求lim(sec5x-tanx)的极限,x→∏/2
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求lim(sec5x-tanx)的极限,x→∏/2
![求lim(sec5x-tanx)的极限,x→∏/2](/uploads/image/z/19084476-12-6.jpg?t=%E6%B1%82lim%EF%BC%88sec5x-tanx%EF%BC%89%E7%9A%84%E6%9E%81%E9%99%90%2Cx%E2%86%92%E2%88%8F%2F2)
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∽5t
sint∽t
代换,得到
原式
=lim (sint - costsin5t)/5t²
=lim (cost - 5costcos5t + sintsin5t)/10t
=∞
=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∽5t
sint∽t
代换,得到
原式
=lim (sint - costsin5t)/5t²
=lim (cost - 5costcos5t + sintsin5t)/10t
=∞
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