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Question Number 100920 by ajfour last updated on 29/Jun/20

$$Findlimit$$$$\underset{x\to +\mathrm{\infty }}{\lim }x(\sqrt{x^2+1}-x)and$$$$\underset{x\to -\mathrm{\infty }}{\lim }x(\sqrt{x^2+1}-x).$$

Commented by bramlex last updated on 29/Jun/20

$$\underset{x\to +\mathrm{\infty }}{\lim }x(x\sqrt{1+\frac{1}{x^2}}-x)$$$$set\frac{1}{x}=z,z\to 0$$$$\underset{z\to 0}{\lim }\frac{1}{z}(\frac{1}{z}\sqrt{1+z^2}-\frac{1}{z})=$$$$\underset{z\to 0}{\lim }\frac{\sqrt{1+z^2}-1}{z^2}=\underset{z\to 0}{\lim }\frac{(1+\frac{z^2}{2})-1}{z^2}$$$$=\frac{1}{2}★$$

Commented by bramlex last updated on 29/Jun/20

$$\underset{x\to -\mathrm{\infty }}{\lim }x(\sqrt{x^2+1}-x)=$$$$\underset{x\to -\mathrm{\infty }}{\lim }x(-x\sqrt{1+\frac{1}{x^2}}-x)=$$$$\underset{x\to -\mathrm{\infty }}{\lim }x(-x\sqrt{1+{\left(\frac{1}{-x}\right)}^2}-x)=$$$$\underset{x\to -\mathrm{\infty }}{\lim }-x^2(\sqrt{1+{(-\frac{1}{x})}^2}+1)$$$$set\frac{1}{-x}=q,q\to 0$$$$\underset{q\to 0}{\lim }-{\left(\frac{1}{q}\right)}^2(\sqrt{1+q^2}+1)=$$$$-\underset{q\to 0}{\lim }\frac{\sqrt{1+q^2}+1}{q^2}=-\mathrm{\infty }.★$$

Answered by mathmax by abdo last updated on 29/Jun/20

$$\mathrm{for}\mathrm{x}>0\mathrm{we}\mathrm{have}\mathrm{x}(\sqrt{{\mathrm{x}}^2+1}-\mathrm{x})=\mathrm{x}(\mathrm{x}\sqrt{1+\frac{1}{{\mathrm{x}}^2}}-\mathrm{x})\sim \mathrm{x}(\mathrm{x}(1+\frac{1}{{2\mathrm{x}}^2})-\mathrm{x})$$$$=\mathrm{x}\left(\frac{1}{2\mathrm{x}}\right)=\frac{1}{2}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{x}(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})=\frac{1}{2}$$$$\mathrm{for}\mathrm{x}<0\mathrm{we}\mathrm{have}\mathrm{x}(\sqrt{{\mathrm{x}}^2+1}-\mathrm{x})=\mathrm{x}\{-\mathrm{x}\sqrt{1+\frac{1}{{\mathrm{x}}^2}}-\mathrm{x}\}$$$$\sim \mathrm{x}\{-\mathrm{x}(1+\frac{1}{{2\mathrm{x}}^2})-\mathrm{x}\}=\mathrm{x}\{-2\mathrm{x}-\frac{1}{2\mathrm{x}}\}=-{2\mathrm{x}}^2-\frac{\mathrm{x}}{2}\to -\mathrm{\infty }$$

Commented by ajfour last updated on 29/Jun/20

$$correctSir,imean\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=\frac{1}{2}$$$$and\underset{x\to -\mathrm{\infty }}{\lim }f\left(x\right)=-\mathrm{\infty }.Thanks.$$

Commented by mathmax by abdo last updated on 29/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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