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Question Number 49800 by maxmathsup by imad last updated on 10/Dec/18

$$find{lim}_{x\to 0}\frac{\sqrt{1+x+x^2}-\sqrt{1+2x+x^3}}{x^2}$$

Answered by afachri last updated on 12/Dec/18

$$\underset{x\to 0}{\lim }\frac{\sqrt{x^2+x+1}-\sqrt{x^3+2x+1}}{x^2}.\frac{\sqrt{x^2+x+1}+\sqrt{x^3+2x+1}}{\sqrt{x^2+x+1}+\sqrt{x^3+2x+1}}$$$$\underset{x\to 0}{\lim }\frac{(x^2+x+1)-(x^3+2x+1)}{x^2(\sqrt{x^2+x+1}+\sqrt{x^3+2x+1})}$$$$\underset{x\to 0}{\lim }\frac{-x^3+x^2-x}{x^2(\sqrt{x^2+x+1}+\sqrt{x^3+2x+1})}$$$$\underset{x\to 0}{\lim }\frac{x^2(-x+1-\frac{1}{x})}{x^2(\sqrt{x^2+x+1}+\sqrt{x^3+2x+1})}$$$$\underset{x\to 0}{\lim }\frac{-x+1-\frac{1}{x}}{\sqrt{x^2+x+1}+\sqrt{x^3+2x+1}}=\mathrm{\infty }$$$$$$

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