求解lim(x->0)(1+x^2)^(1/1-cosx)极限,RT

Question

求解lim(x->0) (1+x^2)^(1/1-cosx)极限,
RT

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