高数等价无穷小问题？
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选C，可以考虑泰勒公式
答案如图所示
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x->0
sinx = x-(1/6)x^3 +o(x^3)
e^(sinx)
=e^[x-(1/6)x^3 +o(x^3)]
=e^x .{e^[-(1/6)x^3 +o(x^3)]}
=e^x. [1-(1/6)x^3 +o(x^3) ]
tanx = x+(1/3)x^3 +o(x^3)
e^(tanx)
=e^[x+(1/3)x^3 +o(x^3)]
=e^x .{e^[(1/3)x^3 +o(x^3)]}
=e^x. [1+(1/3)x^3 +o(x^3)]
e^(sinx)-e^(tanx)
=e^x . { [1-(1/6)x^3 +o(x^3)] -[1+(1/3)x^3 +o(x^3)] }
=e^x .[-(1/2)x^3 +o(x^3)]
=-(1/2)x^3 +o(x^3)
=> n=3
ans : C
sinx = x-(1/6)x^3 +o(x^3)
e^(sinx)
=e^[x-(1/6)x^3 +o(x^3)]
=e^x .{e^[-(1/6)x^3 +o(x^3)]}
=e^x. [1-(1/6)x^3 +o(x^3) ]
tanx = x+(1/3)x^3 +o(x^3)
e^(tanx)
=e^[x+(1/3)x^3 +o(x^3)]
=e^x .{e^[(1/3)x^3 +o(x^3)]}
=e^x. [1+(1/3)x^3 +o(x^3)]
e^(sinx)-e^(tanx)
=e^x . { [1-(1/6)x^3 +o(x^3)] -[1+(1/3)x^3 +o(x^3)] }
=e^x .[-(1/2)x^3 +o(x^3)]
=-(1/2)x^3 +o(x^3)
=> n=3
ans : C
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