Question Number 124273 by bramlexs22 last updated on 02/Dec/20
$$\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{7}+\mathrm{2sin}\:{x}}\:−\sqrt{\mathrm{7}−\mathrm{2sin}\:{x}}}{\mathrm{tan}\:{x}}\:=\:?\: \\ $$
Answered by Dwaipayan Shikari last updated on 02/Dec/20
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\sqrt{\mathrm{7}+\mathrm{2}\left(\pi−{x}\right)}−\sqrt{\mathrm{7}−\mathrm{2}\pi+\mathrm{2}{x}}}{\pi−{x}}=\sqrt{\mathrm{7}}\left(\frac{\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{7}}\left(\pi−{x}\right)}−\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{7}}\left(\pi−{x}\right)}}{\pi−{x}}\right)\:\:\:\: \\ $$$$=\sqrt{\mathrm{7}}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}\left(\pi−{x}\right)−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}\left(\pi−{x}\right)}{\pi−{x}}\right){cosx}=−\sqrt{\mathrm{7}}\:\left(\frac{\mathrm{2}}{\mathrm{7}}\right)=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{7}}} \\ $$
Answered by liberty last updated on 02/Dec/20
$$\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\left(\mathrm{7}+\mathrm{2sin}\:{x}\right)−\left(\mathrm{7}−\mathrm{2sin}\:{x}\right)}{\left(\sqrt{\mathrm{7}+\mathrm{2sin}\:{x}}\:+\sqrt{\mathrm{7}−\mathrm{2sin}\:{x}}\:\right).\mathrm{tan}\:{x}}\:= \\ $$$$\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{2sin}\:{x}}\:+\sqrt{\mathrm{7}−\mathrm{2sin}\:{x}}}\:×\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{4sin}\:{x}}{\mathrm{tan}\:{x}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}}}\:×\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\mathrm{4sin}\:{x}\left(\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}}}\:×\:\left(−\mathrm{4}\right)=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{7}}}. \\ $$