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Question Number 126008 by bramlexs22 last updated on 16/Dec/20

$$Findallasymptotesofthe$$$$functiony=\frac{x}{\sqrt{x^2+2}}.$$

Answered by liberty last updated on 16/Dec/20

$$Horizontalasymptote:y=\underset{x\to \mathrm{\infty }}{\lim }\frac{x}{\sqrt{x^2+2}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt{1+2x^{-2}}}=1$$$$andy=\underset{x\to -\mathrm{\infty }}{\lim }\frac{x}{\sqrt{x^2+2}}=\underset{x\to -\mathrm{\infty }}{\lim }\frac{-1}{\sqrt{1+2x^{-2}}}=-1$$$$soy=-1andy=1arethehorizontalasymptotes$$

Answered by ebi last updated on 16/Dec/20

$$$$$$letf\left(x\right)=\frac{x}{\sqrt{x^2+2}}$$$$verticalasymptotes:$$$$sincethe\sqrt{x^2+2}>0,$$$$\therefore noverticalasymptotesavailable.$$$$$$$$horizontalasymptotes:$$$$\underset{x\to +\mathrm{\infty }}{lim}\left(\frac{x}{\sqrt{x^2+2}}\right)=\underset{x\to \mathrm{\infty }}{lim}\left(\frac{1}{\sqrt{1+\frac{2}{x^2}}}\right)$$$$=\frac{1}{\sqrt{1+0}}=1$$$$\underset{x\to -\mathrm{\infty }}{lim}\left(\frac{x}{\sqrt{x^2+2}}\right)=\underset{x\to \mathrm{\infty }}{lim}(-\frac{x}{\sqrt{x^2+2}})$$$$=\underset{x\to \mathrm{\infty }}{lim}(-\frac{1}{\sqrt{1+\frac{2}{x^2}}})=-\frac{1}{\sqrt{1+0}}=-1$$$$\therefore horizontalasymptotes,y=-1andy=1.$$$$$$$$slankasymptotes:$$$$y=mx+c$$$$m=\frac{f\left(x\right)}{x}=\frac{1}{\sqrt{x^2+2}}$$$$\underset{x\to +\mathrm{\infty }}{lim}\left(\frac{1}{\sqrt{x^2+2}}\right)=0$$$$\underset{x\to -\mathrm{\infty }}{lim}\left(\frac{1}{\sqrt{x^2+2}}\right)=\underset{x\to \mathrm{\infty }}{lim}(-\frac{1}{\sqrt{x^2+2}})=0$$$$\therefore noslankasymptotesavailable.$$

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