0 ln( sin x x ) x 2 = lim x 0 ( x sin x ) · x cos x - sin x x 2 2 x = 1 2 lim x 0 x sin x lim x 0 x cos x - sin x x 3 = 1 2 lim x 0 cos x - x sin x - cos x 3 x 2 = - 1 6 lim x 0 sin x x = - 1 6 .
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