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user7230927 | (Level 1) Salutatorian

Posted June 13, 2013 at 4:31 PM via web

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mvcdc | Student, Graduate | (Level 1) Associate Educator

Posted June 13, 2013 at 4:52 PM (Answer #1)

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Since `cos^-1(x)` is continuous at `x=-1/2` , we can interchange the limit and the inverse cosine to write the problem as:

`cos^-1 (lim_(x rarr infty) ((-3x)/(4+6x)))`

So first, let us evaluate the said limit:

`lim_(x rarr infty) ((-3x)/(4+6x))=-3 lim_(x rarr infty) ((x)/(4+6x))` 

as `x rarr infty` ,both the numerator and the denominator approach infinity - a case of `infty/infty` . Hence, we apply L'Hospital's rule:

`rArr -3 lim_(x rarr infty) (((d/dx)x)/((d/dx)(4+6x)))= - 3 lim_(x rarr infty) (1/6) = -3*(1/6) = -1/2`


`lim_(x rarr infty) (cos^-1((-3x)/(4+6x))) = cos^-1(-1/2) = (2pi)/3`

The answer is `(2pi)/3`  

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