2个回答
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f(t)=e^t (x<t<tanx)
用拉格朗日中值定理:e^x-e^tanx=e^ξ(x-tanx)
x→0 tanx→0 e^ξ→1
∴e^x-e^tanx~x-tanx
lim(x→0)[e^x-e^tanx]/xtan²x]
=lim(x→0)[x-tanx]/xtan²x]
=lim(x→0)[1-sec²x]/[tan²x+2xtanxsec²x]
=lim(x→0)[1-(tan²x+1)]/[tan²x+2xtanxsec²x]
=lim(x→0)[-tan²x]/[tan²x+2xtanxsec²x]
=lim(x→0)[-1]/[1+2sec²x]
=-⅓
用拉格朗日中值定理:e^x-e^tanx=e^ξ(x-tanx)
x→0 tanx→0 e^ξ→1
∴e^x-e^tanx~x-tanx
lim(x→0)[e^x-e^tanx]/xtan²x]
=lim(x→0)[x-tanx]/xtan²x]
=lim(x→0)[1-sec²x]/[tan²x+2xtanxsec²x]
=lim(x→0)[1-(tan²x+1)]/[tan²x+2xtanxsec²x]
=lim(x→0)[-tan²x]/[tan²x+2xtanxsec²x]
=lim(x→0)[-1]/[1+2sec²x]
=-⅓
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