作业帮跪求高数高手!帮小弟解决一道求极限的题目,多谢!
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/05/21 21:22:35
跪求高数高手!帮小弟解决一道求极限的题目,多谢!
e^2 - (1+1/n)^(2n) = ( e + (1+1/n)^n ) * ( e - (1+1/n)^n )
原式 = lim [ {e^2 - (1+1/n)^(2n) } / (1/n), n->∞ ]
= 2e * lim [ ( e - (1+1/n)^n ) / (1/n), n->∞ ] 无穷小比无穷小的极限
= 2e * lim [ ( e - (1+1/x)^x ) / (1/x), x->+∞ ]
y = (1+1/x)^x 的导函数：
lny = x ln(1+1/x)
y ' / y = ln(1+1/x) + x * [ 1/(x+1) - 1/x]
=> y ' = y * [ ln(1+1/x) - 1/(x+1)]
=> lim [ ( e - y ) / (1/x), x->+∞ ] = lim [ - y ' / (-1/x²), x->+∞ ] 罗必塔法则
= lim [ - y * [ ln(1+1/x) - 1/(x+1)] / (-1/x²), x->+∞ ]
= e * lim [ [ ln(1+1/x) - 1/(x+1)] / (1/x²), x->+∞ ] 罗必塔法则
= e * lim [ [ -1/[x(x+1) + 1/(x+1)² ] / (-2/x³), x->+∞ ]
= e/2
原式 = e ²
原式 = lim [ {e^2 - (1+1/n)^(2n) } / (1/n), n->∞ ]
= 2e * lim [ ( e - (1+1/n)^n ) / (1/n), n->∞ ] 无穷小比无穷小的极限
= 2e * lim [ ( e - (1+1/x)^x ) / (1/x), x->+∞ ]
y = (1+1/x)^x 的导函数：
lny = x ln(1+1/x)
y ' / y = ln(1+1/x) + x * [ 1/(x+1) - 1/x]
=> y ' = y * [ ln(1+1/x) - 1/(x+1)]
=> lim [ ( e - y ) / (1/x), x->+∞ ] = lim [ - y ' / (-1/x²), x->+∞ ] 罗必塔法则
= lim [ - y * [ ln(1+1/x) - 1/(x+1)] / (-1/x²), x->+∞ ]
= e * lim [ [ ln(1+1/x) - 1/(x+1)] / (1/x²), x->+∞ ] 罗必塔法则
= e * lim [ [ -1/[x(x+1) + 1/(x+1)² ] / (-2/x³), x->+∞ ]
= e/2
原式 = e ²