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

$$solvewithoutusing{l}^{\prime }hopitalandseries$$$$\underset{x\to 8}{\lim }\frac{x\sqrt[3]{x}-16}{x-8}$$

Answered by Olaf last updated on 22/Feb/21

$$$$$$\mathrm{X}=\frac{x\sqrt[3]{x}-16}{x-8}$$$$\mathrm{X}=\frac{x^{\frac{4}{3}}-2^4}{x^{\frac{3}{3}}-2^3}$$$$\mathrm{Let}u=x^{\frac{1}{3}}$$$$\mathrm{X}=\frac{u^4-2^4}{u^3-2^3}$$$$\mathrm{X}=\frac{(u-2)(u+2)(u^2+2^2)}{(u-2)(u^2+2u+2^2)}$$$$\mathrm{X}=\frac{(u+2)(u^2+2^2)}{u^2+2u+2^2}$$$$\underset{x\to 8}{\lim }\frac{x\sqrt[3]{x}-16}{x-8}=\underset{u\to 2}{\lim }\frac{(u+2)(u^2+2^2)}{u^2+2u+2^2}$$$$=\frac{4\times 8}{12}=\frac{8}{3}$$

Commented by Eric002 last updated on 22/Feb/21

$$welldone$$

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

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

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

$$x=t^3$$$$\underset{t\to 2}{\lim }\frac{t^4-16}{t^3-8}=\frac{(t-2)(t+2)(t^2+4)}{(t-2)(t^2+2t+4)}=\frac{4.8}{12}=\frac{8}{3}$$

Answered by malwan last updated on 22/Feb/21

$$\underset{x\to 8}{lim}\frac{x^{{\left(\frac{2}{3}\right)}^2}-16}{x-8}=\underset{x\to 8}{lim}\frac{(x^{\frac{2}{3}}-4)(x^{\frac{2}{3}}+4)}{x^{{\left(\frac{1}{3}\right)}^3}-2^3}$$$$=\underset{x\to 8}{lim}\frac{(x^{\frac{1}{3}}-2)(x^{\frac{1}{3}}+2)(x^{\frac{2}{3}}+4)}{(x^{\frac{1}{3}}-2)(x^{\frac{2}{3}}+2x^{\frac{1}{3}}+4)}$$$$=\frac{(2+2)(4+4)}{4+4+4}=\frac{4\times 8}{4\times 3}=\frac{8}{3}$$

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