x→0 のとき f(x)>0
log f(x) = (log cosx)/log(1+x^2)
= (log cosx)/x^2 * x^2/log(1+x^2)
= (1/2)(log (1-(sinx)^2))/(sinx)^2 * (sinx)^2/x^2 * x^2/log(1+x^2)
ここで
lim[x→0](log(1+x^2))/x^2=lim[t→0](log(1+t))/t=[(d/dt)(log(1+t))](t=0)=1
lim[x→0](log (1-(sinx)^2))/(sinx)^2=lim[s→0](log(1+s))/(-s)=-1
したがって
log f(x) → -1* 1/2 * 1 * 1 = -1/2
∴　f(x) → 1/√e