極限

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?

No.1 ベストアンサー 回答者： rnakamra 回答日時： 2012/11/06 10:44

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。

No.2 回答者： info22_ 回答日時： 2012/11/06 12:31

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