高数题,求x趋近于无穷时,1+x分之x+3的x+5次方的极限?

2025-12-13 19:42:44
推荐回答(3个)
回答1:

(x+3)/(x+2)

=1 + 2/(x+2)

1/y = 2/(x+2)

x+2 =2y

x=2y-2

lim(x->无穷) [(x+3)/(x+2)]^(x+5)

把(x+3)/(x+2) 变成 =1 + 2/(x+2)

=lim(x->无穷) [1 + 2/(x+2)]^(x+5)

令 1/y = 2/(x+2)

=lim(x->无穷) [1 + 1/y]^(2y-2+5)

=lim(x->无穷) [1 + 1/y]^(2y+3)

=lim(x->无穷) [1 + 1/y]^(2y)

=e^2

得出结果

lim(x->无穷) [(x+3)/(x+2)]^(x+5) =e^2

回答2:

朋友,你好!详细完整清晰过程rt,希望能帮到你解决问题

回答3:

方法如下,
请作参考: