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

If lim_(x→1) ((((ax^3 +b))^(1/3) −2x)/(x^2 −1)) = M then lim_(x→1) ((((ax^3 +b))^(1/3) −2)/(x^2 −3x+2)) =?

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

Answered by benjo_mathlover last updated on 10/Jan/21

i use trick mr Liberty lim_(x→1) ((((ax^3 +b))^(1/3) −2x−2+2x)/((x−1)(x−2))) = lim_(x→1) ((((ax^3 +b))^(1/3) −2x)/(−(x−1))) + lim_(x→1) ((2(x−1))/((x−1)(x−2))) = −lim_(x→1) (((x+1){((ax^3 +b))^(1/3) −2x})/((x^2 −1))) + lim_(x→1) (2/(x−2)) = −2M−2

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

Commented by liberty last updated on 10/Jan/21

hahahaha.....

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

Answered by rydasss last updated on 11/Jan/21

soal utbk 2019

$${soal}\:{utbk}\:\mathrm{2019} \\ $$

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