前に戻る　トップへ戻る　次の問題へ

問題2.4.4

次の極限を漸近展開を用いて求めよ。

（１）${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{(1+x)\sin x-x\cos x}{x^2}}}$

（２）${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{e^{x^2}-\cos x}{x\sin x}}}$

（３）${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sin x-x{e}^x+x^2}{x(\cos x-1)}}}$

（４）${\displaystyle \underset{x\to 0}{lim}\{{\displaystyle \frac{1}{{\sin }^{2}x}}-{\displaystyle \frac{1}{x^2}}\}}$

 

《ポイント》

ランダウの記号の最も基本的な性質が
$${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{o(x^n)}{x^n}}=0}$$
というものです。これを利用して2.4.4では極限値を求め、2.4.5では極値を持つか否かを調べます。

ランダウの記号については当サイトの記事「ランダウの記号と漸近展開の2通りの求め方（漸近展開の合成）」を参照してください。

 

---

 

《解答例》

（１）

$\begin{array}{rl}& {\displaystyle \frac{(1+x)\sin x-x\cos x}{x^2}}\\ & ={\displaystyle \frac{(1+x)(x-o(x^2))-x(1-{\displaystyle \frac{1}{2}}{x}^2+o(x^2))}{x^2}}\\ & ={\displaystyle \frac{x^2+{\displaystyle \frac{1}{2}}{x}^3-(1+2x)o(x^2)}{x^2}}\\ & =1+{\displaystyle \frac{1}{2}}x-(1+2x){\displaystyle \frac{o(x^2)}{x^2}}\\ & \stackrel{(x\to 0)}{\to } 1 \cdots \cdots \text{(答)}\end{array}$

 

（２）

$\begin{array}{rl}& {\displaystyle \frac{e^{x^2}-\cos x}{x\sin x}}\\ & ={\displaystyle \frac{1+(x^2)+o(x^2)-1-{\displaystyle \frac{1}{2}}{x}^2+o(x^2)}{x\{x+o(x)\}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{2}}{x}^2}{x^2+o(x^2)}}(\because x\cdot o(x)=o(x^2))\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{2}}}{1+{\displaystyle \frac{o(x^2)}{x^2}}}}\\ & \stackrel{(x\to 0)}{\to }{\displaystyle \frac{{\displaystyle \frac{3}{2}}}{1+0}}\\ & ={\displaystyle \frac{3}{2}} \cdots \cdots \text{(答)}\end{array}$

 

（３）

$\begin{array}{rl}& {\displaystyle \frac{\sin x-x{e}^x+x^2}{x(\cos x-1)}}\\ & ={\displaystyle \frac{(x-{\displaystyle \frac{1}{3!}}{x}^3+o(x^3))-x(1+x+{\displaystyle \frac{1}{2}}{x}^2+o(x^2))+x^2}{x\{(1-{\displaystyle \frac{1}{2}}{x}^2+o(x^2))-1\}}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{3}{2}}{x}^3+o(x^3)-x\cdot o(x^2)}{-{\displaystyle \frac{1}{2}}{x}^3+x\cdot o(x^2)}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{2}{3}}}{-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{o(x^3)}{x^3}}}}(\because x\cdot o(x^2)=o(x^3))\\ & \stackrel{(x\to 0)}{\to }{\displaystyle \frac{-{\displaystyle \frac{2}{3}}}{-{\displaystyle \frac{1}{2}}+0}}\\ & ={\displaystyle \frac{4}{3}} \cdots \cdots \text{(答)}\end{array}$

 

（４）

$\begin{array}{rl}& {\displaystyle \frac{1}{{\sin }^{2}x}}-{\displaystyle \frac{1}{x^2}}\\ & ={\displaystyle \frac{x^2-{\sin }^{2x}}{x^2{\sin }^{2x}}}\\ & ={\displaystyle \frac{x^2-{(x-{\displaystyle \frac{1}{3!}}{x}^3+o(x^3))}^2}{x^2{(x-{\displaystyle \frac{1}{3!}}{x}^3+o(x^3))}^2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{3}}{x}^4-{\displaystyle \frac{1}{36}}{x}^6+(2x-{\displaystyle \frac{1}{3}}{x}^3+o(x^3))\cdot o(x^3)}{x^2\{x^2-{\displaystyle \frac{1}{3}}{x}^4+{\displaystyle \frac{1}{36}}{x}^6+(2x-{\displaystyle \frac{1}{3}}{x}^3+o(x^3))\cdot o(x^3)\}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{36}}{x}^2+(2-{\displaystyle \frac{1}{3}}{x}^2+{\displaystyle \frac{o(x^3)}{x}})\cdot {\displaystyle \frac{o(x^3)}{x^3}}}{1-{\displaystyle \frac{1}{3}}{x}^2+{\displaystyle \frac{1}{36}}{x}^4+(2-{\displaystyle \frac{1}{3}}{x}^2+{\displaystyle \frac{o(x^3)}{x}}){\displaystyle \frac{o(x^3)}{x^3}}}}\\ & \stackrel{(x\to 0)}{\to }{\displaystyle \frac{{\displaystyle \frac{1}{3}}-0+2\cdot 0}{1-0+0+2\cdot 0}}\\ & ={\displaystyle \frac{1}{3}} \cdots \cdots \text{(答)}\end{array}$

 

 

---

 

復習例題未設定

 

---

前に戻る　トップへ戻る　次の問題へ