求极限lim(x→2) [√(x+2)-2]/√[(x+7)-3]= 求极限lim(x→0)[(1+mx)^n-(1+n
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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
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
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
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