lim x→π/2 ln(tan(x/2)) を求める

$\underset{x\to \frac{\pi }{2}}{lim}\ln (\tan \left(\frac{x}{2}\right))$を求める

高階の微分を求めておく

一階から順に求めておく． $$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}x}\ln (\tan \left(\frac{x}{2}\right))& =& \frac{1}{\tan \left(\frac{x}{2}\right)}(\frac{\mathrm{d}}{\mathrm{d}x}\tan \left(\frac{x}{2}\right))\cdots u=\tan \left(\frac{x}{2}\right),f=\ln \left(u\right),\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}x}\\ & =& \frac{1}{\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}}\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right)\\ & =& \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}(\frac{\mathrm{d}}{\mathrm{d}x}\sin \left(\frac{x}{2}\right){\cos }^{-1}\left(\frac{x}{2}\right))\\ & =& \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}\{(\frac{\mathrm{d}}{\mathrm{d}x}\sin \left(\frac{x}{2}\right)){\cos }^{-1}\left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)(\frac{\mathrm{d}}{\mathrm{d}x}{\cos }^{-1}\left(\frac{x}{2}\right))\}\\ & =& \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}[(\cancel{\cos \left(\frac{x}{2}\right)}\cdot \frac{1}{2})\cancel{{\cos }^{-1}\left(\frac{x}{2}\right)}+\sin \left(\frac{x}{2}\right)\{-{\cos }^{-2}\left(\frac{x}{2}\right)(-\sin \left(\frac{x}{2}\right)\cdot \frac{1}{2})\}]\\ & =& \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}\cdot \frac{1}{2}\{1+{\sin }^2\left(\frac{x}{2}\right){\cos }^{-2}\left(\frac{x}{2}\right)\}\\ & =& \frac{1}{2}(\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}+\frac{\cancel{\cos \left(\frac{x}{2}\right)}}{\cancel{\sin \left(\frac{x}{2}\right)}}{\sin }^{\cancel{2}1}\left(\frac{x}{2}\right){\cos }^{\cancel{-2}-1}\left(\frac{x}{2}\right))\\ & =& \frac{1}{2}(\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}+\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)})\\ & =& \frac{1}{2}\frac{{\cos }^2\left(\frac{x}{2}\right)+{\sin }^2\left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)}\\ & =& \frac{1}{2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)}\\ & =& \frac{1}{\sin (x)}\cdots \sin (x)=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\\ \\ \frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\ln (\tan \left(\frac{x}{2}\right))& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{\sin (x)}\right)\\ & =& -\frac{1}{{\sin }^2(x)}\{\frac{\mathrm{d}}{\mathrm{d}x}\sin (x)\}\\ & =& -\frac{1}{{\sin }^2(x)}\cos (x)\\ & =& -\frac{\cos (x)}{{\sin }^2(x)}\\ \\ \frac{{\mathrm{d}}^3}{\mathrm{d}{x}^3}\ln (\tan \left(\frac{x}{2}\right))& =& \frac{\mathrm{d}}{\mathrm{d}x}(-\frac{\cos (x)}{{\sin }^2(x)})\\ & =& -\frac{\mathrm{d}}{\mathrm{d}x}\cos (x){\sin }^{-2}(x)\\ & =& -\{(\frac{\mathrm{d}}{\mathrm{d}x}\cos (x)){\sin }^{-2}(x)+\cos (x)(\frac{\mathrm{d}}{\mathrm{d}x}{\sin }^{-2}(x))\}\\ & =& -\{(-\sin (x)){\sin }^{-2}(x)+\cos (x)(-2{\sin }^{-3}(x)\cos (x))\}\\ & =& -\{-{\sin }^{-1}(x)-2{\sin }^{-3}(x){\cos }^2(x)\}\\ & =& {\sin }^{-1}(x)+2{\sin }^{-3}(x){\cos }^2(x)\\ & =& \frac{1}{\sin (x)}+\frac{2{\cos }^2(x)}{{\sin }^3(x)}\\ \\ \frac{{\mathrm{d}}^4}{\mathrm{d}{x}^4}\ln (\tan \left(\frac{x}{2}\right))& =& \frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\sin (x)}+\frac{2{\cos }^2(x)}{{\sin }^3(x)})\\ & =& \frac{\mathrm{d}}{\mathrm{d}x}{\sin }^{-1}(x)+2\frac{\mathrm{d}}{\mathrm{d}x}{\cos }^2(x){\sin }^{-3}(x)\\ & =& -{\sin }^{-2}(x)\cos (x)+2\{(\frac{\mathrm{d}}{\mathrm{d}x}{\cos }^2(x)){\sin }^{-3}(x)+{\cos }^2(x)(\frac{\mathrm{d}}{\mathrm{d}x}{\sin }^{-3}(x))\}\\ & =& -{\sin }^{-2}(x)\cos (x)+2[\{2\cos (x)(-\sin (x))\}{\sin }^{-3}(x)+{\cos }^2(x)(-3{\sin }^{-4}(x)\cos (x))]\\ & =& -\frac{\cos (x)}{{\sin }^2(x)}+2[-2\frac{\cos (x)}{{\sin }^2(x)}-3\frac{{\cos }^3(x)}{{\sin }^4(x)}]\\ & =& -\frac{\cos (x)}{{\sin }^2(x)}-4\frac{\cos (x)}{{\sin }^2(x)}-6\frac{{\cos }^3(x)}{{\sin }^4(x)}\\ & =& -5\frac{\cos (x)}{{\sin }^2(x)}-6\frac{{\cos }^3(x)}{{\sin }^4(x)}\\ \\ \frac{{\mathrm{d}}^5}{\mathrm{d}{x}^5}\ln (\tan \left(\frac{x}{2}\right))& =& \frac{\mathrm{d}}{\mathrm{d}x}(-5\frac{\cos (x)}{{\sin }^2(x)}-6\frac{{\cos }^3(x)}{{\sin }^4(x)})\\ & =& -5\frac{\mathrm{d}}{\mathrm{d}x}\cos (x){\sin }^{-2}(x)-6\frac{\mathrm{d}}{\mathrm{d}x}{\cos }^3(x){\sin }^{-4}(x)\\ & =& -5\{(\frac{\mathrm{d}}{\mathrm{d}x}\cos (x)){\sin }^{-2}(x)+\cos (x)\frac{\mathrm{d}}{\mathrm{d}x}{\sin }^{-2}(x)\}-6\{(\frac{\mathrm{d}}{\mathrm{d}x}{\cos }^3(x)){\sin }^{-4}(x)+{\cos }^3(x)\frac{\mathrm{d}}{\mathrm{d}x}{\sin }^{-4}(x)\}\\ & =& -5\{(-\sin (x)){\sin }^{-2}(x)+\cos (x)(-2{\sin }^{-3}(x)\cos (x))\}-6\{(-3{\cos }^2(x)\sin (x)){\sin }^{-4}(x)+{\cos }^3(x)(-4{\sin }^{-5}(x)\cos (x))\}\\ & =& 5\frac{1}{\sin (x)}+28\frac{{\cos }^2(x)}{{\sin }^3(x)}+24\frac{{\cos }^4(x)}{{\sin }^5(x)}\end{array}$$

$x=\frac{\pi }{2}$でのテーラー展開を求めておく

$$\begin{array}{rcl}\ln (\tan \left(\frac{x}{2}\right))& =& \frac{1}{0!}\left[{\frac{{\mathrm{d}}^0}{\mathrm{d}{x}^0}\ln (\tan \left(\frac{x}{2}\right))|}_{x=\frac{\pi }{2}}\right]{(x-\frac{\pi }{2})}^0\\ & & +\frac{1}{1!}\left[{\frac{{\mathrm{d}}^1}{\mathrm{d}{x}^1}\ln (\tan \left(\frac{x}{2}\right))|}_{x=\frac{\pi }{2}}\right]{(x-\frac{\pi }{2})}^1\\ & & +\frac{1}{2!}\left[{\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\ln (\tan \left(\frac{x}{2}\right))|}_{x=\frac{\pi }{2}}\right]{(x-\frac{\pi }{2})}^2\\ & & +\frac{1}{3!}\left[{\frac{{\mathrm{d}}^3}{\mathrm{d}{x}^3}\ln (\tan \left(\frac{x}{2}\right))|}_{x=\frac{\pi }{2}}\right]{(x-\frac{\pi }{2})}^3\\ & & +\frac{1}{4!}\left[{\frac{{\mathrm{d}}^4}{\mathrm{d}{x}^4}\ln (\tan \left(\frac{x}{2}\right))|}_{x=\frac{\pi }{2}}\right]{(x-\frac{\pi }{2})}^4\\ & & +\frac{1}{5!}\left[{\frac{{\mathrm{d}}^5}{\mathrm{d}{x}^5}\ln (\tan \left(\frac{x}{2}\right))|}_{x=\frac{\pi }{2}}\right]{(x-\frac{\pi }{2})}^5\\ & & +\cdots \\ & =& \frac{1}{1}[\ln (\tan \left(\frac{\pi }{4}\right))]{(x-\frac{\pi }{2})}^0\\ & & +\frac{1}{1}\left[\frac{1}{\sin \left(\frac{\pi }{2}\right)}\right]{(x-\frac{\pi }{2})}^1\\ & & +\frac{1}{2}[-\frac{\cos \left(\frac{\pi }{2}\right)}{{\sin }^2\left(\frac{\pi }{2}\right)}]{(x-\frac{\pi }{2})}^2\\ & & +\frac{1}{6}[\frac{1}{\sin \left(\frac{\pi }{2}\right)}+\frac{2{\cos }^2\left(\frac{\pi }{2}\right)}{{\sin }^3\left(\frac{\pi }{2}\right)}]{(x-\frac{\pi }{2})}^3\\ & & +\frac{1}{24}[-5\frac{\cos \left(\frac{\pi }{2}\right)}{{\sin }^2\left(\frac{\pi }{2}\right)}-6\frac{{\cos }^3\left(\frac{\pi }{2}\right)}{{\sin }^4\left(\frac{\pi }{2}\right)}]{(x-\frac{\pi }{2})}^4\\ & & +\frac{1}{120}[5\frac{1}{\sin \left(\frac{\pi }{2}\right)}+28\frac{{\cos }^2\left(\frac{\pi }{2}\right)}{{\sin }^3\left(\frac{\pi }{2}\right)}+24\frac{{\cos }^4\left(\frac{\pi }{2}\right)}{{\sin }^5\left(\frac{\pi }{2}\right)}]{(x-\frac{\pi }{2})}^5\\ & & +\cdots \\ & =& \left[0\right]\cdot 1\\ & & +\left[\frac{1}{1}\right](x-\frac{\pi }{2})\\ & & +\frac{1}{2}[-\frac{0}{1}]{(x-\frac{\pi }{2})}^2\\ & & +\frac{1}{6}[\frac{1}{1}+\frac{2\cdot 0}{1}]{(x-\frac{\pi }{2})}^3\\ & & +\frac{1}{24}[-5\frac{0}{1}-6\frac{0}{1}]{(x-\frac{\pi }{2})}^4\\ & & +\frac{1}{120}[5\frac{1}{1}+28\frac{0}{1}+24\frac{0}{1}]{(x-\frac{\pi }{2})}^5\\ & & +\cdots \\ & =& (x-\frac{\pi }{2})+\frac{1}{6}{(x-\frac{\pi }{2})}^3+\frac{1}{24}{(x-\frac{\pi }{2})}^5+\cdots \end{array}$$

$\underset{x\to \frac{\pi }{2}}{lim}\ln (\tan \left(\frac{x}{2}\right))$を求める

$$\begin{array}{r}\\ \underset{x\to \frac{\pi }{2}}{lim}\ln (\tan \left(\frac{x}{2}\right))& =& \underset{x\to \frac{\pi }{2}}{lim}[(x-\frac{\pi }{2})+\frac{1}{6}{(x-\frac{\pi }{2})}^3+\frac{1}{24}{(x-\frac{\pi }{2})}^5+\cdots ]\\ & =& 0\end{array}$$