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求lim(sec5x-tanx)的极限,x→∏/2
人气:227 ℃ 时间:2020-03-25 19:48:06
解答
lim(sec5x-tanx)=lim (1/cos5x - sinx/cosx)=lim (cosx - sinxcos5x)/cos5xcosx令t=π/2 - x,则t→0原式=lim (sint-costsin5t)/sin5tsint由等价无穷小知sin5t∽5tsint∽t代换,得到原式=lim (sint - costsin5t)/5t&...
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