两道求极限,微积分或数学达人来帮忙已知lim ((ab-cos x)/x^2) =b+1,求a,b的值,x趋向于0lim ((x-2n)/(x+n))^x =e^-6,求自然数n,x趋向于无穷
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![两道求极限,微积分或数学达人来帮忙已知lim ((ab-cos x)/x^2) =b+1,求a,b的值,x趋向于0lim ((x-2n)/(x+n))^x =e^-6,求自然数n,x趋向于无穷](/uploads/image/z/3653696-56-6.jpg?t=%E4%B8%A4%E9%81%93%E6%B1%82%E6%9E%81%E9%99%90%2C%E5%BE%AE%E7%A7%AF%E5%88%86%E6%88%96%E6%95%B0%E5%AD%A6%E8%BE%BE%E4%BA%BA%E6%9D%A5%E5%B8%AE%E5%BF%99%E5%B7%B2%E7%9F%A5lim+%28%28ab-cos+x%29%2Fx%5E2%29+%3Db%2B1%2C%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC%EF%BC%8Cx%E8%B6%8B%E5%90%91%E4%BA%8E0lim+%28%28x-2n%29%2F%28x%2Bn%29%29%5Ex+%3De%5E-6%2C%E6%B1%82%E8%87%AA%E7%84%B6%E6%95%B0n%EF%BC%8Cx%E8%B6%8B%E5%90%91%E4%BA%8E%E6%97%A0%E7%A9%B7)
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两道求极限,微积分或数学达人来帮忙已知lim ((ab-cos x)/x^2) =b+1,求a,b的值,x趋向于0lim ((x-2n)/(x+n))^x =e^-6,求自然数n,x趋向于无穷
两道求极限,微积分或数学达人来帮忙
已知lim ((ab-cos x)/x^2) =b+1,求a,b的值,x趋向于0
lim ((x-2n)/(x+n))^x =e^-6,求自然数n,x趋向于无穷
两道求极限,微积分或数学达人来帮忙已知lim ((ab-cos x)/x^2) =b+1,求a,b的值,x趋向于0lim ((x-2n)/(x+n))^x =e^-6,求自然数n,x趋向于无穷
1.小小的使用一下泰勒展开
2.重要极限公式
1 lim ((ab-cos x)/x^2) =b+1 有ab=1 b+1=1/2 cos x趋近1 所以 ab=1另外 sinx/2x cosx/2=b+1 罗比塔原则 逼近 求解
2 lim ((x-2n)/(x+n))^x =e^-6 n=2 写成 lim[1-3n/(x+n)]^{[(-x-n)/3n]*3n/(-x-n)*x} lim e^[3nx/(-x-n)] =e^-6 有 3n=6