`lim_(x->pi^-) (x-pi)csc(x)` Evaluate the limit.

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Chapter 4, Review - Problem 12 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the limit, hence, you need to replace` pi` for x in expression under limit:

`lim_(x->pi)(x-pi)csc x = (pi-pi)/(sin pi) = 0/0`

Since the limit is indeterminate `0/0` , you may use l'Hospital's theorem:

`lim_(x->pi)(x-pi)csc x =lim_(x->pi)(x-pi)/(sin x) =  lim_(x->pi)((x-pi)')/((sin x)')`

` lim_(x->pi)((x-pi)')/((sin x)') =  lim_(x->pi)(1)/(cos x)`

Replacing by `pi ` yields:

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

Hence, evaluating the limit, using l'Hospital's theorem, yields `lim_(x->pi)(x-pi)csc x = -1.`

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