Question

大神，求极限，

Answer 1

(x²-1)/(x²+1)=(x²+1-1-1)/(x²+1)=1-2/(x²+1)=1+1/(x²+1)/(-2)原式=lim [1+1/(x²+1)/(-2)]^x²=lim [1+1/(x²+1)/(-2)]^(x²+1-1)=lim [1+1/(x²+1)/(-2)]^(x²+1)*[1+1/(x²+1)/(-2)]^(-1)其中lim [1+1/(x²+1)/(-2)]^(-1)=1所以=lim [1+1/(x²+1)/(-2)]^(x²+1)=lim [1+1/(x²+1)/(-2)]^[(-2)*(x²+1)/(-2)]=lim [1+1/(x²+1)/(-2)]^[(x²+1)/(-2)]*(-2)=e^(-2)不懂可以追问大神，求极限，

Answer 2

(x2-1)/(x2+1)
=(x2+1-1-1)/(x2+1)
=1-2/(x2+1)
=1+1/(x2+1)/(-2)
原式=
lim [1+1/(x2+1)/(-2)]^x2
=lim [1+1/(x2+1)/(-2)]^(x2+1-1)
=lim [1+1/(x2+1)/(-2)]^(x2+1)*[1+1/(x2+1)/(-2)]^(-1)
其lim [1+1/(x2+1)/(-2)]^(-1)=1
所
=lim [1+1/(x2+1)/(-2)]^(x2+1)
=lim [1+1/(x2+1)/(-2)]^[(-2)*(x2+1)/(-2)]
=lim [1+1/(x2+1)/(-2)]^[(x2+1)/(-2)]*(-2)
=e^(-2)
懂追问