Question Number 93676 by john santu last updated on 14/May/20
$$\mathrm{solve}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{x}+\mathrm{4}}−\mathrm{1}}{\mathrm{x}}\:? \\ $$
Commented by mathmax by abdo last updated on 14/May/20
$${let}\:{f}\left({x}\right)\:=\frac{\mathrm{5}\sqrt{{x}+\mathrm{1}}−\mathrm{2}\sqrt{{x}+\mathrm{4}}−\mathrm{1}}{{x}}\:\:{we}\:{have}\:\sqrt{\mathrm{1}+{x}}\sim\mathrm{1}+\frac{{x}}{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{4}}=\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}}{\mathrm{4}}}\sim\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{8}}\right)\:\Rightarrow{f}\left({x}\right)\:\sim\frac{\mathrm{5}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}\left(\mathrm{2}+\frac{{x}}{\mathrm{4}}\right)−\mathrm{1}}{{x}} \\ $$$$=\frac{\mathrm{5}+\frac{\mathrm{5}{x}}{\mathrm{2}}−\mathrm{4}−\frac{{x}}{\mathrm{2}}−\mathrm{1}}{{x}}\:=\frac{\mathrm{2}{x}}{{x}}\:=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)\:=\mathrm{2} \\ $$
Answered by i jagooll last updated on 14/May/20
