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Question Number 111351 by bemath last updated on 03/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\underset{x\to -\mathrm{\infty }}{\lim }\frac{\mathrm{x}-\sqrt{{4\mathrm{x}}^2+1}}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+1}}?$$

Answered by ajfour last updated on 03/Sep/20

$$=\underset{x\to -\mathrm{\infty }}{\lim }\frac{\frac{x}{\mid x\mid }-\sqrt{4+\frac{1}{x^2}}}{\sqrt{1+\frac{2x}{\mid x{\mid }^2}+\frac{1}{x^2}}}=\frac{-1-2}{1}$$$$=-3.$$

Answered by bobhans last updated on 03/Sep/20

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

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