極限

質問

lim[x→1](logx/x^2-1)
ロピタルの定理を使わないで求めるにはどうすれば良いのでしょうか?

回答No.1

rnakamra · ベストアンサー

lim[x→1]{logx/(x^2-1)}
=lim[x→1][logx/{(x-1)(x+1)}]
=lim[x→1]1/(x+1)*lim[x→1]{logx/(x-1)}

ここでx-1=hとおくと
=(1/2)*lim[h→0]{log(1+h)/h}
=(1/2)*lim[h→0]log{(1+h)^(1/h)}

lim[h→0](1+h)^(1/h)=eですので(これが使えない場合、t=1/hとおきt→∞とt→-∞でそれぞれ極限を求め両方ともeになることを示せばよい)
lim[h→0]log{(1+h)^(1/h)}=loge=1
これを上の式に入れればよいでしょう。

回答No.2

info22_

L=lim[x→1] log(x)/(x^2-1)
なら

x=t+1とおくと
lim[x→1] log(x)/(x^2-1)
=lim[t→0] log(t+1)/((t+1)^2-1)
=lim[t→0] log(t+1)/(t(t+2))
=lim[t→0] log(t+1)*(1/2){(1/t)-(1/(t+2))}
=(1/2)lim[t→0] log(t+1)/t
-(1/2)lim[t→0] log(t+1)/(t+2)
=(1/2)lim[t→0] {log(t+1)-log(1)}/(t+1-1)
-(1/2)lim[t→0] {log(t+2-1)-log(2-1)/(t+2-0)
t+1=u,t+2=vとおくと
=(1/2)lim[u→1] {log(u)-log(1)}/(u-1)
-(1/2)lim[v→2] {log(v-1)-log(2-1)}/(v-2)*(v-2)/v
=(1/2){log(u)}'|(1)
-(1/2){log(v-1)}'|(2)*0
=(1/2)(1/u)|(u=1)
-(1/2){1/(v-1)|(v=2) *0
=(1/2) -0
=1/2