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Question Number 120720 by bramlexs22 last updated on 02/Nov/20

$$\mathrm{Given}\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\mathrm{x}+2\sqrt{\mathrm{x}}+3}-\sqrt{\mathrm{x}}+\mathrm{b}=4$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{b}}^2+1$$

Answered by Dwaipayan Shikari last updated on 02/Nov/20

$$\sqrt{x}(\sqrt{1+\frac{2}{\sqrt{x}}+\frac{3}{x}}-1+\frac{b}{\sqrt{x}})=4$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x}(1+\frac{1}{\sqrt{x}}+\frac{3}{2x}-1+\frac{b}{\sqrt{x}})=4$$$$\sqrt{x}\left(\frac{1+b}{\sqrt{x}}\right)=4$$$$b=3$$$$b^2+1=10$$

Answered by john santu last updated on 02/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x+2\sqrt{x}+3}-(\sqrt{x}-b)=4$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x+2\sqrt{x}+3}-\sqrt{{(\sqrt{x}-b)}^2}=4$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{x+2\sqrt{x}+3}-\sqrt{x-2b\sqrt{x}+b^2}=4$$$$\frac{2-(-2b)}{2.1}=4\Rightarrow 2+2b=8\Rightarrow b=3\wedge b^2+1=10$$

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