lim(x→π/2)tan3x/tanx
=lim(x→π/2) 3sec^2(3x)/sec^2x
=lim(x→π/2) 3cos^2x/cosx^2(3x)
=lim(y→0) 3sin^2y/sin^2(3y) (令y=π/2-x)
=lim(y→0) 3y^2/(3y)^2
=3/9
=1/3錯(cuò)lim tanx/tan3x=lim ( 1/cosx^2) / (3/cos3x^2)= lim (cos3x^2)/3cosx^2= lim 6cos3x(-sin3x)/6cosx(-sinx)= lim sin6x/sin2x= lim 6cos6x/2cos2x= -6 / -2 = 3lim ( 1/cosx^2) / (3/cos3x^2) 這一步如何得出的 我就這里不明白 詳細(xì)點(diǎn) 謝謝求導(dǎo)求出來(lái)的呀你學(xué)過(guò)羅比達(dá)法則么lim sec^2x/3sec^2(3x)→lim ( 1/cosx^2) / (3/cos3x^2) 我之前學(xué)過(guò)已經(jīng)忘得差不多了你能幫我把前一步到后一步是怎么得出來(lái)的給我解釋明白嗎謝謝是這個(gè)么lim sec^2x/3sec^2(3x)=lim ( 1/cosx^2) / (3/cos3x^2) secx=1/cosx所以 sec^2x=1/cosx^2