Question

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？

Answer 1

(x+3)/(x+2)

=1 + 2/(x+2）

令

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

Answer 2

朋友，你好！详细完整清晰过程rt，希望能帮到你解决问题

Answer 3

方法如下，
请作参考：