何度も解いて体で覚えましょう!
代入してみると
\lim_{x \rightarrow 0}\ \dfrac{\cos{x}-1}{x} = \dfrac{\cos{0}-1}{0} = \dfrac{1-1}{0} = \colorbox{red}{\color{white}$\dfrac{0}{0}$}
不定形 \dfrac{0}{0} ・ \cos{x} - 1 ⇒ \cos{x} + 1 をかける!
【解答】
\begin{align*} & \textcolor{orange}{\phase{\footnotesize\ 分母分子に (\cos{x}+1) をかける!}}\\ \lim_{x \rightarrow 0}\ \dfrac{\cos{x}-1}{x} &= \lim_{x \rightarrow 0}\ \dfrac{(\cos{x}-1)(\cos{x}+1)}{x(\cos{x}+1)}\\\\ &= \lim_{x \rightarrow 0}\ \dfrac{\cos^2{x}-1}{x(\cos{x}+1)}\\ & \textcolor{orange}{\phase{\footnotesize\ \sin^2x+\cos^2x=1 より}}\\ &= \lim_{x \rightarrow 0}\ \dfrac{-\sin^2{x}}{x(\cos{x}+1)}\\\\ &= \lim_{x \rightarrow 0}\ \left(-\colorbox{lavender}{$\dfrac{\sin{x}}{x}$} \times \dfrac{\sin{x}}{\cos{x}+1}\right)\\\\ &= -\colorbox{lavender}{$1$} \times \dfrac{\sin{0}}{\cos{0}+1}\\\\ &= -1 \times \dfrac{0}{1+1} = 0 \end{align*}
代入してみると
\lim_{x \rightarrow 0}\ \dfrac{1-\cos{x}}{x^2} = \dfrac{1-\cos{0}}{0^2} = \dfrac{1-1}{0} = \colorbox{red}{\color{white}$\dfrac{0}{0}$}
不定形 \dfrac{0}{0} ・ 1 - \cos{x} ⇒ 1 + \cos{x} をかける!
【解答】
\begin{align*} & \textcolor{orange}{\phase{\footnotesize\ 分母分子に (1+\cos{x}) をかける!}}\\ \lim_{x \rightarrow 0}\ \dfrac{1-\cos{x}}{x^2} &= \lim_{x \rightarrow 0}\ \dfrac{(1-\cos{x})(1+\cos{x})}{x^2(1+\cos{x})}\\\\ &= \lim_{x \rightarrow 0}\ \dfrac{1-\cos^2{x}}{x^2(1+\cos{x})}\\ & \textcolor{orange}{\phase{\footnotesize\ \sin^2x+\cos^2x=1 より}}\\ &= \lim_{x \rightarrow 0}\ \dfrac{\sin^2{x}}{x^2(1+\cos{x})}\\\\ &= \lim_{x \rightarrow 0}\ \left\{\left(\colorbox{lavender}{$\dfrac{\sin{x}}{x}$}\right)^2 \times \dfrac{1}{1+\cos{x}}\right\}\\\\ &= \colorbox{lavender}{$1$}^2 \times \dfrac{1}{1+\cos{0}}\\\\ &= 1 \times \dfrac{1}{1+1} = \dfrac12 \end{align*}
代入してみると
\lim_{x \rightarrow 0}\ \dfrac{x\sin{x}}{1-\cos{x}} = \dfrac{0\sin{0}}{1-\cos{0}} = \dfrac{0 \cdot 0}{1-1} = \colorbox{red}{\color{white}$\dfrac{0}{0}$}
不定形 \dfrac{0}{0} ・ 1 - \cos{x} ⇒ 1 + \cos{x} をかける!
【解答】
\begin{align*} & \textcolor{orange}{\phase{\footnotesize\ 分母分子に (1+\cos{x}) をかける!}}\\ \lim_{x \rightarrow 0}\ \dfrac{x\sin{x}}{1-\cos{x}} &= \lim_{x \rightarrow 0}\ \dfrac{x\sin{x}(1+\cos{x})}{(1-\cos{x})(1+\cos{x})}\\\\ &= \lim_{x \rightarrow 0}\ \dfrac{x\sin{x}(1+\cos{x})}{1-\cos^2{x}}\\ & \textcolor{orange}{\phase{\footnotesize\ \sin^2x+\cos^2x=1 より}}\\ &= \lim_{x \rightarrow 0}\ \dfrac{x\sin{x}(1+\cos{x})}{\sin^2{x}}\\\\ &= \lim_{x \rightarrow 0}\ \left(\colorbox{lavender}{$\dfrac{x}{\sin{x}}$} \times \dfrac{\sin{x}(1+\cos{x})}{\sin{x}}\right)\\\\ &= \lim_{x \rightarrow 0}\ \left\{\colorbox{lavender}{$\dfrac{x}{\sin{x}}$} \times (1+\cos{x})\right\}\\\\ &= \colorbox{lavender}{$1$} \times (1+\cos{0})\\\\ &= 1 \times (1+1) = 2 \end{align*}
- 20210907…初版公開。問題数3。とりあえず解いただけ。そのうち解説も加える>私