求极限lim(x→2) [√(x+2)-2]/√[(x+7)-3]= 求极限lim(x→0)[(1+mx)^n-(1+nx)^m]/x^2=求极限lim(x→无穷){1/(2!)+2/(3!)+……+n/[(n+1)!]}=3/2 1/2mn(n-m) 1
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![求极限lim(x→2) [√(x+2)-2]/√[(x+7)-3]= 求极限lim(x→0)[(1+mx)^n-(1+nx)^m]/x^2=求极限lim(x→无穷){1/(2!)+2/(3!)+……+n/[(n+1)!]}=3/2 1/2mn(n-m) 1](/uploads/image/z/8511650-26-0.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90lim%EF%BC%88x%E2%86%922%EF%BC%89+%5B%E2%88%9A%28x%2B2%29-2%5D%2F%E2%88%9A%5B%28x%2B7%29-3%5D%3D+%E6%B1%82%E6%9E%81%E9%99%90lim%EF%BC%88x%E2%86%920%EF%BC%89%5B%281%2Bmx%29%5En-%281%2Bnx%29%5Em%5D%2Fx%5E2%3D%E6%B1%82%E6%9E%81%E9%99%90lim%28x%E2%86%92%E6%97%A0%E7%A9%B7%EF%BC%89%7B1%2F%282%21%29%2B2%2F%283%21%29%2B%E2%80%A6%E2%80%A6%2Bn%2F%5B%28n%2B1%29%21%5D%7D%3D3%2F2+1%2F2mn%28n-m%29+1)
求极限lim(x→2) [√(x+2)-2]/√[(x+7)-3]= 求极限lim(x→0)[(1+mx)^n-(1+nx)^m]/x^2=求极限lim(x→无穷){1/(2!)+2/(3!)+……+n/[(n+1)!]}=3/2 1/2mn(n-m) 1
求极限lim(x→2) [√(x+2)-2]/√[(x+7)-3]= 求极限lim(x→0)[(1+mx)^n-(1+nx)^m]/x^2=
求极限lim(x→无穷){1/(2!)+2/(3!)+……+n/[(n+1)!]}=
3/2 1/2mn(n-m) 1
求极限lim(x→2) [√(x+2)-2]/√[(x+7)-3]= 求极限lim(x→0)[(1+mx)^n-(1+nx)^m]/x^2=求极限lim(x→无穷){1/(2!)+2/(3!)+……+n/[(n+1)!]}=3/2 1/2mn(n-m) 1
lim(x→2) [√(x+2)-2]/√[(x+7)-3]=lim(x→2) [√(x+2)-2] [√(x+2)+2]√[(x+7)+3]/√[(x+7)-3][√(x+2)+2]√[(x+7)+3]=lim(x→2)((x+2)-4)√[(x+7)+3]/((x+7)-9)[√(x+2)+2]=lim(x→2)√[(x+7)+3]/[√(x+2)+2]=6/4
第二个把分子用二项式展开,取平方项为1/2mn(n-m) x^2,零次项和一次项为0,三次以上取极限后位0,故极限为 1/2mn(n-m)
lim(n→无穷){1/(2!)+2/(3!)+……+n/[(n+1)!]}=lim(n→无穷){(2-1)/(2!)+(3-1)/(3!)+……+(n+1-1)/[(n+1)!]}=lim(n→无穷){1/1!+1/(2!)+1/(3!)+……+1/n!}-{1/(2!)+1/(3!)+……+1/[(n+1)!]}=lim(n→无穷){1-1/[(n+1)!]}=1
1/3
第一题分子分母同乘以[√(x+2)+2]/√[(x+7)+3] 化解可求的最后结果为3/2
第二题利用罗比达法则分子分母同对x求2阶导数,最后结果为1/2mn(n-m)
第三题?
0;1;第三个没有X怎么算啊,你是不是写错了?