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Question Number 128817 by benjo_mathlover last updated on 10/Jan/21

$$\mathrm{If}\underset{x\to 1}{\lim }\frac{\sqrt[3]{{\mathrm{ax}}^3+\mathrm{b}}-2\mathrm{x}}{{\mathrm{x}}^2-1}=\mathrm{M}\mathrm{then}\underset{x\to 1}{\lim }\frac{\sqrt[3]{{\mathrm{ax}}^3+\mathrm{b}}-2}{{\mathrm{x}}^2-3\mathrm{x}+2}=?$$

Answered by benjo_mathlover last updated on 10/Jan/21

$$\mathrm{i}\mathrm{use}\mathrm{trick}\mathrm{mr}\mathrm{Liberty}$$$$\underset{x\to 1}{\lim }\frac{\sqrt[3]{{\mathrm{ax}}^3+\mathrm{b}}-2\mathrm{x}-2+2\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)}=\underset{x\to 1}{\lim }\frac{\sqrt[3]{{\mathrm{ax}}^3+\mathrm{b}}-2\mathrm{x}}{-(\mathrm{x}-1)}+\underset{x\to 1}{\lim }\frac{2(\mathrm{x}-1)}{(\mathrm{x}-1)(\mathrm{x}-2)}$$$$=-\underset{x\to 1}{\lim }\frac{(\mathrm{x}+1)\{\sqrt[3]{{\mathrm{ax}}^3+\mathrm{b}}-2\mathrm{x}\}}{({\mathrm{x}}^2-1)}+\underset{x\to 1}{\lim }\frac{2}{\mathrm{x}-2}$$$$=-2\mathrm{M}-2$$

Commented by liberty last updated on 10/Jan/21

$$\mathrm{hahahaha}.....$$

Answered by rydasss last updated on 11/Jan/21

$$soalutbk2019$$

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