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Question Number 180106 by cherokeesay last updated on 07/Nov/22

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x^2+x}-x$$

Answered by a.lgnaoui last updated on 07/Nov/22

$$\sqrt{{\mathrm{x}}^2+\mathrm{x}}-\mathrm{x}=\frac{(\sqrt{{\mathrm{x}}^2+\mathrm{x}}-\mathrm{x})(\sqrt{{\mathrm{x}}^2+\mathrm{x}}+\mathrm{x})}{\sqrt{{\mathrm{x}}^2+\mathrm{x}}+\mathrm{x}}(\mathrm{x}\in \left(\right]-\mathrm{\infty },0]\cup [-1,+\mathrm{\infty }\left[\right)$$$$=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}}+\mathrm{x}}=\frac{\mathrm{x}}{\mathrm{x}(\sqrt{1+\frac{1}{\mathrm{x}}}+1)}=\frac{1}{\sqrt{1+\frac{1}{\mathrm{x}}}+1}$$$${\lim }_{\mathrm{x}\to \mathrm{\infty }}\sqrt{{\mathrm{x}}^2+\mathrm{x}}-\mathrm{x}={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{1}{\sqrt{1+\frac{1}{\mathrm{x}}}+1}=\frac{1}{2}$$

Commented by cherokeesay last updated on 07/Nov/22

$$verynice,$$$$thankyousir.$$

Answered by CElcedricjunior last updated on 07/Nov/22

$$\frac{1}{2}$$

Answered by mr W last updated on 07/Nov/22

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\frac{1}{x}}(\sqrt{1+\frac{1}{x}}-1)$$$$=\underset{t\to 0}{\lim }\frac{1}{t}(\sqrt{1+t}-1)$$$$=\underset{t\to 0}{\lim }\frac{1}{t}(1+\frac{t}{2}+o\left(t^2\right)-1)$$$$=\underset{t\to 0}{\lim }(\frac{1}{2}+o\left(t\right))$$$$=\frac{1}{2}$$

Commented by cherokeesay last updated on 07/Nov/22

$$Nice,thankyoumaster.$$

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