Exam_2_Solution

# Exam_2_Solution - Math 1a Test 2 Solutions Fall 2005...

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Unformatted text preview: Math 1a Test 2 Solutions Fall 2005 Problem 1. (Carlos) Find lim x → log(cos2 x ) log(cos x ) . Solution 1. Let f ( x ) = log (cos2 x ) and g ( x ) = log (cos x ). Clearly, lim x → f ( x ) = 0 and lim x → g ( x ) = 0. Further, for some small δ , we have that g ( x ) =- tan x 6 = 0 for 0 < x < δ . Hence, we can apply L’Hopital’s rule and deduce that: lim x → log cos2 x log cos x = lim x →- 2tan(2 x )- tan( x ) (1) = lim x → 2sin(2 x )cos( x ) sin( x )cos(2 x ) (2) = lim x → 4 sin( x )cos( x )cos( x ) sin( x )cos(2 x ) (3) = lim x → 4 cos( x ) 2 cos(2 x ) (4) = 4 (5) Alternatively, one can do the same calculation using little ”o” nota- tion. Note that: cos(2 x ) = 1- 2 x 2 + o ( x 2 ) log(1 + x ) = x + o ( x ) Thus, log cos(2 x ) = log(1 + (- 2 x 2 + o ( x 2 ))) (6) = (- 2 x 2 + o ( x 2 )) + o (- 2 x 2 + o ( x 2 )) (7) =- 2 x 2 + o ( x 2 ) + o ( x 2 ) (8) =- 2 x 2 + o ( x 2 ) (9) 1 2 Similarly, cos( x ) = 1- x 2 2 + o ( x 2 ) (10) log(cos( x )) = log(1 + (- x 2 2...
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Exam_2_Solution - Math 1a Test 2 Solutions Fall 2005...

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