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Question Number 54409 by pooja24 last updated on 03/Feb/19

$$li\underset{x\to \mathrm{\infty }}{m}\sqrt{(x^2+x+1)}-x=?$$$$plssolvethis$$$$$$

Commented by maxmathsup by imad last updated on 03/Feb/19

$$wehave\sqrt{x^2+x+1}=\mid x\mid \sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\sim 1+\frac{1}{2}(\frac{1}{x}+\frac{1}{x^2})(x\to \mathrm{\infty })\Rightarrow$$$$\sqrt{x^2+x+1}-x\sim \mid x\mid \{1+\frac{1}{2x}+\frac{1}{2x^2}\}-x\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}\sqrt{x^2+x+1}-x={lim}_{x\to +\mathrm{\infty }}x\{1+\frac{1}{2x}+\frac{1}{2x^2}\}-x$$$$={lim}_{x\to +\mathrm{\infty }}(\frac{1}{2}+\frac{1}{2x})=\frac{1}{2}$$$${lim}_{x\to -\mathrm{\infty }}\sqrt{x^2+x+1}-x={lim}_{x\to -\mathrm{\infty }}-x\{1+\frac{1}{2x}+\frac{1}{2x^2}\}-x$$$$={lim}_{x\to -\mathrm{\infty }}-2x-\frac{1}{2}-\frac{1}{2x}=+\mathrm{\infty }.$$

Answered by kaivan.ahmadi last updated on 03/Feb/19

$$\times \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\mathrm{x}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+1}=\mathrm{li}\underset{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{m}}\frac{\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\mathrm{x}}\approx \mathrm{li}\underset{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{m}}\frac{\mathrm{x}}{2\mathrm{x}}=\frac{1}{2}$$$$$$$$\mathrm{or}$$$$$$$$\approx \mathrm{li}\underset{\mathrm{x}\to \mathrm{\infty }}{\mathrm{m}}\mid \mathrm{x}+\frac{1}{2}\mid -\mathrm{x}=\frac{1}{2}$$

Answered by Prithwish sen last updated on 03/Feb/19

$$\mathrm{let}\mathrm{x}=\frac{1}{\mathrm{h}}$$$$\therefore \mathrm{x}\to \mathrm{\infty }\Rightarrow \mathrm{h}\to 0$$$$\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}\frac{\sqrt{{\mathrm{h}}^2+\mathrm{h}+1}-1}{\mathrm{h}}(\because \mathrm{form}\frac{0}{0},\mathrm{we}\mathrm{apply}$$$${\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{rule})$$$$=\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}\frac{2\mathrm{h}+1}{2\sqrt{{\mathrm{h}}^2+\mathrm{h}+1}}$$$$=\frac{1}{2}$$

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