【微分積分】2-3-2 三角関数｜問題集

\(\displaystyle =\lim_{x\to0}\frac{2(\sin\frac{x}{2})^2}{x^2}\)
\(\displaystyle =\lim_{x\to0}\frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\)
\(\displaystyle =\frac{1}{2}\)

\(\displaystyle =\lim_{x\to0}\frac{-\sin^2x}{x^2}\)
\(\displaystyle =-\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\)
\(=-1\)

対数をとると、
\(\displaystyle \lim_{x\to0}\log(1+\sin x)^{\frac{1}{x}}\)
\(\displaystyle =\lim_{x\to0}\frac{\sin x}{x}・\frac{\log(1+\sin x)}{\sin x}\)
\(=1・1\)
\(=1\)
よって、
\(\displaystyle \lim_{x\to0}(1+\sin x)^{\frac{1}{x}}=\lim_{x\to0}e^{\log(1+\sin x)^{\frac{1}{x}}}=e\)

\(\displaystyle =\lim_{x\to0}\frac{\frac{\sin2x}{2x}}{\frac{\sin3x}{3x}}・\frac{2x}{3x}\)
\(\displaystyle =\frac{2}{3}\)

\(\displaystyle =\lim_{x\to0}\frac{\cos^2x-1}{x(\cos x+1)}\)
\(\displaystyle =-\lim_{x\to0}\frac{\sin^2x}{x(\cos x+1)}\)
\(\displaystyle =-\lim_{x\to0}\frac{\sin x}{x}・\frac{\sin x}{\cos x+1}\)
\(=-1・0\)
\(=0\)

\(t=x-\pi\)とおくと、\(x\to\pi\)のとき\(t\to0\)
\(\displaystyle =\lim_{t\to0}\frac{\sin(\pi+t)}{t}\)
\(\displaystyle =\lim_{t\to0}\frac{-\sin t}{t}\)
\(=-1\)

\(\displaystyle =\lim_{x\to0}\frac{\sin3x}{3x}・\frac{3x}{x}\)
\(=1・3\)
\(=3\)

\(\displaystyle =\lim_{x\to0}\frac{2x}{\frac{\sin x}{x}}・\frac{1}{x}\)
\(=2\)

\(\displaystyle 0\leqq\left|x\sin\frac{1}{x}\right|\leqq1\)より、はさみうちの定理から
\(\displaystyle \lim_{x\to0}\left|x\sin\frac{1}{x}\right|=0\)
よって、
\(\displaystyle \lim_{x\to0}x\sin\frac{1}{x}=0\)

\(\displaystyle 0\leqq\left|x\sin\frac{1}{x}\right|\leqq1\)より、はさみうちの定理から
\(\displaystyle \lim_{x\to0}x\sin\frac{1}{x}=0\) また、\(f(0)=0\)より、
\(\displaystyle \lim_{x\to0}f(x)=f(0)\)
よって、\(f(x)\)は\(x=0\)で連続