Question #eae4a

1 Answer
Konstantinos Michailidis
Jun 26, 2017

We have that

#(cot2x)/(cscx)=((cos2x)/(sin2x))/(1/sinx)=(sinx*cos2x)/(sin2x)= ((sinx*cos2x)/(2sinxcosx))=1/2*((cos2x)/(cosx))#

Hence

#lim_(x->0) ((cot2x)/(cscx))=1/2*lim_(x->0) ((cos2x)/cosx)=1/2*(lim_(x->0) cos2x)/(lim_(x->0) cosx)=1/2*1/1=1/2#

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