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

$$letf\left(x\right)=\sqrt{1+n{x}^2}-nx+3withnintegr$$$$1)calculate{lim}_{x\to +\mathrm{\infty }}and{lim}_{x\to -\mathrm{\infty }}f\left(x\right)$$$$2)calculate{f}^{{}^{\prime }}\left(x\right)$$$$3)givetheequationofassymptoteoffat$$$$pointA(1,f\left(1\right)).$$$$4)calculate{lim}_{x\to +\mathrm{\infty }}\frac{f\left(x\right)}{x}and{lim}_{x\to -\mathrm{\infty }}\frac{f\left(x\right)}{x}.$$

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

$$1)wehave{lim}_{x\to -\mathrm{\infty }}(-nx+3)=+\mathrm{\infty }and$$$${lim}_{x\to -\mathrm{\infty }}\sqrt{1+{nx}^2}=+\mathrm{\infty }\Rightarrow {lim}_{x\to -\mathrm{\infty }}f\left(x\right)=+\mathrm{\infty }$$$$forx>0\sqrt{1+{nx}^2}=\sqrt{{nx}^2(1+\frac{1}{{nx}^2}})$$$$=x\sqrt{n}\sqrt{1+\frac{1}{{nx}^2}}\sim x\sqrt{n}\{1+\frac{1}{2{nx}^2}\}=x\sqrt{n}+\frac{\sqrt{n}}{2nx}$$$$=x\sqrt{n}+\frac{1}{2x\sqrt{n}}(x\to +\mathrm{\infty })\Rightarrow$$$$f\left(x\right)\sim (\sqrt{n}-n)x+\frac{1}{2x\sqrt{n}}-3(x\to +\mathrm{\infty })\Rightarrow$$$$f\left(x\right)\sim -\sqrt{n}(\sqrt{n}-1)x\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=-\mathrm{\infty }$$$$$$

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

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

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

$$3)wehavef\left(1\right)=\sqrt{1+n}-n+3and$$$$f^{{}^{\prime }}\left(x\right)=\frac{nx}{\sqrt{1+{nx}^2}}-n\Rightarrow f^{{}^{\prime }}\left(1\right)=\frac{n}{\sqrt{1+n}}-nsothe$$$$equationofassymptoteatpointAis$$$$y=f^{{}^{\prime }}\left(1\right)(x-1)+f\left(1\right)\Rightarrow$$$$y=(\frac{n}{\sqrt{1+n}}-n)(x-1)+\sqrt{1+n}-n+3.$$$$$$

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

$$4)wehaveforx>0$$$$\frac{f\left(x\right)}{x}=\sqrt{\frac{1+{nx}^2}{x^2}}-n+\frac{3}{x}$$$$=\sqrt{n+\frac{1}{x^2}}-n+\frac{3}{x}\Rightarrow {lim}_{x\to +\mathrm{\infty }}\frac{f\left(x\right)}{x}=\sqrt{n}-n$$$$forx<0\frac{f\left(x\right)}{x}=-\sqrt{\frac{1+{nx}^2}{x^2}}-n+\frac{3}{x}$$$$=-\sqrt{n+\frac{1}{x^2}}-n+\frac{3}{x}\Rightarrow {lim}_{n\to -\mathrm{\infty }}\frac{f\left(x\right)}{x}=-\sqrt{n}-n.$$

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

$$3)forgiveequationoftangente....$$

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