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What is `lim_(x->pi)` `{1 - sin (x/2)} / (pi - x)`      = ?

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user8235304 | Student, Grade 11 | (Level 1) Valedictorian

Posted July 2, 2013 at 2:36 PM via web

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What is `lim_(x->pi)` `{1 - sin (x/2)} / (pi - x)`      = ?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 2, 2013 at 2:50 PM (Answer #1)

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The limit `lim_(x->pi)(1 - sin(x/2))/(pi - x)` has to be determined.

Substituting `x = pi` gives a relation of the form `0/0` which is indeterminate. Using l'Hopital's rule, the numerator and denominator can be replaced with their derivatives

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

=> `lim_(x->pi)(-(1/2)*cos(x/2))/ (- 1)`

Substituting `x = pi`

=> 0

The limit `lim_(x->pi)(1 - sin(x/2))/(pi - x) = 0`

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