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Question Number 128329 by bramlexs22 last updated on 06/Jan/21

$$\left(2\right)\mathrm{Solution}\mathrm{set}:x\mid 2x-6\mid <3x$$ $$\mathrm{is}\mathrm{\_}$$ $$\left(2\right)\mathrm{If}\underset{x\to 2}{\lim }\frac{2-\sqrt{a+{bx}^3}}{x-2}=H,then$$ $$\underset{x\to 2}{\lim }\frac{x^2-4}{\sqrt{a+{bx}^3}-1}=\mathrm{\_}$$

Answered by liberty last updated on 06/Jan/21

$$\left(2\right)\underset{x\to 2}{\lim }\frac{(\mathrm{x}-2)(\mathrm{x}+2)}{\sqrt{\mathrm{a}+{\mathrm{bx}}^3}-2}=-\underset{x\to 2}{\lim }\frac{1}{\left(\frac{2-\sqrt{\mathrm{a}+{\mathrm{bx}}^3}}{\mathrm{x}-2}\right)}\times \underset{x\to 2}{\lim }(\mathrm{x}+2)$$ $$=-\frac{1}{\mathrm{H}}\times 4=-\frac{4}{\mathrm{H}}$$

Commented bybramlexs22 last updated on 06/Jan/21

$$creative$$

Answered by liberty last updated on 06/Jan/21

$$\left(1\right)\mathrm{x}⩾3\Rightarrow \mathrm{x}\{2\mathrm{x}-6-3\}<0;\mathrm{we}\mathrm{get}3⩽\mathrm{x}<\frac{9}{2}$$ $$\mathrm{x}<3\Rightarrow \mathrm{x}\{6-2\mathrm{x}-3\}<0;\mathrm{we}\mathrm{get}\mathrm{x}<0\cup \frac{3}{2}<\mathrm{x}<3$$ $$\mathrm{then}\mathrm{the}\mathrm{solution}\mathrm{set}\mathrm{is}\mathrm{x}<0\cup \frac{3}{2}<\mathrm{x}<\frac{9}{2}$$ $$$$

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