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Question Number 133494 by Eric002 last updated on 22/Feb/21

solve without using l′hopital and series lim_(x→8) ((x (x)^(1/3) −16)/(x−8))

$${solve}\:{without}\:{using}\:{l}'{hopital}\:{and}\:{series}\: \\ $$$$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\frac{{x}\:\sqrt[{\mathrm{3}}]{{x}}−\mathrm{16}}{{x}−\mathrm{8}} \\ $$

Answered by Olaf last updated on 22/Feb/21

X = ((x(x)^(1/3) −16)/(x−8)) X = ((x^(4/3) −2^4 )/(x^(3/3) −2^3 )) Let u = x^(1/3) X = ((u^4 −2^4 )/(u^3 −2^3 )) X = (((u−2)(u+2)(u^2 +2^2 ))/((u−2)(u^2 +2u+2^2 ))) X = (((u+2)(u^2 +2^2 ))/(u^2 +2u+2^2 )) lim_(x→8) ((x(x)^(1/3) −16)/(x−8)) = lim_(u→2) (((u+2)(u^2 +2^2 ))/(u^2 +2u+2^2 )) = ((4×8)/(12)) = (8/3)

$$ \\ $$$$\mathrm{X}\:=\:\frac{{x}\sqrt[{\mathrm{3}}]{{x}}−\mathrm{16}}{{x}−\mathrm{8}} \\ $$$$\mathrm{X}\:=\:\frac{{x}^{\frac{\mathrm{4}}{\mathrm{3}}} −\mathrm{2}^{\mathrm{4}} }{{x}^{\frac{\mathrm{3}}{\mathrm{3}}} −\mathrm{2}^{\mathrm{3}} } \\ $$$$\mathrm{Let}\:{u}\:=\:{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{X}\:=\:\frac{{u}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} }{{u}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} } \\ $$$$\mathrm{X}\:=\:\frac{\left({u}−\mathrm{2}\right)\left({u}+\mathrm{2}\right)\left({u}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)}{\left({u}−\mathrm{2}\right)\left({u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{2}^{\mathrm{2}} \right)} \\ $$$$\mathrm{X}\:=\:\frac{\left({u}+\mathrm{2}\right)\left({u}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{2}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\:\frac{{x}\sqrt[{\mathrm{3}}]{{x}}−\mathrm{16}}{{x}−\mathrm{8}}\:=\:\underset{{u}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\left({u}+\mathrm{2}\right)\left({u}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{2}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{4}×\mathrm{8}}{\mathrm{12}}\:=\:\frac{\mathrm{8}}{\mathrm{3}} \\ $$

Commented by Eric002 last updated on 22/Feb/21

well done

$${well}\:{done} \\ $$

Commented by otchereabdullai@gmail.com last updated on 14/Mar/21

nice one

$$\mathrm{nice}\:\mathrm{one} \\ $$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

x=t^3 lim_(t→2) ((t^4 −16)/(t^3 −8))=(((t−2)(t+2)(t^2 +4))/((t−2)(t^2 +2t+4)))=((4.8)/(12))=(8/3)

$${x}={t}^{\mathrm{3}} \\ $$$$\underset{{t}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{t}^{\mathrm{4}} −\mathrm{16}}{{t}^{\mathrm{3}} −\mathrm{8}}=\frac{\left({t}−\mathrm{2}\right)\left({t}+\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{4}\right)}{\left({t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{4}\right)}=\frac{\mathrm{4}.\mathrm{8}}{\mathrm{12}}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$

Answered by malwan last updated on 22/Feb/21

lim_(x→8) ((x^(((2/3))^2 ) −16)/(x−8)) = lim_(x→8) (((x^(2/3) −4)(x^(2/3) +4))/(x^(((1/3))^3 ) −2^3 )) = lim_(x→8) (((x^(1/3) −2)(x^(1/3) +2)(x^(2/3) +4))/((x^(1/3) −2)(x^(2/3) +2x^(1/3) +4))) = (((2+2)(4+4))/(4+4+4)) = ((4×8)/(4×3)) = (8/3)

$$\underset{{x}\rightarrow\mathrm{8}} {{lim}}\:\frac{{x}^{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} } −\mathrm{16}}{{x}−\mathrm{8}}\:=\:\underset{{x}\rightarrow\mathrm{8}} {{lim}}\:\frac{\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{4}\right)\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{4}\right)}{{x}^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} } −\mathrm{2}^{\mathrm{3}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{8}} {{lim}}\:\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}\right)\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}\right)\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{4}\right)}{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}\right)\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4}\right)} \\ $$$$=\:\frac{\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{4}+\mathrm{4}\right)}{\mathrm{4}+\mathrm{4}+\mathrm{4}}\:=\:\frac{\mathrm{4}×\mathrm{8}}{\mathrm{4}×\mathrm{3}}\:=\:\frac{\mathrm{8}}{\mathrm{3}} \\ $$

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