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Do i have to use trigonometry? What is limit ln(sin2x)/ln(sinx) with x ->0? 

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eelfy | Student, Undergraduate | (Level 1) eNoter

Posted June 26, 2013 at 1:00 PM via web

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Do i have to use trigonometry? What is limit ln(sin2x)/ln(sinx) with x ->0? 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 26, 2013 at 1:10 PM (Answer #1)

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You need to evaluate the limit by replacing 0 for x in the given quotient, such that:

`lim_(x->0) (ln(sin2x))/(ln(sinx)) = (ln(sin0))/(ln(sin0))`

Since `sin 0 = 0` yields:

`lim_(x->0) ln(sin2x)/ln(sinx) = ln 0/ln 0 = (-oo)/(-oo)`

The indetermination `oo/oo` requests for you to use l'Hospital's theorem, such that:

 

`lim_(x->0) ln(sin2x)/ln(sinx) = lim_(x->0) ((ln(sin2x))')/((ln(sinx))') `

`lim_(x->0) ((ln(sin2x))')/((ln(sinx))') = lim_(x->0) (1/(sin 2x)*(sin 2x)')/(1/sin x*(sin x)')`

 

`lim_(x->0) ((2cos 2x)/(sin 2x))/(cos x/sin x) `

You may use trigonometric identity `sin 2x = 2 sin x*cos x` such that:

`lim_(x->0) ((2cos 2x)/(sin 2x))/(cos x/sin x) = lim_(x->0) ((2cos 2x)/(2 sin x*cos x))/(cos x/sin x) `

Reducing duplicate factors yields:

`lim_(x->0) (cos 2x)/(cos^2 x) `

Replacing 0 for x yields:

`lim_(x->0) (cos 2x)/(cos^2 x) = (cos 0)/(cos^2 0) `

Since `cos 0 = 1` yields:

`lim_(x->0) (cos 2x)/(cos^2 x) = 1/1 = 1`

Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->0) (ln(sin2x))/(ln(sinx)) = 1` .

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