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Question Number 190347 by mustafazaheen last updated on 01/Apr/23
how is solution  lim_(x→sinπ ) ((sin(π/2))/(sinx))=?
howissolutionlimxsinπsinπ2sinx=?
Answered by JDamian last updated on 01/Apr/23
L=lim_(x→0)  (1/(sin x)) =  { ((+∞   x→0^+ )),((−∞  x→0^− )) :}  ∄L
L=limx01sinx={+x0+x0L

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