Question

求极限lim(x→0)（1+x²)^(1/1-cosx)

求极限lim(x→0)（1+x²)^(1/1-cosx)

Answer 1

设y=(1+x^2)^[1/(1-cosx)]
lny==ln{(1+x^2)^[1/(1-cosx)]}
=[1/(1-cosx)]*ln(1+x^2)
=ln(1+x^2)/(1-cosx)
lim(x→0)
lny
＝lim(x→0)
ln(1+x^2)/(1-cosx)
【注：是0/0型，故运用洛必达法则,即对分子分母求导，】
＝lim(x→0)
[ln(1+x^2)]'/(1-cosx)'
＝lim(x→0)
[2x/(1+x^2)]/sinx
＝{lim(x→0)
[2/(1+x^2)]}*{lim(x→0)
x/sinx}
＝2*{lim(x→0)
1/(sinx/x)}
＝2/{lim(x→0)
sinx/x}
＝2/1
＝2
lim(x→0)（1+x²)^[1/(1-cosx)]=lim(x→0)
y
=lim(x→0)
e^(lny)
=e^{lim(x→0)
lny}
=e^2