求证一道高数题~~内详 求解一道大学高数题,谢谢~
\u6c42\u89e3\u4e00\u9053\u5927\u5b66\u9ad8\u6570\u9898\u76ee,\u8c22\u8c22~\u4f5c\u66ff\u6362 t=1+xy\uff0c\u5219
(x,y)\u2192(2,-1/2) ==> (t,y)\u2192(0,-1/2)\uff0c
\u4e8e\u662f\uff0c
\u539f\u6781\u9650 = lim[(t,y)\u2192(0,-1/2)][(1+t)^(1/t)]^y = e^(-1/2)\u3002
\u53ef\u4ee5\u4f7f\u7528\u4e0d\u540c\u8def\u5f84\u53bb\u6c42\u6781\u9650\uff0c\u770b\u770b\u76f8\u4e0d\u76f8\u7b49\uff0c\u4e0d\u76f8\u7b49\u5c31\u6ca1\u6709\u6781\u9650\u3002
这是书上课后的原题啊！！
条件得是x趋于0的情况下
证明当x→0时secx-1与x²/2是等价的无穷小。
证明：x→0lim(secx-1)/(x²/2)=x→0lim[2(secx-1)/x²]=x→0lim[(2secxtanx)/(2x)]
=x→0lim[(secxtanx)/x]=x→0lim(secxtan²x+sec³x)=1
故当x→0时secx-1与x²/2是等价的无穷小。
两者相除求极限，用罗比达法则计算出极限值为1
所以这两个是等价无穷小
lim (secx-1)/(x²/2) (0/0)=lim (-sinxsecx/xcosx)=lim (-cosxsecx-sinx*sinx/cos²x)/(cosx-xsinx)=1
x->0 x->0 x->0
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