Question Number 169147 by mathlove last updated on 25/Apr/22
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2}{cosx}}{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right)}=? \\ $$
Commented by cortano1 last updated on 25/Apr/22
$$\:{let}\:{x}−\frac{\pi}{\mathrm{3}}={h} \\ $$$$\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{3}}+{h}\right)}{\mathrm{sin}\:{h}} \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{3}}+{h}\right)}{\mathrm{cos}\:{h}}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$
Commented by infinityaction last updated on 25/Apr/22
$$\:\:\:\:\:\:\:{p}\:\:\:=\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\mathrm{2}\left(\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{3}}−\mathrm{cos}\:{x}}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)}\right) \\ $$$$\:\:\:\:\:\:{p}\:\:=\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}2}\left(\frac{\mathrm{2sin}\:\left(\frac{{x}+\pi/\mathrm{3}}{\mathrm{2}}\right)\mathrm{sin}\left(\:\frac{{x}−\pi/\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2sin}\:\left(\frac{{x}−\pi/\mathrm{3}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}−\pi/\mathrm{3}}{\mathrm{2}}\right)}\right) \\ $$$$\:\:\:\:\:{p}\:\:\:\:\:=\:\:\:\mathrm{2sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:{p}\:\:\:=\:\:\:\:\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:=\:\:\:\:\:\sqrt{\mathrm{3}} \\ $$$$ \\ $$