Question Number 35986 by abdo mathsup 649 cc last updated on 26/May/18
$${let}\:{f}\left({x}\right)=\:\sqrt{\mathrm{1}\:+{n}\:{x}^{\mathrm{2}} }\:\:\:−{nx}\:+\mathrm{3}\:\:{with}\:{n}\:{integr} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{lim}_{{x}\rightarrow+\infty} \:{and}\:{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{give}\:{the}\:{equation}\:{of}\:{assymptote}\:{of}\:{f}\:{at} \\ $$$${point}\:\:{A}\left(\mathrm{1},{f}\left(\mathrm{1}\right)\right)\:. \\ $$$$\left.\mathrm{4}\right){calculate}\:{lim}_{{x}\rightarrow+\infty} \:\frac{{f}\left({x}\right)}{{x}}\:{and}\:{lim}_{{x}\rightarrow−\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:. \\ $$
Commented by prof Abdo imad last updated on 31/Aug/18
$$\left.\mathrm{1}\right){we}\:{have}\:{lim}_{{x}\rightarrow−\infty} \left(−{nx}\:+\mathrm{3}\right)=+\infty\:{and} \\ $$$${lim}_{{x}\rightarrow−\infty} \sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }=+\infty\:\Rightarrow{lim}_{{x}\rightarrow−\infty} \:\:\:{f}\left({x}\right)\:=+\infty \\ $$$$\left.{for}\:{x}>\mathrm{0}\:\:\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }=\sqrt{{nx}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{nx}^{\mathrm{2}} }\right.}\right) \\ $$$$={x}\sqrt{{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{nx}^{\mathrm{2}} }}\:\sim{x}\sqrt{{n}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{nx}^{\mathrm{2}} }\right\}={x}\sqrt{{n}}\:+\frac{\sqrt{{n}}}{\mathrm{2}{nx}} \\ $$$$={x}\sqrt{{n}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{n}}}\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:\sim\left(\sqrt{{n}}−{n}\right){x}\:\:+\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{n}}}\:−\mathrm{3}\:\left({x}\rightarrow+\infty\right)\Rightarrow \\ $$$${f}\left({x}\right)\:\sim−\sqrt{{n}}\left(\sqrt{{n}}−\mathrm{1}\right){x}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=−\infty \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 31/Aug/18
$$\left.\mathrm{2}\right)\forall{x}\in{R}\:\:\:\:\:{f}^{'} \left({x}\right)=\frac{\mathrm{2}{nx}}{\mathrm{2}\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−{n} \\ $$$$=\:\:\frac{{nx}}{\:\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−{n}\:={n}\left\{\frac{{x}}{\:\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−\mathrm{1}\right\}. \\ $$
Commented by prof Abdo imad last updated on 31/Aug/18
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left(\mathrm{1}\right)\:=\sqrt{\mathrm{1}+{n}}−{n}+\mathrm{3}\:\:{and} \\ $$$${f}^{'} \left({x}\right)=\:\frac{{nx}}{\:\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−{n}\:\Rightarrow{f}^{'} \left(\mathrm{1}\right)=\:\frac{{n}}{\:\sqrt{\mathrm{1}+{n}}}\:−{n}\:{so}\:{the} \\ $$$${equation}\:{of}\:{assymptote}\:{at}\:{point}\:{A}\:{is} \\ $$$${y}\:={f}^{'} \left(\mathrm{1}\right)\left({x}−\mathrm{1}\right)\:+{f}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${y}\:=\left(\frac{{n}}{\:\sqrt{\mathrm{1}+{n}}}\:−{n}\right)\left({x}−\mathrm{1}\right)\:+\sqrt{\mathrm{1}+{n}}\:−{n}\:+\mathrm{3}\:. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 31/Aug/18
$$\left.\mathrm{4}\right)\:{we}\:{have}\:{for}\:{x}>\mathrm{0}\: \\ $$$$\:\frac{{f}\left({x}\right)}{{x}}\:=\sqrt{\frac{\mathrm{1}+{nx}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\:−{n}\:+\frac{\mathrm{3}}{{x}} \\ $$$$=\sqrt{{n}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−{n}+\frac{\mathrm{3}}{{x}}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:=\sqrt{{n}}−{n} \\ $$$${for}\:{x}<\mathrm{0}\:\:\:\frac{{f}\left({x}\right)}{{x}}\:\:=−\sqrt{\frac{\mathrm{1}+{nx}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\:−{n}\:+\frac{\mathrm{3}}{{x}} \\ $$$$=−\sqrt{{n}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:−{n}\:+\frac{\mathrm{3}}{{x}}\:\Rightarrow{lim}_{{n}\rightarrow−\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:=−\sqrt{{n}}−{n}\:. \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
$$\left.\mathrm{3}\right)\:{forgive}\:{equation}\:{of}\:{tangente}…. \\ $$