Question Number 78324 by jagoll last updated on 16/Jan/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{2}{x}}{\sqrt{{x}+\mathrm{1}}−\sqrt{\mathrm{1}−{x}}}= \\ $$
Commented by john santu last updated on 16/Jan/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{1}−{x}}\right)×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} {x}+\mathrm{2sin}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}{x}} \\ $$$$=\:\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}\left\{\mathrm{sin}\:{x}+\mathrm{2cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\right\}}{\mathrm{2}{x}} \\ $$$$=\:\mathrm{2}×\left\{\mathrm{2}\right\}=\mathrm{4} \\ $$
Commented by mathmax by abdo last updated on 16/Jan/20
$${let}\:{f}\left({x}\right)=\frac{\mathrm{1}+{sin}\left(\mathrm{4}{x}\right)−{cos}\left(\mathrm{2}{x}\right)}{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}\:\:{we}\:{have}\: \\ $$$$\sqrt{\mathrm{1}+{x}}\sim\mathrm{1}+\frac{{x}}{\mathrm{2}}\:\:{and}\:\sqrt{\mathrm{1}−{x}}\sim\mathrm{1}−\frac{{x}}{\mathrm{2}}\:\:{also}\:{son}\left(\mathrm{4}{x}\right)\sim\mathrm{4}{x}\:\: \\ $$$${cos}\left(\mathrm{2}{x}\right)\sim\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:\:\:\left({x}\:\rightarrow\mathrm{0}\right)\:\Rightarrow{f}\left({x}\right)\sim\frac{\mathrm{1}+\mathrm{4}{x}−\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+\frac{{x}}{\mathrm{2}}−\mathrm{1}+\frac{{x}}{\mathrm{2}}} \\ $$$$\Rightarrow{f}\left({x}\right)\sim\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}}{{x}}\:\Rightarrow{f}\left({x}\right)\sim\mathrm{2}{x}\:+\mathrm{4}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{4} \\ $$