Is that infinite ?

Question

$lim x-> pi/2 = ((x-pi/2))/((1-sinx)cosx)$

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Answer 1 · 2017-03-01T14:36:09

$-oo$

Calling $y=x-pi/2$ we have

$((x-pi/2))/((1-sinx)cosx) = -y/((1-cos(y))sin(y))$

and also

$lim_( x-> pi/2) ((x-pi/2))/((1-sinx)cosx) =lim_(y->0) -y/((1-cos(y))sin(y))=$ $= lim_(y->0)1/(1-cos(y))lim_(y->0)(-y/siny) = oo (-1) = -oo$