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

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### 2 Answers

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.**

`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`