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

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

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violy's profile pic

Posted (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.


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's profile pic

Posted (Answer #2)

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