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Question Number 108456 by Khalmohmmad last updated on 17/Aug/20

Commented by Rizwankhan last updated on 17/Aug/20

$$howdidyouaddedthispagesir?$$

Answered by bemath last updated on 17/Aug/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{2x^2+nx+2}-\sqrt{2x^2-x+2}=2$$$$\iff \frac{n-(-1)}{2\sqrt{2}}=2$$$$n+1=4\sqrt{2}$$$$n=4\sqrt{2}-1$$

Answered by Dwaipayan Shikari last updated on 17/Aug/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{9x^2+nx+2-9x^2+x-2}{\sqrt{9x^2+nx+2}+\sqrt{9x^2-x+2}}=\frac{x(n+1)}{x(\sqrt{9+\frac{n}{x}+\frac{2}{x^2}}+\sqrt{9-\frac{1}{x}+\frac{2}{x^2}})}$$$$=\frac{n+1}{6}=2$$$$\Rightarrow n=11$$

Commented by bemath last updated on 17/Aug/20

$$this9or2inthequestion?$$

Commented by Aziztisffola last updated on 17/Aug/20

$$\mathrm{It}\mathrm{looks}\mathrm{like}9\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{clear}.$$

Answered by Aziztisffola last updated on 17/Aug/20

$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }(\sqrt{{9\mathrm{x}}^2+\mathrm{nx}+2}-\sqrt{{9\mathrm{x}}^2-\mathrm{x}+2})=2$$$$\Rightarrow \frac{\mathrm{n}+1}{2\times 3}=2\iff \mathrm{n}+1=12\iff \mathrm{n}=11$$

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