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Q.Evaluate: `lim_(x-->pi/2)` `(2x - pi) / cosx`.` `

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user8235304 | Student, Grade 11 | Valedictorian

Posted June 29, 2013 at 12:37 AM via web

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Q.Evaluate: `lim_(x-->pi/2)` `(2x - pi) / cosx`.` `

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violy | High School Teacher | (Level 3) Assistant Educator

Posted June 29, 2013 at 1:06 AM (Answer #1)

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If we take the limit directly, we will have an indeterminate form.

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

So, we will apply L'hospital Rule here. 

Take the derivative of top and bottom separately first.

`2/-sinx`

So, we will have:

`lim_(x->pi/2) = 2/(-sin(pi/2)) = 2/-(1) = -2.`

Hence, final answer is -2.

 

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aruv | High School Teacher | Valedictorian

Posted June 30, 2013 at 9:07 AM (Answer #2)

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`lim_{x->pi/2}(2x-pi)/cos(x)=lim_{x->pi/2}(2(x-pi/2))/sin(pi/2-x)`

`=2lim_{x->pi/2}(x-pi/2)/(-sin(x-pi/2))`

`=-2lim_{x->pi/2}(x-pi/2)/sin(x-pi/2)`

`=-2`

`because`

`lim_{x->0}x/sin(x)=1`

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