Thursday, November 29, 2018

Limit of goniometric function without l'Hospital's rule




I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?



limxπ21sinx(π2x)2


Answer



Set t=π2x,
limxπ21sinx(π2x)2=limt01costt2=limt02sin2(t/2)4(t/2)2=12


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