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Question Number 35986 by abdo mathsup 649 cc last updated on 26/May/18

let f(x)= (√(1 +n x^2 )) −nx +3 with n integr 1) calculate lim_(x→+∞) and lim_(x→−∞) f(x) 2) calculate f^′ (x) 3) give the equation of assymptote of f at point A(1,f(1)) . 4)calculate lim_(x→+∞) ((f(x))/x) and lim_(x→−∞) ((f(x))/x) .

$${let}\:{f}\left({x}\right)=\:\sqrt{\mathrm{1}\:+{n}\:{x}^{\mathrm{2}} }\:\:\:−{nx}\:+\mathrm{3}\:\:{with}\:{n}\:{integr} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{lim}_{{x}\rightarrow+\infty} \:{and}\:{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{give}\:{the}\:{equation}\:{of}\:{assymptote}\:{of}\:{f}\:{at} \\ $$$${point}\:\:{A}\left(\mathrm{1},{f}\left(\mathrm{1}\right)\right)\:. \\ $$$$\left.\mathrm{4}\right){calculate}\:{lim}_{{x}\rightarrow+\infty} \:\frac{{f}\left({x}\right)}{{x}}\:{and}\:{lim}_{{x}\rightarrow−\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:. \\ $$

Commented by prof Abdo imad last updated on 31/Aug/18

1)we have lim_(x→−∞) (−nx +3)=+∞ and lim_(x→−∞) (√(1+nx^2 ))=+∞ ⇒lim_(x→−∞) f(x) =+∞ for x>0 (√(1+nx^2 ))=(√(nx^2 (1+(1/(nx^2 ))))) =x(√n)(√(1+(1/(nx^2 )))) ∼x(√n){1+(1/(2nx^2 ))}=x(√n) +((√n)/(2nx)) =x(√n) +(1/(2x(√n))) (x→+∞) ⇒ f(x) ∼((√n)−n)x +(1/(2x(√n))) −3 (x→+∞)⇒ f(x) ∼−(√n)((√n)−1)x ⇒lim_(x→+∞) f(x) =−∞

$$\left.\mathrm{1}\right){we}\:{have}\:{lim}_{{x}\rightarrow−\infty} \left(−{nx}\:+\mathrm{3}\right)=+\infty\:{and} \\ $$$${lim}_{{x}\rightarrow−\infty} \sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }=+\infty\:\Rightarrow{lim}_{{x}\rightarrow−\infty} \:\:\:{f}\left({x}\right)\:=+\infty \\ $$$$\left.{for}\:{x}>\mathrm{0}\:\:\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }=\sqrt{{nx}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{nx}^{\mathrm{2}} }\right.}\right) \\ $$$$={x}\sqrt{{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{nx}^{\mathrm{2}} }}\:\sim{x}\sqrt{{n}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{nx}^{\mathrm{2}} }\right\}={x}\sqrt{{n}}\:+\frac{\sqrt{{n}}}{\mathrm{2}{nx}} \\ $$$$={x}\sqrt{{n}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{n}}}\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:\sim\left(\sqrt{{n}}−{n}\right){x}\:\:+\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{n}}}\:−\mathrm{3}\:\left({x}\rightarrow+\infty\right)\Rightarrow \\ $$$${f}\left({x}\right)\:\sim−\sqrt{{n}}\left(\sqrt{{n}}−\mathrm{1}\right){x}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=−\infty \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 31/Aug/18

2)∀x∈R f^′ (x)=((2nx)/(2(√(1+nx^2 )))) −n = ((nx)/(√(1+nx^2 ))) −n =n{(x/(√(1+nx^2 ))) −1}.

$$\left.\mathrm{2}\right)\forall{x}\in{R}\:\:\:\:\:{f}^{'} \left({x}\right)=\frac{\mathrm{2}{nx}}{\mathrm{2}\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−{n} \\ $$$$=\:\:\frac{{nx}}{\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−{n}\:={n}\left\{\frac{{x}}{\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−\mathrm{1}\right\}. \\ $$

Commented by prof Abdo imad last updated on 31/Aug/18

3) we have f(1) =(√(1+n))−n+3 and f^′ (x)= ((nx)/(√(1+nx^2 ))) −n ⇒f^′ (1)= (n/(√(1+n))) −n so the equation of assymptote at point A is y =f^′ (1)(x−1) +f(1) ⇒ y =((n/(√(1+n))) −n)(x−1) +(√(1+n)) −n +3 .

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left(\mathrm{1}\right)\:=\sqrt{\mathrm{1}+{n}}−{n}+\mathrm{3}\:\:{and} \\ $$$${f}^{'} \left({x}\right)=\:\frac{{nx}}{\sqrt{\mathrm{1}+{nx}^{\mathrm{2}} }}\:−{n}\:\Rightarrow{f}^{'} \left(\mathrm{1}\right)=\:\frac{{n}}{\sqrt{\mathrm{1}+{n}}}\:−{n}\:{so}\:{the} \\ $$$${equation}\:{of}\:{assymptote}\:{at}\:{point}\:{A}\:{is} \\ $$$${y}\:={f}^{'} \left(\mathrm{1}\right)\left({x}−\mathrm{1}\right)\:+{f}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${y}\:=\left(\frac{{n}}{\sqrt{\mathrm{1}+{n}}}\:−{n}\right)\left({x}−\mathrm{1}\right)\:+\sqrt{\mathrm{1}+{n}}\:−{n}\:+\mathrm{3}\:. \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 31/Aug/18

4) we have for x>0 ((f(x))/x) =(√((1+nx^2 )/x^2 )) −n +(3/x) =(√(n+(1/x^2 )))−n+(3/x) ⇒lim_(x→+∞) ((f(x))/x) =(√n)−n for x<0 ((f(x))/x) =−(√((1+nx^2 )/x^2 )) −n +(3/x) =−(√(n+(1/x^2 ))) −n +(3/x) ⇒lim_(n→−∞) ((f(x))/x) =−(√n)−n .

$$\left.\mathrm{4}\right)\:{we}\:{have}\:{for}\:{x}>\mathrm{0}\: \\ $$$$\:\frac{{f}\left({x}\right)}{{x}}\:=\sqrt{\frac{\mathrm{1}+{nx}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\:−{n}\:+\frac{\mathrm{3}}{{x}} \\ $$$$=\sqrt{{n}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−{n}+\frac{\mathrm{3}}{{x}}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:=\sqrt{{n}}−{n} \\ $$$${for}\:{x}<\mathrm{0}\:\:\:\frac{{f}\left({x}\right)}{{x}}\:\:=−\sqrt{\frac{\mathrm{1}+{nx}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\:−{n}\:+\frac{\mathrm{3}}{{x}} \\ $$$$=−\sqrt{{n}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:−{n}\:+\frac{\mathrm{3}}{{x}}\:\Rightarrow{lim}_{{n}\rightarrow−\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:=−\sqrt{{n}}−{n}\:. \\ $$

Commented by maxmathsup by imad last updated on 31/Aug/18

3) forgive equation of tangente....

$$\left.\mathrm{3}\right)\:{forgive}\:{equation}\:{of}\:{tangente}.... \\ $$

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