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Question Number 128593 by Eric002 last updated on 08/Jan/21

$$\underset{x\to 0}{\lim }\frac{\mid 3x-1\mid -\mid 3x+1\mid }{x}$$

Commented by MJS_new last updated on 08/Jan/21

$$\frac{\mid 3x-1\mid -\mid 3x+1\mid }{x}=\frac{3}{x}(\mid x-\frac{1}{3}\mid -\mid x+\frac{1}{3}\mid )=$$$$[-\frac{1}{3}⩽x⩽\frac{1}{3}]$$$$=\frac{3}{x}(-2x)=-\frac{6x}{x}=-6\mathrm{\forall }x\in [-\frac{1}{3};\frac{1}{3}]\wedge x\ne 0$$$$\Rightarrow \mathrm{limit}\mathrm{is}-6$$

Commented by john_santu last updated on 08/Jan/21

$$\underset{x\to 0}{\lim }\frac{\sqrt{{9\mathrm{x}}^2-6\mathrm{x}+1}-\sqrt{{9\mathrm{x}}^2+6\mathrm{x}+1}}{\mathrm{x}}=$$$$\underset{x\to 0}{\lim }\frac{-12\mathrm{x}}{\mathrm{x}[\sqrt{{9\mathrm{x}}^2-6\mathrm{x}+1}+\sqrt{{9\mathrm{x}}^2+6\mathrm{x}+1}]}=$$$$\underset{x\to 0}{\lim }\frac{-12}{\sqrt{{9\mathrm{x}}^2-6\mathrm{x}+1}+\sqrt{{9\mathrm{x}}^2+6\mathrm{x}+1}}=\frac{-12}{2}=-6$$

Answered by Olaf last updated on 08/Jan/21

$$\underset{x\to 0}{\lim }\frac{\mid 3x-1\mid -\mid 3x+1\mid }{x}$$$$=\underset{x\to 0}{\lim }\frac{(-3x+1)-(3x+1)}{x}$$$$=\underset{x\to 0}{\lim }\frac{-6x}{x}=-6$$

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