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Question Number 144816 by liberty last updated on 29/Jun/21

$$\underset{x\to 0}{\lim }\frac{\sqrt{1+{6\mathrm{x}}^2}-(1+7\mathrm{x})}{{\mathrm{x}}^2(\mathrm{x}-3)}=?$$

Answered by imjagoll last updated on 29/Jun/21

$$=-\frac{1}{3}\underset{x\to 0}{\lim }\frac{(1+{6\mathrm{x}}^2)-(1+14\mathrm{x}+{49\mathrm{x}}^2)}{{2\mathrm{x}}^2}$$$$=-\frac{1}{6}\underset{x\to 0}{\lim }\frac{-{43\mathrm{x}}^2-14\mathrm{x}}{{\mathrm{x}}^2}$$$$=\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 29/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{1+{6\mathrm{x}}^2}-1-7\mathrm{x}}{{\mathrm{x}}^2(\mathrm{x}-3)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{1+{3\mathrm{x}}^2-1-7\mathrm{x}}{{\mathrm{x}}^2(\mathrm{x}-3)}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{(3\mathrm{x}-7)}{\mathrm{x}(\mathrm{x}-3)}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\infty }$$

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