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

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Unformatted text preview: MA 1505 Mathematics I Tutorial 2 Solutions 1. (a) lim x → π/ 2 1- sin x 1 + cos2 x = lim x → π/ 2- cos x- 2sin2 x = lim x → π/ 2 sin x- 4cos2 x = 1 4 . (b) lim x → ln(cos ax ) ln(cos bx ) = lim x →- a sin ax cos ax- b sin bx cos bx = lim x → 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 = 1. (g) lim x → ln " sin x x ¶ 1 x 2 # = lim x → ln( sin x x ) x 2 = lim x → ( x sin x ) · x cos x- sin x x 2 2 x = 1 2 lim x → x sin x lim x → x cos...
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Solution_2 - MA 1505 Mathematics I Tutorial 2 Solutions 1....

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