Question Number 107710 by bemath last updated on 12/Aug/20
$$\:\:\:\checkmark\frac{\mathcal{B}{e}\mathcal{M}{ath}\checkmark}{{cooll}} \\ $$$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{3}}\right)^{\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{cos}\:\mathrm{2}{x}}} \:? \\ $$
Answered by Dwaipayan Shikari last updated on 12/Aug/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{8}}{{x}−\mathrm{3}}\right)^{\frac{\mathrm{8}}{{x}−\mathrm{3}}.\frac{{x}−\mathrm{3}}{\mathrm{8}}.\frac{\mathrm{3}{x}+\mathrm{1}}{{cos}\mathrm{2}{x}}} ={y} \\ $$$${e}^{\frac{\mathrm{8}}{{x}−\mathrm{3}}.\frac{\mathrm{3}{x}+\mathrm{1}}{{cos}\mathrm{2}{x}}} ={e}^{\frac{\mathrm{8}}{\mathrm{1}−\frac{\mathrm{3}}{{x}}}.\frac{\mathrm{3}+\frac{\mathrm{1}}{{x}}}{{cos}\mathrm{2}{x}}} ={e}^{\frac{\mathrm{24}}{{cos}\mathrm{2}{x}}} \\ $$$${As}\:−\mathrm{1}\leqslant{cos}\mathrm{2}{x}\leqslant\mathrm{1} \\ $$