Question

求当x趋于0时lim(1+sinx)^(2/x);lim1-cos2x/(sin^2）3x的极限；当x趋于无穷时lim(x-1)(sin1/x-1)的极限

要过程，谢谢！

Answer 1

①考察两个重要极限
lim(1+sinx)^2/x
=lim(1+sinx)^(1/sinx*sinx*2/x)
=e^lim(2sinx/x)
=e²
②考察等价无穷小
x→0时，1-cosx~x²/2,sinx~x
∴lim(1-cos2x)/sin²3x
=2x²/9x²=2/9
③x→∞时，1/(x-1)→0
∴lim(x-1)(sin1/x-1)
=limsin(1/(x-1))/(1/(x-1)
=1

Answer 2

lim(1+sinx)^(2/x)=e^2

lim1-cos2x/(sin^2）3x=1/18

lim(x-1)(sin1/x-1)=1

追问

可以写下过程吗？谢谢啦!

追答

lim(x→0) (1+sinx)^(2/x)

=lim(x→0) [(1+sinx)^(1/sinx)]^(2sinx/x)
=lim(x→0) [(1+sinx)^(1/sinx)]^[lim(x→0) (2sinx/x)]
=e^2
lim(x→0) 1-cos2x/(sin^2）3x
=lim(x→0) (x^2/2)/(3x)^2
=1/18

lim(x→∞) (x-1)(sin1/x-1)
=lim(x→∞) (sin1/x-1)/[1/(x-1)]
=1