- 554 名前：１３２人目の素数さん mailto:sage [2012/09/11(火) 03:33:32.61 ]
- >>512
(1) lim_[x→π/2] (π-2x ) / ( ( 1-sin(x) ) tan(x) )
π/2-x＝h とおくと x→π/2 のとき h→0
分子は 2h
sin(x)＝sin(π/2-h)＝cos(h)
tan(x)＝sin(x) / cos(x)＝ sin(π/2-h) / cos(π/2-h)＝cos(h) / sin(h)
与式
＝lim_[h→0] (2h) / ( (1-cos(h)) (cos(h)/sin(h)) )
＝lim_[h→0] (2h) tan(h) / ( 1-cos(h) )
有理化して
＝lim_[h→0] (2h) tan(h) (1+cos(h)) / ( 1-cos(h)^2 )
＝lim_[h→0] (2h) tan(h) (1+cos(h)) / ( sin(h)^2 )
分母分子に (1/h^2) を掛けて
＝lim_[h→0] 2(tan(h)/h) (1+cos(h)) / ( (sin(h)/h)^2 )
h→0 ならば tan(h)/h＝1、cos(h)＝1、sin(h)/h＝1
＝ 2 (1) (1+1) / 1
＝4
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