関数の極限②無限÷無限

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何度も解いて体で覚えましょう！

次の極限を求めよ。

\displaystyle\lim_{x \rightarrow \infty}\ \dfrac{2x^2-5}{3x^2-4x+1}

代入してみると

\lim_{x \rightarrow \infty}\ \dfrac{2x^2-5}{3x^2-4x+1} = \dfrac{2\infty^2-5}{3\infty^2-4\infty+1} = \colorbox{red}{\textcolor{white}{$\dfrac{\infty}{\infty}$}}

不定形 $\dfrac{\infty}{\infty}$ ⇒ 分母で次数の高い項で割る！

\begin{align*} & 　\textcolor{orange}{\phase{\footnotesize\ 分母で次数の髙い x^2 で割る！}}\\ \lim_{x \rightarrow \infty}\ \dfrac{2x^2-5}{3x^2-4x+1} &\textcolor{orange}{= \lim_{x \rightarrow \infty}\ \dfrac{\dfrac{2x^2}{x^2}\ \dfrac{-5}{x^2}}{\dfrac{3x^2}{x^2}\ \dfrac{-4x}{x^2}\ \dfrac{+1}{x^2}}}\\\\ &= \lim_{x \rightarrow \infty}\ \dfrac{2-\dfrac{5}{x^2}}{3-\dfrac{4}{x}+\dfrac{1}{x^2}}\\\\ &\textcolor{orange}{= \dfrac{2-\dfrac{5}{\infty^2}}{3-\dfrac{4}{\infty}+\dfrac{1}{\infty^2}}}\\\\ &\textcolor{orange}{= \dfrac{2-\colorbox{lightcyan}{$\dfrac{5}{\infty}$}}{3-\colorbox{lightcyan}{$\dfrac{4}{\infty}$}+\colorbox{lightcyan}{$\dfrac{1}{\infty}$}}}\\\\ &\textcolor{orange}{= \dfrac{2-\colorbox{lightcyan}{$0$}}{3-\colorbox{lightcyan}{$0$}+\colorbox{lightcyan}{$0$}}} = \dfrac23 \end{align*}

\displaystyle\lim_{x \rightarrow -\infty}\ \dfrac{x^2+3}{x-2}

代入してみると

\lim_{x \rightarrow -\infty}\ \dfrac{x^2+3}{x-2} = \dfrac{(-\infty)^2+3}{-\infty-2} = -\colorbox{red}{\textcolor{white}{$\dfrac{\infty}{\infty}$}}

不定形 $\dfrac{\infty}{\infty}$ ⇒ 分母で次数の高い項で割る！

\begin{align*} & 　\textcolor{orange}{\phase{\footnotesize\ 分母で次数の髙い x で割る！}}\\ \lim_{x \rightarrow -\infty}\ \dfrac{x^2+3}{x-2} &\textcolor{orange}{= \lim_{x \rightarrow -\infty}\ \dfrac{\ \dfrac{x^2}{x}\ \dfrac{+3}{x}\ }{\dfrac{x}{x}\ \dfrac{-2}{x}}}\\\\ &= \lim_{x \rightarrow -\infty}\ \dfrac{x+\dfrac{3}{x}}{1-\dfrac{2}{x}}\\\\ &\textcolor{orange}{= \dfrac{-\infty+\colorbox{lightcyan}{$\dfrac{3}{-\infty}$}}{1-\colorbox{lightcyan}{$\dfrac{2}{-\infty}$}}}\\\\ &\textcolor{orange}{= \dfrac{-\infty+\colorbox{lightcyan}{$0$}}{1-\colorbox{lightcyan}{$0$}}} = -\infty \end{align*}

\displaystyle\lim_{x \rightarrow \infty}\ \dfrac{2x-1}{4x+3}

代入してみると

\lim_{x \rightarrow \infty}\ \dfrac{2x-1}{4x+3} = \dfrac{2\infty-1}{4\infty+3} = \colorbox{red}{\textcolor{white}{$\dfrac{\infty}{\infty}$}}

不定形 $\dfrac{\infty}{\infty}$ ⇒ 分母で次数の高い項で割る！

\begin{align*} & 　\textcolor{orange}{\phase{\footnotesize\ 分母で次数の髙い x で割る！}}\\ \lim_{x \rightarrow \infty}\ \dfrac{2x-1}{4x+3} &\textcolor{orange}{= \lim_{x \rightarrow \infty}\ \dfrac{\dfrac{2x}{x}\ \dfrac{-1}{x}}{\ \dfrac{4x}{x}\ \dfrac{+3}{x}\ }}\\\\ &= \lim_{x \rightarrow \infty}\ \dfrac{2-\dfrac{1}{x}}{4+\dfrac{3}{x}}\\\\ &\textcolor{orange}{= \dfrac{2-\colorbox{lightcyan}{$\dfrac{1}{\infty}$}}{4+\colorbox{lightcyan}{$\dfrac{3}{\infty}$}}}\\\\ &\textcolor{orange}{= \dfrac{2-\colorbox{lightcyan}{$0$}}{4+\colorbox{lightcyan}{$0$}}} = \dfrac24 = \dfrac12 \end{align*}

\displaystyle\lim_{x \rightarrow -\infty}\ \dfrac{5x^2+4}{2x^2-3x}

代入してみると

\lim_{x \rightarrow -\infty}\ \dfrac{5x^2+4}{2x^2-3x} = \dfrac{5(-\infty)^2+4}{2(-\infty)^2-3(-\infty)} = \colorbox{red}{\textcolor{white}{$\dfrac{\infty}{\infty}$}}

不定形 $\dfrac{\infty}{\infty}$ ⇒ 分母で次数の高い項で割る！

\begin{align*} & 　\textcolor{orange}{\phase{\footnotesize\ 分母で次数の髙い x^2 で割る！}}\\ \lim_{x \rightarrow -\infty}\ \dfrac{5x^2+4}{2x^2-3x} &\textcolor{orange}{= \lim_{x \rightarrow -\infty}\ \dfrac{\dfrac{5x^2}{x^2}\ \dfrac{+4}{x^2}}{\ \dfrac{2x^2}{x^2}\ \dfrac{-3x}{x^2}\ }}\\\\ &= \lim_{x \rightarrow -\infty}\ \dfrac{\ 5+\dfrac{4}{x^2}\ }{2-\dfrac{3}{x}}\\\\ &\textcolor{orange}{= \dfrac{\ 5+\dfrac{4}{(-\infty)^2}\ }{2-\dfrac{3}{-\infty}}}\\\\ &\textcolor{orange}{= \dfrac{5+\colorbox{lightcyan}{$\dfrac{4}{\infty}$}}{2-\colorbox{lightcyan}{$\dfrac{3}{-\infty}$}}}\\\\ &\textcolor{orange}{= \dfrac{5+\colorbox{lightcyan}{$0$}}{2-\colorbox{lightcyan}{$0$}}} = \dfrac52 \end{align*}

\displaystyle\lim_{x \rightarrow \infty}\ \dfrac{4-x^2}{3x+2}

代入してみると

\lim_{x \rightarrow \infty}\ \dfrac{4-x^2}{3x+2} = \dfrac{4-\infty^2}{3\infty+2} = -\colorbox{red}{\textcolor{white}{$\dfrac{\infty}{\infty}$}}

不定形 $\dfrac{\infty}{\infty}$ ⇒ 分母で次数の高い項で割る！

\begin{align*} & 　\textcolor{orange}{\phase{\footnotesize\ 分母で次数の髙い x で割る！}}\\ \lim_{x \rightarrow \infty}\ \dfrac{4-x^2}{3x+2} &\textcolor{orange}{= \lim_{x \rightarrow \infty}\ \dfrac{\ \dfrac{4}{x}\ \dfrac{-x^2}{x}\ }{\dfrac{3x}{x}\ \dfrac{+2}{x}}}\\\\ &= \lim_{x \rightarrow \infty}\ \dfrac{\dfrac{4}{x}-x}{3+\dfrac{2}{x}}\\\\ &\textcolor{orange}{= \dfrac{\colorbox{lightcyan}{$\dfrac{4}{\infty}$}-\infty}{3+\colorbox{lightcyan}{$\dfrac{2}{\infty}$}}}\\\\ &\textcolor{orange}{= \dfrac{\colorbox{lightcyan}{$0$}-\infty}{3+\colorbox{lightcyan}{$0$}}} = -\infty \end{align*}

\displaystyle\lim_{x \rightarrow -\infty}\ \dfrac{3-2x}{x^2-4x+1}

代入してみると

\lim_{x \rightarrow -\infty}\ \dfrac{3-2x}{x^2-4x+1} = \dfrac{3-2(-\infty)}{(-\infty)^2-4(-\infty)+1} = \colorbox{red}{\textcolor{white}{$\dfrac{\infty}{\infty}$}}

不定形 $\dfrac{\infty}{\infty}$ ⇒ 分母で次数の高い項で割る！

\begin{align*} & 　\textcolor{orange}{\phase{\footnotesize\ 分母で次数の髙い x^2 で割る！}}\\ \lim_{x \rightarrow -\infty}\ \dfrac{3-2x}{x^2-4x+1} &\textcolor{orange}{= \lim_{x \rightarrow -\infty}\ \dfrac{\ \dfrac{3}{x^2}\ \dfrac{-2x}{x^2}\ }{\dfrac{x^2}{x^2}\ \dfrac{-4x}{x^2}\ \dfrac{+1}{x^2}}}\\\\ &= \lim_{x \rightarrow -\infty}\ \dfrac{\dfrac{3}{x^2}-\dfrac{2}{x}}{1-\dfrac{4}{x}+\dfrac{1}{x^2}}\\\\ &\textcolor{orange}{= \dfrac{\dfrac{3}{\infty^2}-\dfrac{2}{\infty}}{1-\dfrac{4}{\infty}+\dfrac{1}{\infty^2}}}\\\\ &\textcolor{orange}{= \dfrac{\colorbox{lightcyan}{$\dfrac{3}{\infty}$}-\colorbox{lightcyan}{$\dfrac{2}{\infty}$}}{1-\colorbox{lightcyan}{$\dfrac{4}{\infty}$}+\colorbox{lightcyan}{$\dfrac{1}{\infty}$}}}\\\\ &\textcolor{orange}{= \dfrac{\colorbox{lightcyan}{$0$}-\colorbox{lightcyan}{$0$}}{1-\colorbox{lightcyan}{$0$}+\colorbox{lightcyan}{$0$}}} = 0 \end{align*}

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