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Question Number 114028 by AbhishekBasnet last updated on 16/Sep/20

Commented by bemath last updated on 17/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x^2+x}-\sqrt{x}=\frac{1-0}{2.1}=\frac{1}{2}$$

Answered by Olaf last updated on 16/Sep/20

$$\sqrt{x}(\sqrt{x+1}-\sqrt{x})=x(\sqrt{1+\frac{1}{x}}-1)$$$$\underset{\mathrm{\infty }}{\sim }x(1+\frac{1}{2x}-1)\underset{x\to \mathrm{\infty }}{\to }\frac{1}{2}$$

Answered by Olaf last updated on 16/Sep/20

$$\sqrt{x}(\sqrt{x+1}-\sqrt{x})=\sqrt{x}\frac{1}{\sqrt{x+1}+\sqrt{x}}$$$$\underset{\mathrm{\infty }}{\sim }\frac{\sqrt{x}}{\sqrt{x}+\sqrt{x}}=\frac{1}{2}$$

Answered by Dwaipayan Shikari last updated on 16/Sep/20

$$\sqrt{x}\left(\frac{x+1-x}{\sqrt{x+1}+\sqrt{x}}\right)=\frac{\sqrt{x}}{\sqrt{x}}\left(\frac{1}{\sqrt{1+\frac{1}{x}}+\sqrt{1}}\right)=\frac{1}{2}$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\{\sqrt{\mathrm{x}+1}-\sqrt{\mathrm{x}}\}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}}}=\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}\{\sqrt{1+\frac{1}{\mathrm{x}}}+1\}}$$$$=\frac{1}{\sqrt{1+\frac{1}{\sqrt{\mathrm{x}}}}+1}\sim \frac{1}{1+\frac{1}{2\sqrt{\mathrm{x}}}+1}=\frac{1}{2+\frac{1}{2\sqrt{\mathrm{x}}}}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}$$

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