Question #eae4a

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Question #eae4a

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Answer 1 · 2017-06-26T06:31:30

We have that

$\frac{cot 2 x}{csc x} = \frac{\frac{cos 2 x}{sin 2 x}}{\frac{1}{sin x}} = \frac{sin x \cdot cos 2 x}{sin 2 x} = (\frac{sin x \cdot cos 2 x}{2 sin x cos x}) = \frac{1}{2} \cdot (\frac{cos 2 x}{cos x})$ $(cot2x)/(cscx)=((cos2x)/(sin2x))/(1/sinx)=(sinx*cos2x)/(sin2x)= ((sinx*cos2x)/(2sinxcosx))=1/2*((cos2x)/(cosx))$

Hence

$lim x \to 0 (\frac{cot 2 x}{csc x}) = \frac{1}{2} \cdot lim x \to 0 (\frac{cos 2 x}{cos x}) = \frac{1}{2} \cdot \frac{lim x \to 0 cos 2 x}{lim x \to 0 cos x} = \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2}$ $lim_(x->0) ((cot2x)/(cscx))=1/2*lim_(x->0) ((cos2x)/cosx)=1/2*(lim_(x->0) cos2x)/(lim_(x->0) cosx)=1/2*1/1=1/2$

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