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What is `lim_(x->0) (1-(cos(2x))^n)/(x^2)`

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pourjour | Student, Undergraduate | eNoter

Posted April 30, 2012 at 1:14 PM via web

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What is `lim_(x->0) (1-(cos(2x))^n)/(x^2)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 30, 2012 at 2:47 PM (Answer #1)

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The limit `lim_(x->0) (1-(cos(2x))^n)/(x^2)` has to be determined.

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

Substituting x = 0 gives the indeterminate form `0/0` . This allows the use of l'Hopital's rule and the denominator and numerator can be substituted with their derivatives.

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

Again, substituting x = 0 gives 0/0,

=> `lim_(x->0) (2*n*sin 2x*(n - 1)*(cos 2x)^(n - 2)*(-sin 2x)*x + 2*n*(cos 2x)^(n - 1)*cos 2x*2)/2`

substituting x = 0 gives

`(0 + 2*n*1*1*2)/2`

=> 2*n

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

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