>>486
弧PA=θ,PQ=1-cosθだろ
lim[θ→+0](1-cosθ)/(θ^2)
=lim[θ→+0](1-cos2θ)/(2θ)^2
=lim[θ→+0]{1-(cos^2θ-sin^2θ)}/(4θ^2)
=lim[θ→+0](2sin^2θ)/(4θ^2)
=lim[θ→+0](1/2)(sinθ/θ)^2
=(1/2)*1^2=1/2